Processamento de Linguagem Natural

Gustavo C. A Corradi

robotreport.me

@gustavocorradi, @data4sci

# Lesson I - Text representation

In this lesson we will see in some details how we can best represent text in our application. Let's start by importing the modules we will be using:

```python
import string
from collections import Counter
from pprint import pprint
import gzip
import matplotlib.pyplot as plt
import numpy as np

%matplotlib inline
```

We choose a well known nursery rhyme, that has the added distinction of having been the first audio ever recorded, to be the short snippet of text that we will use in our examples:

```python
text = """Mary had a little lamb, little lamb,
    little lamb. Mary had a little lamb
    whose fleece was white as snow.
    And everywhere that Mary went
    Mary went, Mary went. Everywhere
    that Mary went,
    The lamb was sure to go"""
```

## Tokenization

The first step in any analysis is to tokenize the text. What this means is that we will extract all the individual words in the text. For the sake of simplicity, we will assume that our text is well formed and that our words are delimited either by white space or punctuation characters.

```python
def extract_words(text):
    temp = text.split() # Split the text on whitespace
    text_words = []

    for word in temp:
        # Remove any punctuation characters present in the beginning of the word
        while word[0] in string.punctuation:
            word = word[1:]

        # Remove any punctuation characters present in the end of the word
        while word[-1] in string.punctuation:
            word = word[:-1]

        # Append this word into our list of words.
        text_words.append(word.lower())

    return text_words
```

After this step we now have our text represented as an array of individual, lowercase, words:

```python
text_words = extract_words(text)
print(text_words)
```

    ['mary', 'had', 'a', 'little', 'lamb', 'little', 'lamb', 'little', 'lamb', 'mary', 'had', 'a', 'little', 'lamb', 'whose', 'fleece', 'was', 'white', 'as', 'snow', 'and', 'everywhere', 'that', 'mary', 'went', 'mary', 'went', 'mary', 'went', 'everywhere', 'that', 'mary', 'went', 'the', 'lamb', 'was', 'sure', 'to', 'go']

As we saw during the video, this is a wasteful way to represent text. We can be much more efficient by representing each word by a number

```python
word_dict = {}
word_list = []
vocabulary_size = 0
text_tokens = []

for word in text_words:
    # If we are seeing this word for the first time, create an id for it and added it to our word dictionary
    if word not in word_dict:
        word_dict[word] = vocabulary_size
        word_list.append(word)
        vocabulary_size += 1

    # add the token corresponding to the current word to the tokenized text.
    text_tokens.append(word_dict[word])
```

When we were tokenizing our text, we also generated a dictionary **word_dict** that maps words to integers and a **word_list** that maps each integer to the corresponding word.

```python
print("Word list:", word_list, "\n\n Word dictionary:")
pprint(word_dict)
```

    Word list: ['mary', 'had', 'a', 'little', 'lamb', 'whose', 'fleece', 'was', 'white', 'as', 'snow', 'and', 'everywhere', 'that', 'went', 'the', 'sure', 'to', 'go']

     Word dictionary:
    {'a': 2,
     'and': 11,
     'as': 9,
     'everywhere': 12,
     'fleece': 6,
     'go': 18,
     'had': 1,
     'lamb': 4,
     'little': 3,
     'mary': 0,
     'snow': 10,
     'sure': 16,
     'that': 13,
     'the': 15,
     'to': 17,
     'was': 7,
     'went': 14,
     'white': 8,
     'whose': 5}

These two datastructures already proved their usefulness when we converted our text to a list of tokens.

```python
print(text_tokens)
```

    [0, 1, 2, 3, 4, 3, 4, 3, 4, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 0, 14, 0, 14, 0, 14, 12, 13, 0, 14, 15, 4, 7, 16, 17, 18]

Unfortunately, while this representation is convenient for memory reasons it has some severe limitations. Perhaps the most important of which is the fact that computers naturally assume that numbers can be operated on mathematically (by addition, subtraction, etc) in a way that doesn't match our understanding of words.

## One-hot encoding

One typical way of overcoming this difficulty is to represent each word by a one-hot encoded vector where every element is zero except the one corresponding to a specific word.

```python
def one_hot(word, word_dict):
    """
        Generate a one-hot encoded vector corresponding to *word*
    """

    vector = np.zeros(len(word_dict))
    vector[word_dict[word]] = 1

    return vector
```

So, for example, the word "fleece" would be represented by:

```python
fleece_hot = one_hot("fleece", word_dict)
print(fleece_hot)
```

    [0. 0. 0. 0. 0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]

This vector has every element set to zero, except element 6, since:

```python
print(word_dict["fleece"])
fleece_hot[6] == 1
```

    6

    True

## Bag of words

We can now use the one-hot encoded vector for each word to produce a vector representation of our original text, by simply adding up all the one-hot encoded vectors:

```python
text_vector1 = np.zeros(vocabulary_size)

for word in text_words:
    hot_word = one_hot(word, word_dict)
    text_vector1 += hot_word

print(text_vector1)
```

    [6. 2. 2. 4. 5. 1. 1. 2. 1. 1. 1. 1. 2. 2. 4. 1. 1. 1. 1.]

In practice, we can also easily skip the encoding step at the word level by using the *word_dict* defined above:

```python
text_vector = np.zeros(vocabulary_size)

for word in text_words:
    text_vector[word_dict[word]] += 1

print(text_vector)
```

    [6. 2. 2. 4. 5. 1. 1. 2. 1. 1. 1. 1. 2. 2. 4. 1. 1. 1. 1.]

Naturally, this approach is completely equivalent to the previous one and has the added advantage of being more efficient in terms of both speed and memory requirements.

This is known as the __bag of words__ representation of the text. It should be noted that these vectors simply contains the number of times each word appears in our document, so we can easily tell that the word *mary* appears exactly 6 times in our little nursery rhyme.

```python
text_vector[word_dict["mary"]]
```

    6.0

A more pythonic (and efficient) way of producing the same result is to use the standard __Counter__ module:

```python
word_counts = Counter(text_words)
pprint(word_counts)
```

    Counter({'mary': 6,
             'lamb': 5,
             'little': 4,
             'went': 4,
             'had': 2,
             'a': 2,
             'was': 2,
             'everywhere': 2,
             'that': 2,
             'whose': 1,
             'fleece': 1,
             'white': 1,
             'as': 1,
             'snow': 1,
             'and': 1,
             'the': 1,
             'sure': 1,
             'to': 1,
             'go': 1})

From which we can easily generate the __text_vector__ and __word_dict__ data structures:

```python
items = list(word_counts.items())

# Extract word dictionary and vector representation
word_dict2 = dict([[items[i][0], i] for i in range(len(items))])
text_vector2 = [items[i][1] for i in range(len(items))]
```

And let's take a look at them:

```python
print("Text vector:", text_vector2, "\n\nWord dictionary:")
pprint(word_dict2)
```

    Text vector: [6, 2, 2, 4, 5, 1, 1, 2, 1, 1, 1, 1, 2, 2, 4, 1, 1, 1, 1]

    Word dictionary:
    {'a': 2,
     'and': 11,
     'as': 9,
     'everywhere': 12,
     'fleece': 6,
     'go': 18,
     'had': 1,
     'lamb': 4,
     'little': 3,
     'mary': 0,
     'snow': 10,
     'sure': 16,
     'that': 13,
     'the': 15,
     'to': 17,
     'was': 7,
     'went': 14,
     'white': 8,
     'whose': 5}

The results using this approach are slightly different than the previous ones, because the words are mapped to different integer ids but the corresponding values are the same:

```python
for word in word_dict.keys():
    if text_vector[word_dict[word]] != text_vector2[word_dict2[word]]:
        print("Error!")
```

As expected, there are no differences!

## Term Frequency

The bag of words vector representation introduced above relies simply on the frequency of occurence of each word. Following a long tradition of giving fancy names to simple ideas, this is known as __Term Frequency__.

Intuitively, we expect the the frequency with which a given word is mentioned should correspond to the relevance of that word for the piece of text we are considering. For example, **Mary** is a pretty important word in our little nursery rhyme and indeed it is the one that occurs the most often:

```python
sorted(items, key=lambda x:x[1], reverse=True)
```

    [('mary', 6),
     ('lamb', 5),
     ('little', 4),
     ('went', 4),
     ('had', 2),
     ('a', 2),
     ('was', 2),
     ('everywhere', 2),
     ('that', 2),
     ('whose', 1),
     ('fleece', 1),
     ('white', 1),
     ('as', 1),
     ('snow', 1),
     ('and', 1),
     ('the', 1),
     ('sure', 1),
     ('to', 1),
     ('go', 1)]

However, it's hard to draw conclusions from such a small piece of text. Let us consider a significantly larger piece of text, the first 100 MB of the english Wikipedia from: http://mattmahoney.net/dc/textdata. For the sake of convenience, text8.gz has been included in this repository in the **data/** directory. We start by loading it's contents into memory as an array of words:

```python
data = []

for line in gzip.open("data/text8.gz", 'rt'):
    data.extend(line.strip().split())
```

Now let's take a look at the most common words in this large corpus:

```python
counts = Counter(data)

sorted_counts = sorted(list(counts.items()), key=lambda x:x[1], reverse=True)

for word, count in sorted_counts[:10]:
    print(word, count)
```

    the 1061396
    of 593677
    and 416629
    one 411764
    in 372201
    a 325873
    to 316376
    zero 264975
    nine 250430
    two 192644

Surprisingly, we find that the most common words are not particularly meaningful. Indeed, this is a common occurence in Natural Language Processing. The most frequent words are typically auxiliaries required due to gramatical rules.

On the other hand, there is also a large number of words that occur very infrequently as can be easily seen by glancing at the word freqency distribution.

```python
dist = Counter(counts.values())
dist = list(dist.items())
dist.sort(key=lambda x:x[0])
dist = np.array(dist)

norm = np.dot(dist.T[0], dist.T[1])

plt.loglog(dist.T[0], dist.T[1]/norm)
plt.xlabel("count")
plt.ylabel("P(count)")
plt.title("Word frequency distribution")
```

## Stopwords

One common technique to simplify NLP tasks is to remove what are known as Stopwords, words that are very frequent but not meaningful. If we simply remove the most common 100 words, we significantly reduce the amount of data we have to consider while losing little information.

```python
stopwords = set([word for word, count in sorted_counts[:100]])

clean_data = []

for word in data:
    if word not in stopwords:
        clean_data.append(word)

print("Original size:", len(data))
print("Clean size:", len(clean_data))
print("Reduction:", 1-len(clean_data)/len(data))
```

    Original size: 17005207
    Clean size: 9006229
    Reduction: 0.470384041782026

Wow, our dataset size was reduced almost in half!

In practice, we don't simply remove the most common words in our corpus but rather a manually curate list of stopwords. Lists for dozens of languages and applications can easily be found online.

## Term Frequency/Inverse Document Frequency

One way of determining of the relative importance of a word is to see how often it appears across multiple documents. Words that are relevant to a specific topic are more likely to appear in documents about that topic and much less in documents about other topics. On the other hand, less meaningful words (like **the**) will be common across documents about any subject.

To measure the document frequency of a word we will need to have multiple documents. For the sake of simplicity, we will treat each sentence of our nursery rhyme as an individual document:

```python
corpus_text = text.split('.')
corpus_words = []

for document in corpus_text:
    doc_words = extract_words(document)
    corpus_words.append(doc_words)
```

Now our corpus is represented as a list of word lists, where each list is just the word representation of the corresponding sentence:

```python
pprint(corpus_words)
```

    [['mary', 'had', 'a', 'little', 'lamb', 'little', 'lamb', 'little', 'lamb'],
     ['mary',
      'had',
      'a',
      'little',
      'lamb',
      'whose',
      'fleece',
      'was',
      'white',
      'as',
      'snow'],
     ['and', 'everywhere', 'that', 'mary', 'went', 'mary', 'went', 'mary', 'went'],
     ['everywhere',
      'that',
      'mary',
      'went',
      'the',
      'lamb',
      'was',
      'sure',
      'to',
      'go']]

Let us now calculate the number of documents in which each word appears:

```python
document_count = {}

for document in corpus_words:
    word_set = set(document)

    for word in word_set:
        document_count[word] = document_count.get(word, 0) + 1

pprint(document_count)
```

    {'a': 2,
     'and': 1,
     'as': 1,
     'everywhere': 2,
     'fleece': 1,
     'go': 1,
     'had': 2,
     'lamb': 3,
     'little': 2,
     'mary': 4,
     'snow': 1,
     'sure': 1,
     'that': 2,
     'the': 1,
     'to': 1,
     'was': 2,
     'went': 2,
     'white': 1,
     'whose': 1}

As we can see, the word __Mary__ appears in all 4 of our documents, making it useless when it comes to distinguish between the different sentences. On the other hand, words like __white__ which appear in only one document are very discriminative. Using this approach we can define a new quantity, the ___Inverse Document Frequency__ that tells us how frequent a word is across the documents in a specific corpus:

```python
def inv_doc_freq(corpus_words):
    number_docs = len(corpus_words)

    document_count = {}

    for document in corpus_words:
        word_set = set(document)

        for word in word_set:
            document_count[word] = document_count.get(word, 0) + 1

    IDF = {}

    for word in document_count:
        IDF[word] = np.log(number_docs/document_count[word])

    return IDF
```

Where we followed the convention of using the logarithm of the inverse document frequency. This has the numerical advantage of avoiding to have to handle small fractional numbers.

We can easily see that the IDF gives a smaller weight to the most common words and a higher weight to the less frequent:

```python
IDF = inv_doc_freq(corpus_words)

pprint(IDF)
```

    {'a': 0.6931471805599453,
     'and': 1.3862943611198906,
     'as': 1.3862943611198906,
     'everywhere': 0.6931471805599453,
     'fleece': 1.3862943611198906,
     'go': 1.3862943611198906,
     'had': 0.6931471805599453,
     'lamb': 0.28768207245178085,
     'little': 0.6931471805599453,
     'mary': 0.0,
     'snow': 1.3862943611198906,
     'sure': 1.3862943611198906,
     'that': 0.6931471805599453,
     'the': 1.3862943611198906,
     'to': 1.3862943611198906,
     'was': 0.6931471805599453,
     'went': 0.6931471805599453,
     'white': 1.3862943611198906,
     'whose': 1.3862943611198906}

As expected **Mary** has the smallest weight of all words 0, meaning that it is effectively removed from the dataset. You can consider this as a way of implicitly identify and remove stopwords. In case you do want to keep even the words that appear in every document, you can just add a 1. to the argument of the logarithm above:

\begin{equation}
\log\left[1+\frac{N_d}{N_d\left(w\right)}\right]
\end{equation}

When we multiply the term frequency of each word by it's inverse document frequency, we have a good way of quantifying how relevant a word is to understand the meaning of a specific document.

```python
def tf_idf(corpus_words):
    IDF = inv_doc_freq(corpus_words)

    TFIDF = []

    for document in corpus_words:
        TFIDF.append(Counter(document))

    for document in TFIDF:
        for word in document:
            document[word] = document[word]*IDF[word]

    return TFIDF
```

```python
tf_idf(corpus_words)
```

    [Counter({'mary': 0.0,
              'had': 0.6931471805599453,
              'a': 0.6931471805599453,
              'little': 2.0794415416798357,
              'lamb': 0.8630462173553426}),
     Counter({'mary': 0.0,
              'had': 0.6931471805599453,
              'a': 0.6931471805599453,
              'little': 0.6931471805599453,
              'lamb': 0.28768207245178085,
              'whose': 1.3862943611198906,
              'fleece': 1.3862943611198906,
              'was': 0.6931471805599453,
              'white': 1.3862943611198906,
              'as': 1.3862943611198906,
              'snow': 1.3862943611198906}),
     Counter({'and': 1.3862943611198906,
              'everywhere': 0.6931471805599453,
              'that': 0.6931471805599453,
              'mary': 0.0,
              'went': 2.0794415416798357}),
     Counter({'everywhere': 0.6931471805599453,
              'that': 0.6931471805599453,
              'mary': 0.0,
              'went': 0.6931471805599453,
              'the': 1.3862943611198906,
              'lamb': 0.28768207245178085,
              'was': 0.6931471805599453,
              'sure': 1.3862943611198906,
              'to': 1.3862943611198906,
              'go': 1.3862943611198906})]

Now we finally have a vector representation of each of our documents that takes the informational contributions of each word into account. Each of these vectors provides us with a unique representation of each document, in the context (corpus) in which it occurs, making it posssible to define the similarity of two documents, etc.

## Porter Stemmer

There is still, however, one issue with our approach to representing text. Since we treat each word as a unique token and completely independently from all others, for large documents we will end up with many variations of the same word such as verb conjugations, the corresponding adverbs and nouns, etc.

One way around this difficulty is to use stemming algorithm to reduce words to their root (or stem) version. The most famous Stemming algorithm is known as the **Porter Stemmer** and was introduced by Martin Porter in 1980 [Program 14, 130 (1980)](https://dl.acm.org/citation.cfm?id=275705)

The algorithm starts by defining consonants (C) and vowels (V):

```python
V = set('aeiouy')
C = set('bcdfghjklmnpqrstvwxz')
```

The stem of a word is what is left of that word after a speficic ending has been removed. A function to do this is easy to implement:

```python
def get_stem(suffix, word):
    """
        Extract the stem of a word
    """

    if word.lower().endswith(suffix.lower()): # Case insensitive comparison
        return word[:-len(suffix)]

    return None
```

It also defines words (or stems) to be sequences of vowels and consonants of the form:

\begin{equation}
[C](VC)^m[V]
\end{equation}

where $m$ is called the **measure** of the word and [] represent optional sections.

```python
def measure(orig_word):
    """
        Calculate the "measure" m of a word or stem, according to the Porter Stemmer algorthim
    """

    word = orig_word.lower()

    optV = False
    optC = False
    VC = False
    m = 0

    pos = 0

    # We can think of this implementation as a simple finite state machine
    # looks for sequences of vowels or consonants depending of the state
    # in which it's in, while keeping track of how many VC sequences it
    # has encountered.
    # The presence of the optional V and C portions is recorded in the
    # optV and optC booleans.

    # We're at the initial state.
    # gobble up all the optional consonants at the beginning of the word
    while pos < len(word) and word[pos] in C:
        pos += 1
        optC = True

    while pos < len(word):
        # Now we know that the next state must be a vowel
        while pos < len(word) and word[pos] in V:
            pos += 1
            optV = True

        # Followd by a consonant
        while pos < len(word) and word[pos] in C:
            pos += 1
            optV = False

        # If a consonant was found, then we matched VC
        # so we should increment m by one. Otherwise,
        # optV remained true and we simply had a dangling
        # V sequence.
        if not optV:
            m += 1

    return m
```

Let's consider a simple example. The word __crepusculars__ should have measure 4:

[cr] (ep) (usc) (ul) (ars)

and indeed it does.

```python
word = "crepusculars"
print(measure(word))
```

    4

The Porter algorithm sequentially applies a series of transformation rules over a series of 5 steps (step 1 is divided in 3 substeps and step 5 in 2). The rules are only applied if a certain condition is true.

In addition to possibily specifying a requirement on the measure of a word, conditions can make use of different boolean functions as well:

```python
def ends_with(char, stem):
    """
        Checks the ending of the word
    """
    return stem[-1] == char

def double_consonant(stem):
    """
        Checks the ending of a word for a double consonant
    """
    if len(stem) < 2:
        return False

    if stem[-1] in C and stem[-2] == stem[-1]:
        return True

    return False

def contains_vowel(stem):
    """
        Checks if a word contains a vowel or not
    """
    return len(set(stem) & V) > 0
```

Finally, we define a function to apply a specific rule to a word or stem:

```python
def apply_rule(condition, suffix, replacement, word):
    """
        Apply Porter Stemmer rule.
        if "condition" is True replace "suffix" by "replacement" in "word"
    """

    stem = get_stem(suffix, word)

    if stem is not None and condition is True:
        # Remove the suffix
        word = stem

        # Add the replacement suffix, if any
        if replacement is not None:
            word += replacement

    return word
```

Now we can see how rules can be applied. For example, this rule, from step 1b is successfully applied to __pastered__:

```python
word = "plastered"
suffix = "ed"
stem = get_stem(suffix, word)
apply_rule(contains_vowel(stem), suffix, None, word)
```

    'plaster'

While try applying the same rule to **bled** will fail to pass the condition resulting in no change.

```python
word = "bled"
suffix = "ed"
stem = get_stem(suffix, word)
apply_rule(contains_vowel(stem), suffix, None, word)
```

    'bled'

For a more complex example, we have, in Step 4:

```python
word = "adoption"
suffix = "ion"
stem = get_stem(suffix, word)
apply_rule(measure(stem) > 1 and (ends_with("s", stem) or ends_with("t", stem)), suffix, None, word)
```

    'adopt'

In total, the Porter Stemmer algorithm (for the English language) applies several dozen rules (see https://tartarus.org/martin/PorterStemmer/def.txt for a complete list). Implementing all of them is both tedious and error prone, so we abstain from providing a full implementation of the algorithm here. High quality implementations can be found in all major NLP libraries such as [NLTK](http://www.nltk.org/howto/stem.html).

The dificulties of defining matching rules to arbitrary text cannot be fully resolved without the use of Regular Expressions (typically implemented as Finite State Machines like our __measure__ implementation above), a more advanced topic that is beyond the scope of this course.


# The Jupyter Notebook

It is a **```web application```** that allows you to create and share documents that contain
- live code
- equations
- visualizations
- explanatory text

Table of Contents

The Jupyter Notebook

搜狗输入法表情和符号

一级标题

二级标题

运行C代码

Jupyter 魔术命令

END

```python
# my first python script
print("hello world! \n I am Cheng-Jun Wang.")
```

    hello world!
     I am Cheng-Jun Wang.

Uses include:
- data cleaning and transformation,
- numerical simulation,
- statistical modeling,
- machine learning
- and much more.

```python
print('hello world')
```

    hello world

```python
1 + 1
```

    2

$E = MC^2$

\begin{align}
\dot{x} & = \sigma(y-x) \\
\dot{y} & = \rho x - y - xz \\
\dot{z} & = -\beta z + xy
\end{align}

# 搜狗输入法表情和符号

- ∫ ∑ ※ ➕➖✖➗ ❎ √ ×
- 😪😠😡😎☺😁📚🌲
- 👌👍👎👂👃👀✋❌💰🌂
- 0⃣1⃣2⃣3⃣4⃣5⃣6⃣7⃣8⃣9⃣②🔟
- 🐶🐱🐔🐷🐖🐴🐎🐂🐑🐯🐧🐺🐒🐵🐻🐦🐲
- 💻 🌈🌎☁❄🏃♀👩👱✨
- 🆚🔥🌹✈🌉🎄

(✿◡‿◡)害羞 ⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄ d=====(￣▽￣*)b厉害

我是我，不一样花火。～(￣▽￣～)(～￣▽￣)～ 矜持

```python
%matplotlib inline
import matplotlib.pyplot as plt

xi = [1, 2, 3, 4, 5]
y = [3, 5, 9, 13, 16]

plt.plot(xi, y, 'g-s')
plt.xlabel('$x_i$', fontsize = 20)
plt.ylabel('$y$', fontsize = 20)
plt.title('$Scatter\,Plot$', fontsize = 20)
plt.show()
```

# 一级标题
## 二级标题
[复旦大学](http://www.fdu.edu.cn)是一个*非常棒*的大学！

1. point 1
1. point 2
1. point 3

# 运行C代码

C functions are typically split into header files (.h) where things are declared but not defined, and implementation files (.c) where they are defined. http://people.duke.edu/~ccc14/sta-663/CrashCourseInC.html#a-tutorial-example-coding-a-fibonacci-function-in-c

```python
%%file hello.c
#include <stdio.h>

int main() {
    printf("Hello, world!");
}
```

    Overwriting hello.c

```python
! gcc hello.c -o hello # 编译
```

    xcrun: error: invalid active developer path (/Library/Developer/CommandLineTools), missing xcrun at: /Library/Developer/CommandLineTools/usr/bin/xcrun

```python
! ./hello # 执行
```

    Hello, world!

# Jupyter 魔术命令

```python
%lsmagic
```

    Available line magics:
    %alias  %alias_magic  %autocall  %automagic  %autosave  %bookmark  %cat  %cd  %clear  %colors  %config  %connect_info  %cp  %debug  %dhist  %dirs  %doctest_mode  %ed  %edit  %env  %gui  %hist  %history  %killbgscripts  %ldir  %less  %lf  %lk  %ll  %load  %load_ext  %loadpy  %logoff  %logon  %logstart  %logstate  %logstop  %ls  %lsmagic  %lx  %macro  %magic  %man  %matplotlib  %mkdir  %more  %mv  %notebook  %page  %pastebin  %pdb  %pdef  %pdoc  %pfile  %pinfo  %pinfo2  %popd  %pprint  %precision  %profile  %prun  %psearch  %psource  %pushd  %pwd  %pycat  %pylab  %qtconsole  %quickref  %recall  %rehashx  %reload_ext  %rep  %rerun  %reset  %reset_selective  %rm  %rmdir  %run  %save  %sc  %set_env  %store  %sx  %system  %tb  %time  %timeit  %unalias  %unload_ext  %who  %who_ls  %whos  %xdel  %xmode

    Available cell magics:
    %%!  %%HTML  %%SVG  %%bash  %%capture  %%debug  %%file  %%html  %%javascript  %%js  %%latex  %%markdown  %%perl  %%prun  %%pypy  %%python  %%python2  %%python3  %%ruby  %%script  %%sh  %%svg  %%sx  %%system  %%time  %%timeit  %%writefile

    Automagic is ON, % prefix IS NOT needed for line magics.

>  pip install version_information

```python
!pip install version_information
```

    Requirement already satisfied: version_information in /Users/datalab/Applications/anaconda/lib/python3.5/site-packages (1.0.3)
    You are using pip version 19.0.3, however version 19.1.1 is available.
    You should consider upgrading via the 'pip install --upgrade pip' command.

```python
# install version_information in the terminal first.
%reload_ext version_information
%version_information numpy, matplotlib, pandas, scipy, statsmodels
```

Software	Version

Python	3.5.4 64bit [GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)]

IPython	6.2.1

OS	Darwin 18.6.0 x86_64 i386 64bit

numpy	1.16.3

matplotlib	3.0.1

pandas	0.23.4

scipy	1.1.0

statsmodels	0.9.0

Fri Jun 07 14:29:42 2019 CST

# END

This is the end.

```python

```


|
Name
               |
Date
 |
| ---------------------------------------------------| ------------------------------------- |
|
Diaaeldin SHALABY
   | 18.06.2021                            |

Hands-on AI II

Unit 7 — Introduction to Reinforcement Learning (Assignment)

Authors: B. Schäfl, S. Lehner, J. Brandstetter

Date: 11-06-2021

This file is part of the "Hands-on AI II" lecture material. The following copyright statement applies to all code within this file.

Copyright statement:

This  material,  no  matter  whether  in  printed  or  electronic  form,  may  be  used  for personal  and non-commercial educational use only.  Any reproduction of this manuscript, no matter whether as a whole or in parts, no matter whether in printed or in electronic form, requires explicit prior acceptance of the authors.

Table of contents

Dissection of an Environment

States and actions

Tackling the Environment with Random Exploration

Implementing random search

The problem with random search

Tackling the Environment with $Q$-Learning

First approach of learning a $Q$-table

The problem with missing exploration

Evaluate the agent's performance

The role of randomness in the environment

How to use this notebook

This notebook is designed to run from start to finish. There are different tasks (displayed in orange boxes) which require your contribution (in form of code, plain text, ...). Most/All of the supplied functions are imported from the file u7_utils.py which can be seen and treated as a black box. However, for further understanding, you can look at the implementations of the helper functions. In order to run this notebook, the packages which are imported at the beginning of u7_utils.py need to be installed.

```python
# Import pre-defined utilities specific to this notebook.
import u7_utils as u7

# Import additional utilities needed in this notebook.
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import seaborn as sns
import sys
import time

from IPython import display
from typing import Any, Dict, Tuple

# Setup Jupyter notebook (warning: this may affect all Jupyter notebooks running on the same Jupyter server).
u7.setup_jupyter()
```

Setting up notebook ... finished.

Module versions

As mentioned in the introductory slides, specific minimum versions of Python itself as well as of used modules is recommended.

```python
u7.check_module_versions()
```

    Installed Python version: 3.8 (✓)
    Installed numpy version: 1.19.1 (✓)
    Installed pandas version: 1.1.3 (✓)
    Installed PyTorch version: 1.7.1 (✓)
    Installed scikit-learn version: 0.23.2 (✓)
    Installed scipy version: 1.5.0 (✓)
    Installed matplotlib version: 3.3.1 (✓)
    Installed seaborn version: 0.11.0 (✓)
    Installed PIL version: 8.0.0 (✓)
    Installed rdkit version: 2020.09.1 (✓)
    Installed gym version: 0.18.3 (✓)

Dissection of an Environment

All exercises in this assignment are referring to the FrozenLake-v0 environment of OpenAI Gym. This environment is descibed according to its official OpenAI Gym website as follows:

    Winter is here. You and your friends were tossing around a frisbee at the park when you made a wild throw that left the frisbee out in the middle of the lake. The water is mostly frozen, but there are a few holes where the ice has melted. If you step into one of those holes, you'll fall into the freezing water. At this time, there's an international frisbee shortage, so it's absolutely imperative that you navigate across the lake and retrieve the disc. However, the ice is slippery, so you won't always move in the direction you intend.

There are four types of surfaces described in this environment:

S $\rightarrow$ starting point (safe)

F $\rightarrow$ frozen surface (safe)

H $\rightarrow$ hole (fall to your doom)

G $\rightarrow$ goal (frisbee location)

If not already done, more information on how to install and import the gym module is available in the lecture's notebook.

    Execute the notebook until here and try to solve the following tasks:

Create a new instance of <tt>FrozenLakeEnv</tt> with the seed set to <tt>42</tt> and render the current state in a human-readable way.

Gather and print the amount of different actions as well as states of the <tt>FrozenLakeEnv</tt> instance. Discuss the results.

Display the reward table entry for the current state. Discuss the different elements of the resulting dictionary.

```python
enviroment_lake = u7.FrozenLakeEnv()
u7.set_environment_seed(environment=enviroment_lake, seed=42)
```

```python
enviroment_lake.render(mode=r'human')
current_state_id = enviroment_lake.s
print(f'Current state ID: {current_state_id}')
```

    SFFF
    FHFH
    FFFH
    HFFG
    Current state ID: 0

The first and naïve approach to solve this task is by simply brute forcing it applying random search. The outline of this approach is the following:

        Step
        Description

        0
        Choose a random action with respect to the current state.

        1
        Execute previously chosen action and transition into a new state.

        2
        Repeat the previous steps as long as the current episode is still ongoing.

For such an approach to be at least remotely applicable, the number of possible actions and states is of utter importance. Otherwise, we are lost in the depth of combinatorial explosion. The property n of the action_space and observation_space of the respective environment gives the amount of actions as well as states.

```python
num_actions = enviroment_lake.action_space.n
num_states = enviroment_lake.observation_space.n
print(f'The FrozenLake-v0 environment comprises <{num_actions}> actions and <{num_states}> states.')
```

    The FrozenLake-v0 environment comprises <4> actions and <16> states.

```python
current_state_id = enviroment_lake.s
enviroment_lake.P[current_state_id]
```

    {0: [(0.3333333333333333, 0, 0.0, False),
      (0.3333333333333333, 0, 0.0, False),
      (0.3333333333333333, 4, 0.0, False)],
     1: [(0.3333333333333333, 0, 0.0, False),
      (0.3333333333333333, 4, 0.0, False),
      (0.3333333333333333, 1, 0.0, False)],
     2: [(0.3333333333333333, 4, 0.0, False),
      (0.3333333333333333, 1, 0.0, False),
      (0.3333333333333333, 0, 0.0, False)],
     3: [(0.3333333333333333, 1, 0.0, False),
      (0.3333333333333333, 0, 0.0, False),
      (0.3333333333333333, 0, 0.0, False)]}

Each entry of the reward table contains a dictionary of the form s: {a: [] for a in range(nA)} for s in range(nS).

        Element
        Description

        nS=nrows*ncols
        The number of possible moves.

        nA=4
        Number of surfaces. [S,F,H,G]

        a
        Potential state based on the surface of the next move.

Tackling the Environment with Random Exploration

Previously, we talked about solving this task in a naïve way by simply applying brute force: using random search. In the meantime we analyzed the action as well as the state space and came to the conclusion, that such an approach is more than feasible. To repeat the outline of such an approach:

I $\rightarrow$ choose a random action with respect to the current state.

II $\rightarrow$ execute previously chosen action and transition into a new state.

III $\rightarrow$ if the episode is finished, but the goal not reached, reset the position of the disc retrieving entity.

This procedure is repeated as long as the task is not solved or a defined maximum of steps is reached, whatever triggers first (IV). Adapt the function apply_random_search as discussed during the lecture. Mark the corresponding sections of the code using I, II, III and IV. Note that our random search is not guaranteed to find  the solution of a task in finite time, hence an upper border on the runtime is often applied as a safety net (in our case the number of allowed steps).

    Execute the notebook until here and try to solve the following tasks:

Implement the random search algorithm as outlined above (equivalently to the one discussed during the exercise).

Apply your random search implementation on a freshly seeded <tt>FrozenLakeEnv</tt> instance, with an animation delay of $0.1$.

How many steps are necessary to reach the goal at least once and how often did an involuntary dive happen?

```python
def apply_random_search(environment: u7.FrozenLakeEnv, animate: bool = False,
                        delay: float = 0.01, max_steps: int = 1000) -> Tuple[int, int, Dict[str, Any]]:
    """
    Solve specified environment by applying random search.

    :param environment: the environment on which to apply random search
    :param animate: animate the random search process
    :param delay: the minimum delay in milliseconds between each rendered frame (ignored if not animated)
    :param max_steps: maximum amount of steps to perform
    :return: amount of steps performed, involuntary dives and captured frames
    """
    num_steps, num_penalties, final_reward, captured_frames = 0, 0, 0, []

    # <III>: repeat random search procedure as long as the episode is still ongoing.

    u7.set_environment_seed(environment=enviroment_lake, seed=42)

    done = False
    while not done:
        if num_steps == max_steps:
            break

        # <I>: choose a random action with respect to the current state.
        current_action = environment.action_space.sample()

        # <II>: execute previously chosen action and transition into a new state.
        current_state, current_reward, done, info = environment.step(current_action)

        # Update counter for inflicted penalties.
        final_reward += current_reward
        if current_reward <= 0:
            num_penalties += 1
        num_steps += 1

        # Save rendering of current state.
        captured_frames.append({
            r'frame': environment.render(mode=r'ansi'),
            r'state': current_state,
            r'action': current_action,
            r'reward': current_reward
        })

        # Optionally display current state.
        if animate:
            display.clear_output(wait=True)
            print(captured_frames[-1][r'frame'])
            print(f'Step No.: {num_steps}'
                  f'\nState ID: {current_state}'
                  f'\nAction ID: {current_action}'
                  f'\nReward: {current_reward}')
            time.sleep(delay)

        if not done and current_reward == 0:
            enviroment_lake.reset()

    return num_steps, num_penalties, final_reward, captured_frames
```

```python
u7.set_environment_seed(environment=enviroment_lake, seed=42)
num_steps, num_penalties, final_reward, _ = apply_random_search(
    environment=enviroment_lake,
    animate=True,
    delay=0.1
)
```

      (Right)
    SFFF
    FHFH
    FFFH
    HFFG

    Step No.: 1000
    State ID: 1
    Action ID: 2
    Reward: 0.0

Unfortunatly the function did not find a solution.

To drill down on the drawbacks of plain random search, we are designing the following experimental setup (hint: it is actually the same experimental setup as already discussed during the exercise, so you might orient yourself on the implementation presented during class):

Repeat the previous random search procedure a specified amount of times.

Aggregate the results of each run for later analysis.

Visualise the aggegrated results using box- and swarm-plots.

Once again, we are setting the random seed, but take care of setting it outside the loop, otherwise the same result is reported with each iteration (and an aggregation of the results would not give us any more insights).

    Execute the notebook until here and try to solve the following tasks:

Conduct a random search experiment as outlined above, using $100$ repetitions and the random seed set to $42$.

Interpret the visualization (e.g. the span of the boxes) and keep the scaling of the x-axis in mind.

In comparison with the <tt>Taxi-v3</tt> environment, what might be the problem with <tt>FrozenLakeEnv</tt> w.r.t. random exploration?

```python
u7.set_environment_seed(environment=enviroment_lake, seed=42)
num_steps_total, num_penalties_total, final_reward_total = [], [], []
num_repetitions = 100

# Collect information over multiple repetitions.
for repetition in range(num_repetitions):
    enviroment_lake.reset()
    num_steps, num_penalties, final_reward, _ = apply_random_search(environment=enviroment_lake)
    num_steps_total.append(num_steps)
    num_penalties_total.append(num_penalties)
    final_reward_total.append(final_reward)

# Combine collected information to a data frame for further downstream analysis.
collected_experiment_info = pd.DataFrame(zip(
    num_steps_total, num_penalties_total, final_reward_total
), columns=(r'Steps performed', r'Penalties inflicted', r'Final reward'))
```

```python
# Set default plotting style.
sns.set()

# Visualize aggregated results of the random search procedure.
fig, ax = plt.subplots(nrows=1, ncols=1, squeeze=False, figsize=(20, 7))
ax[0, 0].set_xscale(r'symlog')
_ = sns.boxplot(data=collected_experiment_info, ax=ax[0, 0], orient=r'h')
_ = sns.swarmplot(data=collected_experiment_info, ax=ax[0, 0], color=r'0.3', orient=r'h')
```

Tackling the Environment with $Q$-Learning

In a simplified version of $Q$-learning, the $\boldsymbol{Q}$-value
\begin{equation}
    Q(s,a)
\end{equation}

is the expected future reward of being in state $s$ and taking action $a$. Intuitively, if the the $Q$-values are learned correctly, a good policy would be to take the action which maximizes the expected future reward. This is what $Q$-learning is doing. $Q$-learning lets the agent use the environment's rewards to learn, over time, the best action to take in a given state. $Q$-values are initialized to an arbitrary value, and as the agent exposes itself to the environment and receives different rewards by executing different actions, the $Q$-values are updated using the equation:
\begin{equation}
    Q(s_t,a_t) \leftarrow (1 - \alpha) \cdot Q(s_t,a_t) + \alpha \cdot \left( r + \max_{a_{t+1}} Q(s_{t+1}, a_{t+1})\right)
\end{equation}

We are assigning $\leftarrow$, or updating, the $Q$-value of the agent's current state and action, denoted as $Q(s_t,a_t)$ with $\alpha$ as the learning rate, i.e the extent to which our $Q$-values are being updated in every iteration.

The $\boldsymbol{Q}$-table is a matrix where we have a row for every state and a column for every action – $500$ and $6$, respectively, when referring to the Taxi-v3 example, as discussed during class. It's first initialized to $0$, and then values are updated after training.

Previously, we talked about solving this task in a naïve way by simply applying brute force: using random search. This time we want to apply a more sophisticated algorithm – $Q$-learning:

I $\rightarrow$ Choose action $a_t$.

II $\rightarrow$ Go from state $s_t$ to state $s_{t+1}$ by taking action $a_{t}$.

III $\rightarrow$ For all possible $Q$-values from the state $s_{t+1}$, select the highest.

IV $\rightarrow$ Update $Q$-table values using the equation from above.

V $\rightarrow$ Set the next state as the current state.

This procedure is repeated for as many episodes as specified (VI).

    Execute the notebook until here and try to solve the following tasks:

Implement $Q$-learning as outlined above (equivalently to the one discussed during the exercise).

Apply $Q$-learning on a freshly seeded <tt>FrozenLakeEnv</tt> instance for $10^4$ episodes, with $10^3$ delay steps and $\alpha{}=0.1$.

Interpret the visualization of the resulting $Q$-table. What do you observe?

    The following code snippet is taken from the accompanying exercise notebook. You do not need to modify it for this assignment.

```python
def visualize_q_table(q_table: np.ndarray, title: str = r'') -> None:
    """
    Visualize Q-table using a heatmap plot.

    :param q_table: Q-table to visualize
    :return: None
    """
    sns.set()
    fig, ax = plt.subplots(nrows=1, ncols=1, squeeze=False, figsize=(20, 7))
    _ = sns.heatmap(data=q_table, ax=ax[0, 0])
    _ = ax[0, 0].set(xlabel=r'Action', ylabel=r'State', title=title)
    display.clear_output(wait=True)
    display.display(fig)
    plt.close(fig=fig)

def apply_q_learning(environment: u7.FrozenLakeEnv, num_episodes: int = 1000, alpha: float = 0.1,
                     animate: bool = False, delay_steps: int = 100) -> np.ndarray:
    """
    Solve specified environment by applying Q-learning.

    :param environment: the environment on which to apply Q-learning
    :param num_episodes: the total amount of episodes used to adapt the Q-table
    :param alpha: the learning rate to be applied by Q-learning
    :param animate: animate the Q-learning process
    :param delay_steps: the steps between each Q-table visualization (ignored if not animated)
    """
    q_table = np.zeros(shape=(environment.observation_space.n, environment.action_space.n))

    # <VI>: repeat Q-learning as long as the total amount of episodes is not yet reached.
    for episode in range(num_episodes):
        state = environment.reset()

        done = False
        while not done:

            # <I>: choose next action according to current Q-table.
            action = np.argmax(q_table[state])

            # <II>: go from the current state to the next by applying chosen action.
            next_state, reward, done, info = environment.step(action)

            # <III>: from all possible Q-values w.r.t. the new state, select the highest.
            next_max = np.max(q_table[next_state])

            # <IV>: update the Q-table accordingly.
            old_value = q_table[state, action]
            new_value = (1 - alpha) * old_value + alpha * (reward + next_max)
            q_table[state, action] = new_value

            # <V>: update the next step with the current one.
            state = next_state

        # Optionally visualize the current Q-table.
        if animate and any(((episode + 1) % delay_steps == 0, (episode + 1) == num_episodes)):
            visualize_q_table(q_table=q_table, title=f'Episode {episode + 1}')

    return q_table
```

```python
u7.set_environment_seed(environment=enviroment_lake, seed=42)
q_table = apply_q_learning(
    environment=enviroment_lake,
    num_episodes=10000,
    alpha=0.1,
    animate=True,
    delay_steps=1000
)
```

I obsere that it's all the same color which indicates that it's the same level.

Very likely the $Q$-table of the previous experiment looked a little bit odd. Try to add exploration to your algorithm by adapting your $Q$-learning implementation:

I $\rightarrow$ Throw a random uniform number between $0$ and $1$.

II $\rightarrow$ If the number is smaller than $0.1$, sample a random action.

III $\rightarrow$ Otherwise, choose your action as usual.

    Execute the notebook until here and try to solve the following tasks:

Modify the $Q$-learning implementation from the previous tasks as outlined above (mark the corresponding code sections).

Apply $Q$-learning on a freshly seeded <tt>FrozenLakeEnv</tt> instance for $10^4$ episodes, with $10^3$ delay steps and $\alpha{}=0.1$.

Interpret the visualization of the resulting $Q$-table. What do you observe (compare with the previous visualization)?

```python
def apply_q_learning(environment: u7.FrozenLakeEnv, num_episodes: int = 1000, alpha: float = 0.1,
                    animate: bool = False, delay_steps: int = 100, threshold: float = 0.125) -> np.ndarray:
    """
    Solve specified environment by applying Q-learning.

    :param environment: the environment on which to apply Q-learning
    :param num_episodes: the total amount of episodes used to adapt the Q-table
    :param alpha: the learning rate to be applied by Q-learning
    :param animate: animate the Q-learning process
    :param delay_steps: the steps between each Q-table visualization (ignored if not animated)
    :param threshold: threshold for randomly sampling next action
    :return: adapted Q-table
    """
    import random

    q_table = np.zeros(shape=(environment.observation_space.n, environment.action_space.n))

    # <VI>: repeat Q-learning as long as the total amount of episodes is not yet reached.
    for episode in range(num_episodes):
        state = environment.reset()

        done = False
        while not done:
            rnd_num = random.uniform(0, 1)
            action = np.argmax(q_table[state])

            if rnd_num < 0.1:
                action = environment.action_space.sample()
            else:
                action = np.argmax(q_table[state])

            # <II>: go from the current state to the next by applying chosen action.
            next_state, reward, done, info = environment.step(action)

            # <III>: from all possible Q-values w.r.t. the new state, select the highest.
            next_max = np.max(q_table[next_state])

            # <IV>: update the Q-table accordingly.
            old_value = q_table[state, action]
            new_value = (1 - alpha) * old_value + alpha * (reward + next_max)
            q_table[state, action] = new_value

            # <V>: update the next step with the current one.
            state = next_state

        # Optionally visualize the current Q-table.
        if animate and any(((episode + 1) % delay_steps == 0, (episode + 1) == num_episodes)):
            visualize_q_table(q_table=q_table, title=f'Episode {episode + 1}')

    return q_table
```

```python
u7.set_environment_seed(environment=enviroment_lake, seed=42)
q_table = apply_q_learning(
    environment=enviroment_lake,
    num_episodes=10000,
    alpha=0.1,
    animate=True,
    delay_steps=1000
)

```

    Execute the notebook until here and try to solve the following tasks:

Implement a function for applying a pre-trained $Q$-table on a <tt>FrozenLakeEnv</tt> instance (like discussed during class).

Conduct a $Q$-table guided search on a freshly seeded <tt>FrozenLakeEnv</tt> instance, with an animation delay of $0.1$.

How many steps are necessary to reach the goal at least once and how often did an involuntary dive happen?

```python
def apply_q_table(environment: u7.FrozenLakeEnv, q_table: np.ndarray, animate: bool = False,
                  delay: float = 0.01, max_steps: int = 1000) -> Tuple[int, int, Dict[str, Any]]:
    """
    Solve specified environment by applying specified Q-table.

    :param environment: the environment on which to apply Q-table guided search
    :param q_table: the Q-table used during Q-table guided search
    :param animate: animate the Q-table guided search process
    :param delay: the minimum delay in milliseconds between each rendered frame (ignored if not animated)
    :param max_steps: maximum amount of steps to perform
    :return: amount of steps performed, involuntary dives and captured frames
    """
    raise NotImplementedError(r'Exchange this error with your implementation.')
```

```python

```

```python

```

    Execute the notebook until here and try to solve the following tasks:

Conduct a $Q$-table guided search experiment as outlined previously, using $100$ repetitions and the random seed set to $42$.

Interpret the visualization (e.g. the span of the boxes) and keep the scaling of the x-axis in mind.

In comparison with the random search experiment, how does the $Q$-table guided search perform? Discuss the results.

```python

```

```python

```

Introduction to Reinforcement Learning.ipynb


Probabilistic Programming and Bayesian Methods for Hackers
========

Welcome to *Bayesian Methods for Hackers*. The full Github repository is available at [github/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers](https://github.com/CamDavidsonPilon/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers). The other chapters can be found on the project's [homepage](https://camdavidsonpilon.github.io/Probabilistic-Programming-and-Bayesian-Methods-for-Hackers/). We hope you enjoy the book, and we encourage any contributions!

#### Looking for a printed version of Bayesian Methods for Hackers?

_Bayesian Methods for Hackers_ is now a published book by Addison-Wesley, available on [Amazon](http://www.amazon.com/Bayesian-Methods-Hackers-Probabilistic-Addison-Wesley/dp/0133902838)!

Chapter 1
======
***

The Philosophy of Bayesian Inference
------

> You are a skilled programmer, but bugs still slip into your code. After a particularly difficult implementation of an algorithm, you decide to test your code on a trivial example. It passes. You test the code on a harder problem. It passes once again. And it passes the next, *even more difficult*, test too! You are starting to believe that there may be no bugs in this code...

If you think this way, then congratulations, you already are thinking Bayesian! Bayesian inference is simply updating your beliefs after considering new evidence. A Bayesian can rarely be certain about a result, but he or she can be very confident. Just like in the example above, we can never be 100% sure that our code is bug-free unless we test it on every possible problem; something rarely possible in practice. Instead, we can test it on a large number of problems, and if it succeeds we can feel more *confident* about our code, but still not certain.  Bayesian inference works identically: we update our beliefs about an outcome; rarely can we be absolutely sure unless we rule out all other alternatives.

### The Bayesian state of mind

Bayesian inference differs from more traditional statistical inference by preserving *uncertainty*. At first, this sounds like a bad statistical technique. Isn't statistics all about deriving *certainty* from randomness? To reconcile this, we need to start thinking like Bayesians.

The Bayesian world-view interprets probability as measure of *believability in an event*, that is, how confident we are in an event occurring. In fact, we will see in a moment that this is the natural interpretation of probability.

For this to be clearer, we consider an alternative interpretation of probability: *Frequentist*, known as the more *classical* version of statistics, assumes that probability is the long-run frequency of events (hence the bestowed title). For example, the *probability of plane accidents* under a frequentist philosophy is interpreted as the *long-term frequency of plane accidents*. This makes logical sense for many probabilities of events, but becomes more difficult to understand when events have no long-term frequency of occurrences. Consider: we often assign probabilities to outcomes of presidential elections, but the election itself only happens once! Frequentists get around this by invoking alternative realities and saying across all these realities, the frequency of occurrences defines the probability.

Bayesians, on the other hand, have a more intuitive approach. Bayesians interpret a probability as measure of *belief*, or confidence, of an event occurring. Simply, a probability is a summary of an opinion. An individual who assigns a belief of 0 to an event has no confidence that the event will occur; conversely, assigning a belief of 1 implies that the individual is absolutely certain of an event occurring. Beliefs between 0 and 1 allow for weightings of other outcomes. This definition agrees with the probability of a plane accident example, for having observed the frequency of plane accidents, an individual's belief should be equal to that frequency, excluding any outside information. Similarly, under this definition of probability being equal to beliefs, it is meaningful to speak about probabilities (beliefs) of presidential election outcomes: how confident are you candidate *A* will win?

Notice in the paragraph above, I assigned the belief (probability) measure to an *individual*, not to Nature. This is very interesting, as this definition leaves room for conflicting beliefs between individuals. Again, this is appropriate for what naturally occurs: different individuals have different beliefs of events occurring, because they possess different *information* about the world. The existence of different beliefs does not imply that anyone is wrong. Consider the following examples demonstrating the relationship between individual beliefs and probabilities:

- I flip a coin, and we both guess the result. We would both agree, assuming the coin is fair, that the probability of Heads is 1/2. Assume, then, that I peek at the coin. Now I know for certain what the result is: I assign probability 1.0 to either Heads or Tails (whichever it is). Now what is *your* belief that the coin is Heads? My knowledge of the outcome has not changed the coin's results. Thus we assign different probabilities to the result.

-  Your code either has a bug in it or not, but we do not know for certain which is true, though we have a belief about the presence or absence of a bug.

-  A medical patient is exhibiting symptoms $x$, $y$ and $z$. There are a number of diseases that could be causing all of them, but only a single disease is present. A doctor has beliefs about which disease, but a second doctor may have slightly different beliefs.

This philosophy of treating beliefs as probability is natural to humans. We employ it constantly as we interact with the world and only see partial truths, but gather evidence to form beliefs. Alternatively, you have to be *trained* to think like a frequentist.

To align ourselves with traditional probability notation, we denote our belief about event $A$ as $P(A)$. We call this quantity the *prior probability*.

John Maynard Keynes, a great economist and thinker, said "When the facts change, I change my mind. What do you do, sir?" This quote reflects the way a Bayesian updates his or her beliefs after seeing evidence. Even — especially — if the evidence is counter to what was initially believed, the evidence cannot be ignored. We denote our updated belief as $P(A |X )$, interpreted as the probability of $A$ given the evidence $X$. We call the updated belief the *posterior probability* so as to contrast it with the prior probability. For example, consider the posterior probabilities (read: posterior beliefs) of the above examples, after observing some evidence $X$:

1\. $P(A): \;\;$ the coin has a 50 percent chance of being Heads. $P(A | X):\;\;$ You look at the coin, observe a Heads has landed, denote this information $X$, and trivially assign probability 1.0 to Heads and 0.0 to Tails.

2\.   $P(A): \;\;$  This big, complex code likely has a bug in it. $P(A | X): \;\;$ The code passed all $X$ tests; there still might be a bug, but its presence is less likely now.

3\.  $P(A):\;\;$ The patient could have any number of diseases. $P(A | X):\;\;$ Performing a blood test generated evidence $X$, ruling out some of the possible diseases from consideration.

It's clear that in each example we did not completely discard the prior belief after seeing new evidence $X$, but we *re-weighted the prior* to incorporate the new evidence (i.e. we put more weight, or confidence, on some beliefs versus others).

By introducing prior uncertainty about events, we are already admitting that any guess we make is potentially very wrong. After observing data, evidence, or other information, we update our beliefs, and our guess becomes *less wrong*. This is the alternative side of the prediction coin, where typically we try to be *more right*.

### Bayesian Inference in Practice

 If frequentist and Bayesian inference were programming functions, with inputs being statistical problems, then the two would be different in what they return to the user. The frequentist inference function would return a number, representing an estimate (typically a summary statistic like the sample average etc.), whereas the Bayesian function would return *probabilities*.

For example, in our debugging problem above, calling the frequentist function with the argument "My code passed all $X$ tests; is my code bug-free?" would return a *YES*. On the other hand, asking our Bayesian function "Often my code has bugs. My code passed all $X$ tests; is my code bug-free?" would return something very different: probabilities of *YES* and *NO*. The function might return:

>    *YES*, with probability 0.8; *NO*, with probability 0.2

This is very different from the answer the frequentist function returned. Notice that the Bayesian function accepted an additional argument:  *"Often my code has bugs"*. This parameter is the *prior*. By including the prior parameter, we are telling the Bayesian function to include our belief about the situation. Technically this parameter in the Bayesian function is optional, but we will see excluding it has its own consequences.

#### Incorporating evidence

As we acquire more and more instances of evidence, our prior belief is *washed out* by the new evidence. This is to be expected. For example, if your prior belief is something ridiculous, like "I expect the sun to explode today", and each day you are proved wrong, you would hope that any inference would correct you, or at least align your beliefs better. Bayesian inference will correct this belief.

Denote $N$ as the number of instances of evidence we possess. As we gather an *infinite* amount of evidence, say as $N \rightarrow \infty$, our Bayesian results (often) align with frequentist results. Hence for large $N$, statistical inference is more or less objective. On the other hand, for small $N$, inference is much more *unstable*: frequentist estimates have more variance and larger confidence intervals. This is where Bayesian analysis excels. By introducing a prior, and returning probabilities (instead of a scalar estimate), we *preserve the uncertainty* that reflects the instability of statistical inference of a small $N$ dataset.

One may think that for large $N$, one can be indifferent between the two techniques since they offer similar inference, and might lean towards the computationally-simpler, frequentist methods. An individual in this position should consider the following quote by Andrew Gelman (2005)[1], before making such a decision:

> Sample sizes are never large. If $N$ is too small to get a sufficiently-precise estimate, you need to get more data (or make more assumptions). But once $N$ is "large enough," you can start subdividing the data to learn more (for example, in a public opinion poll, once you have a good estimate for the entire country, you can estimate among men and women, northerners and southerners, different age groups, etc.). $N$ is never enough because if it were "enough" you'd already be on to the next problem for which you need more data.

### Are frequentist methods incorrect then?

**No.**

Frequentist methods are still useful or state-of-the-art in many areas. Tools such as least squares linear regression, LASSO regression, and expectation-maximization algorithms are all powerful and fast. Bayesian methods complement these techniques by solving problems that these approaches cannot, or by illuminating the underlying system with more flexible modeling.

#### A note on *Big Data*
Paradoxically, big data's predictive analytic problems are actually solved by relatively simple algorithms [2][4]. Thus we can argue that big data's prediction difficulty does not lie in the algorithm used, but instead on the computational difficulties of storage and execution on big data. (One should also consider Gelman's quote from above and ask "Do I really have big data?" )

The much more difficult analytic problems involve *medium data* and, especially troublesome, *really small data*. Using a similar argument as  Gelman's above, if big data problems are *big enough* to be readily solved, then we should be more interested in the *not-quite-big enough* datasets.

### Our Bayesian framework

We are interested in beliefs, which can be interpreted as probabilities by thinking Bayesian. We have a *prior* belief in event $A$, beliefs formed by previous information, e.g., our prior belief about bugs being in our code before performing tests.

Secondly, we observe our evidence. To continue our buggy-code example: if our code passes $X$ tests, we want to update our belief to incorporate this. We call this new belief the *posterior* probability. Updating our belief is done via the following equation, known as Bayes' Theorem, after its discoverer Thomas Bayes:

\begin{align}
 P( A | X ) = & \frac{ P(X | A) P(A) } {P(X) } \\\\[5pt]
& \propto P(X | A) P(A)\;\; (\propto \text{is proportional to } )
\end{align}

The above formula is not unique to Bayesian inference: it is a mathematical fact with uses outside Bayesian inference. Bayesian inference merely uses it to connect prior probabilities $P(A)$ with an updated posterior probabilities $P(A | X )$.

##### Example: Mandatory coin-flip example

Every statistics text must contain a coin-flipping example, I'll use it here to get it out of the way. Suppose, naively, that you are unsure about the probability of heads in a coin flip (spoiler alert: it's 50%). You believe there is some true underlying ratio, call it $p$, but have no prior opinion on what $p$ might be.

We begin to flip a coin, and record the observations: either $H$ or $T$. This is our observed data. An interesting question to ask is how our inference changes as we observe more and more data? More specifically, what do our posterior probabilities look like when we have little data, versus when we have lots of data.

Below we plot a sequence of updating posterior probabilities as we observe increasing amounts of data (coin flips).

```python
"""
The book uses a custom matplotlibrc file, which provides the unique styles for
matplotlib plots. If executing this book, and you wish to use the book's
styling, provided are two options:
    1. Overwrite your own matplotlibrc file with the rc-file provided in the
       book's styles/ dir. See http://matplotlib.org/users/customizing.html
    2. Also in the styles is  bmh_matplotlibrc.json file. This can be used to
       update the styles in only this notebook. Try running the following code:

        import json, matplotlib
        s = json.load( open("../styles/bmh_matplotlibrc.json") )
        matplotlib.rcParams.update(s)

"""

# The code below can be passed over, as it is currently not important, plus it
# uses advanced topics we have not covered yet. LOOK AT PICTURE, MICHAEL!
%matplotlib inline
from IPython.core.pylabtools import figsize
import numpy as np
from matplotlib import pyplot as plt
figsize(11, 9)

import scipy.stats as stats

dist = stats.beta
n_trials = [0, 1, 2, 3, 4, 5, 8, 15, 50, 500]
data = stats.bernoulli.rvs(0.5, size=n_trials[-1])
x = np.linspace(0, 1, 100)

# For the already prepared, I'm using Binomial's conj. prior.
for k, N in enumerate(n_trials):
    sx = plt.subplot(len(n_trials) / 2, 2, k + 1)
    plt.xlabel("$p$, probability of heads") \
        if k in [0, len(n_trials) - 1] else None
    plt.setp(sx.get_yticklabels(), visible=False)
    heads = data[:N].sum()
    y = dist.pdf(x, 1 + heads, 1 + N - heads)
    plt.plot(x, y, label="observe %d tosses,\n %d heads" % (N, heads))
    plt.fill_between(x, 0, y, color="#348ABD", alpha=0.4)
    plt.vlines(0.5, 0, 4, color="k", linestyles="--", lw=1)

    leg = plt.legend()
    leg.get_frame().set_alpha(0.4)
    plt.autoscale(tight=True)

plt.suptitle("Bayesian updating of posterior probabilities",
             y=1.02,
             fontsize=14)

plt.tight_layout()
```

The posterior probabilities are represented by the curves, and our uncertainty is proportional to the width of the curve. As the plot above shows, as we start to observe data our posterior probabilities start to shift and move around. Eventually, as we observe more and more data (coin-flips), our probabilities will tighten closer and closer around the true value of $p=0.5$ (marked by a dashed line).

Notice that the plots are not always *peaked* at 0.5. There is no reason it should be: recall we assumed we did not have a prior opinion of what $p$ is. In fact, if we observe quite extreme data, say 8 flips and only 1 observed heads, our distribution would look very biased *away* from lumping around 0.5 (with no prior opinion, how confident would you feel betting on a fair coin after observing 8 tails and 1 head). As more data accumulates, we would see more and more probability being assigned at $p=0.5$, though never all of it.

The next example is a simple demonstration of the mathematics of Bayesian inference.

##### Example: Bug, or just sweet, unintended feature?

Let $A$ denote the event that our code has **no bugs** in it. Let $X$ denote the event that the code passes all debugging tests. For now, we will leave the prior probability of no bugs as a variable, i.e. $P(A) = p$.

We are interested in $P(A|X)$, i.e. the probability of no bugs, given our debugging tests $X$ pass. To use the formula above, we need to compute some quantities.

What is $P(X | A)$, i.e., the probability that the code passes $X$ tests *given* there are no bugs? Well, it is equal to 1, for code with no bugs will pass all tests.

$P(X)$ is a little bit trickier: The event $X$ can be divided into two possibilities, event $X$ occurring even though our code *indeed has* bugs (denoted $\sim A\;$, spoken *not $A$*), or event $X$ without bugs ($A$). $P(X)$ can be represented as:

\begin{align}
P(X ) & = P(X \text{ and } A) + P(X \text{ and } \sim A) \\\\[5pt]
 & = P(X|A)P(A) + P(X | \sim A)P(\sim A)\\\\[5pt]
& = P(X|A)p + P(X | \sim A)(1-p)
\end{align}

We have already computed $P(X|A)$ above. On the other hand, $P(X | \sim A)$ is subjective: our code can pass tests but still have a bug in it, though the probability there is a bug present is reduced. Note this is dependent on the number of tests performed, the degree of complication in the tests, etc. Let's be conservative and assign $P(X|\sim A) = 0.5$. Then

\begin{align}
P(A | X) & = \frac{1\cdot p}{ 1\cdot p +0.5 (1-p) } \\\\
& = \frac{ 2 p}{1+p}
\end{align}
This is the posterior probability. What does it look like as a function of our prior, $p \in [0,1]$?

```python
figsize(12.5, 4)
p = np.linspace(0, 1, 50)
plt.plot(p, 2 * p / (1 + p), color="#348ABD", lw=3)
# plt.fill_between(p, 2*p/(1+p), alpha=.5, facecolor=["#A60628"])
plt.scatter(0.2, 2 * (0.2) / 1.2, s=140, c="#348ABD")
plt.xlim(0, 1)
plt.ylim(0, 1)
plt.xlabel("Prior, $P(A) = p$")
plt.ylabel("Posterior, $P(A|X)$, with $P(A) = p$")
plt.title("Is my code bug-free?")
```

We can see the biggest gains if we observe the $X$ tests passed when the prior probability, $p$, is low. Let's settle on a specific value for the prior. I'm a strong programmer (I think), so I'm going to give myself a realistic prior of 0.20, that is, there is a 20% chance that I write code bug-free. To be more realistic, this prior should be a function of how complicated and large the code is, but let's pin it at 0.20. Then my updated belief that my code is bug-free is 0.33.

Recall that the prior is a probability: $p$ is the prior probability that there *are no bugs*, so $1-p$ is the prior probability that there *are bugs*.

Similarly, our posterior is also a probability, with $P(A | X)$ the probability there is no bug *given we saw all tests pass*, hence $1-P(A|X)$ is the probability there is a bug *given all tests passed*. What does our posterior probability look like? Below is a chart of both the prior and the posterior probabilities.

```python
figsize(12.5, 4)
colours = ["#348ABD", "#A60628"]

prior = [0.20, 0.80]
posterior = [1. / 3, 2. / 3]
plt.bar([0, .7], prior, alpha=0.70, width=0.25,
        color=colours[0], label="prior distribution",
        lw="3", edgecolor=colours[0])

plt.bar([0 + 0.25, .7 + 0.25], posterior, alpha=0.7,
        width=0.25, color=colours[1],
        label="posterior distribution",
        lw="3", edgecolor=colours[1])

plt.ylim(0,1)
plt.xticks([0.20, .95], ["Bugs Absent", "Bugs Present"])
plt.title("Prior and Posterior probability of bugs present")
plt.ylabel("Probability")
plt.legend(loc="upper left");
```

Notice that after we observed $X$ occur, the probability of bugs being absent increased. By increasing the number of tests, we can approach confidence (probability 1) that there are no bugs present.

This was a very simple example of Bayesian inference and Bayes rule. Unfortunately, the mathematics necessary to perform more complicated Bayesian inference only becomes more difficult, except for artificially constructed cases. We will later see that this type of mathematical analysis is actually unnecessary. First we must broaden our modeling tools. The next section deals with *probability distributions*. If you are already familiar, feel free to skip (or at least skim), but for the less familiar the next section is essential.

_______

## Probability Distributions

**Let's quickly recall what a probability distribution is:** Let $Z$ be some random variable. Then associated with $Z$ is a *probability distribution function* that assigns probabilities to the different outcomes $Z$ can take. Graphically, a probability distribution is a curve where the probability of an outcome is proportional to the height of the curve. You can see examples in the first figure of this chapter.

We can divide random variables into three classifications:

-   **$Z$ is discrete**: Discrete random variables may only assume values on a specified list. Things like populations, movie ratings, and number of votes are all discrete random variables. Discrete random variables become more clear when we contrast them with...

-   **$Z$ is continuous**: Continuous random variable can take on arbitrarily exact values. For example, temperature, speed, time, color are all modeled as continuous variables because you can progressively make the values more and more precise.

- **$Z$ is mixed**: Mixed random variables assign probabilities to both discrete and continuous random variables, i.e. it is a combination of the above two categories.

#### Expected Value
Expected value (EV) is one of the most important concepts in probability. The EV for a given probability distribution can be described as "the mean value in the long run for many repeated samples from that distribution." To borrow a metaphor from physics, a distribution's EV acts like its "center of mass." Imagine repeating the same experiment many times over, and taking the average over each outcome. The more you repeat the experiment, the closer this average will become to the distributions EV. (side note: as the number of repeated experiments goes to infinity, the difference between the average outcome and the EV becomes arbitrarily small.)

### Discrete Case
If $Z$ is discrete, then its distribution is called a *probability mass function*, which measures the probability $Z$ takes on the value $k$, denoted $P(Z=k)$. Note that the probability mass function completely describes the random variable $Z$, that is, if we know the mass function, we know how $Z$ should behave. There are popular probability mass functions that consistently appear: we will introduce them as needed, but let's introduce the first very useful probability mass function. We say $Z$ is *Poisson*-distributed if:

$$P(Z = k) =\frac{ \lambda^k e^{-\lambda} }{k!}, \; \; k=0,1,2, \dots, \; \; \lambda \in \mathbb{R}_{>0} $$

$\lambda$ is called a parameter of the distribution, and it controls the distribution's shape. For the Poisson distribution, $\lambda$ can be any positive number. By increasing $\lambda$, we add more probability to larger values, and conversely by decreasing $\lambda$ we add more probability to smaller values. One can describe $\lambda$ as the *intensity* of the Poisson distribution.

Unlike $\lambda$, which can be any positive number, the value $k$ in the above formula must be a non-negative integer, i.e., $k$ must take on values 0,1,2, and so on. This is very important, because if you wanted to model a population you could not make sense of populations with 4.25 or 5.612 members.

If a random variable $Z$ has a Poisson mass distribution, we denote this by writing

$$Z \sim \text{Poi}(\lambda) $$

One useful property of the Poisson distribution is that its expected value is equal to its parameter, i.e.:

$$E\large[ \;Z\; | \; \lambda \;\large] = \lambda $$

We will use this property often, so it's useful to remember. Below, we plot the probability mass distribution for different $\lambda$ values. The first thing to notice is that by increasing $\lambda$, we add more probability of larger values occurring. Second, notice that although the graph ends at 15, the distributions do not. They assign positive probability to every non-negative integer.

```python
figsize(12.5, 4)

import scipy.stats as stats
a = np.arange(16)
poi = stats.poisson
lambda_ = [1.5, 4.25]
colours = ["#348ABD", "#A60628"]

plt.bar(a, poi.pmf(a, lambda_[0]), color=colours[0],
        label="$\lambda = %.1f$" % lambda_[0], alpha=0.60,
        edgecolor=colours[0], lw="3")

plt.bar(a, poi.pmf(a, lambda_[1]), color=colours[1],
        label="$\lambda = %.1f$" % lambda_[1], alpha=0.60,
        edgecolor=colours[1], lw="3")

plt.xticks(a + 0.4, a)
plt.legend()
plt.ylabel("probability of $k$")
plt.xlabel("$k$")
plt.title("Probability mass function of a Poisson random variable; differing \
$\lambda$ values")
```

### Continuous Case
Instead of a probability mass function, a continuous random variable has a *probability density function*. This might seem like unnecessary nomenclature, but the density function and the mass function are very different creatures. An example of continuous random variable is a random variable with *exponential density*. The density function for an exponential random variable looks like this:

$$f_Z(z | \lambda) = \lambda e^{-\lambda z }, \;\; z\ge 0$$

Like a Poisson random variable, an exponential random variable can take on only non-negative values. But unlike a Poisson variable, the exponential can take on *any* non-negative values, including non-integral values such as 4.25 or 5.612401. This property makes it a poor choice for count data, which must be an integer, but a great choice for time data, temperature data (measured in Kelvins, of course), or any other precise *and positive* variable. The graph below shows two probability density functions with different $\lambda$ values.

When a random variable $Z$ has an exponential distribution with parameter $\lambda$, we say *$Z$ is exponential* and write

$$Z \sim \text{Exp}(\lambda)$$

Given a specific $\lambda$, the expected value of an exponential random variable is equal to the inverse of $\lambda$, that is:

$$E[\; Z \;|\; \lambda \;] = \frac{1}{\lambda}$$

```python
a = np.linspace(0, 4, 100)
expo = stats.expon
lambda_ = [0.5, 1]

for l, c in zip(lambda_, colours):
    plt.plot(a, expo.pdf(a, scale=1. / l), lw=3,
             color=c, label="$\lambda = %.1f$" % l)
    plt.fill_between(a, expo.pdf(a, scale=1. / l), color=c, alpha=.33)

plt.legend()
plt.ylabel("PDF at $z$")
plt.xlabel("$z$")
plt.ylim(0, 1.2)
plt.title("Probability density function of an Exponential random variable;\
 differing $\lambda$");
```

### But what is $\lambda \;$?

**This question is what motivates statistics**. In the real world, $\lambda$ is hidden from us. We see only $Z$, and must go backwards to try and determine $\lambda$. The problem is difficult because there is no one-to-one mapping from $Z$ to $\lambda$. Many different methods have been created to solve the problem of estimating $\lambda$, but since $\lambda$ is never actually observed, no one can say for certain which method is best!

Bayesian inference is concerned with *beliefs* about what $\lambda$ might be. Rather than try to guess $\lambda$ exactly, we can only talk about what $\lambda$ is likely to be by assigning a probability distribution to $\lambda$.

This might seem odd at first. After all, $\lambda$ is fixed; it is not (necessarily) random! How can we assign probabilities to values of a non-random variable? Ah, we have fallen for our old, frequentist way of thinking. Recall that under Bayesian philosophy, we *can* assign probabilities if we interpret them as beliefs. And it is entirely acceptable to have *beliefs* about the parameter $\lambda$.

##### Example: Inferring behaviour from text-message data

Let's try to model a more interesting example, one that concerns the rate at which a user sends and receives text messages:

>  You are given a series of daily text-message counts from a user of your system. The data, plotted over time, appears in the chart below. You are curious to know if the user's text-messaging habits have changed over time, either gradually or suddenly. How can you model this? (This is in fact my own text-message data. Judge my popularity as you wish.)

```python
figsize(12.5, 3.5)
count_data = np.loadtxt("data/txtdata.csv")
n_count_data = len(count_data)
plt.bar(np.arange(n_count_data), count_data, color="#348ABD")
plt.xlabel("Time (days)")
plt.ylabel("count of text-msgs received")
plt.title("Did the user's texting habits change over time?")
plt.xlim(0, n_count_data);
```

Before we start modeling, see what you can figure out just by looking at the chart above. Would you say there was a change in behaviour during this time period?

How can we start to model this? Well, as we have conveniently already seen, a Poisson random variable is a very appropriate model for this type of *count* data. Denoting day $i$'s text-message count by $C_i$,

$$ C_i \sim \text{Poisson}(\lambda)  $$

We are not sure what the value of the $\lambda$ parameter really is, however. Looking at the chart above, it appears that the rate might become higher late in the observation period, which is equivalent to saying that $\lambda$ increases at some point during the observations. (Recall that a higher value of $\lambda$ assigns more probability to larger outcomes. That is, there is a higher probability of many text messages having been sent on a given day.)

How can we represent this observation mathematically? Let's assume that on some day during the observation period (call it $\tau$), the parameter $\lambda$ suddenly jumps to a higher value. So we really have two $\lambda$ parameters: one for the period before $\tau$, and one for the rest of the observation period. In the literature, a sudden transition like this would be called a *switchpoint*:

$$
\lambda =
\begin{cases}
\lambda_1  & \text{if } t \lt \tau \cr
\lambda_2 & \text{if } t \ge \tau
\end{cases}
$$

If, in reality, no sudden change occurred and indeed $\lambda_1 = \lambda_2$, then the $\lambda$s posterior distributions should look about equal.

We are interested in inferring the unknown $\lambda$s. To use Bayesian inference, we need to assign prior probabilities to the different possible values of $\lambda$. What would be good prior probability distributions for $\lambda_1$ and $\lambda_2$? Recall that $\lambda$ can be any positive number. As we saw earlier, the *exponential* distribution provides a continuous density function for positive numbers, so it might be a good choice for modeling $\lambda_i$. But recall that the exponential distribution takes a parameter of its own, so we'll need to include that parameter in our model. Let's call that parameter $\alpha$.

\begin{align}
&\lambda_1 \sim \text{Exp}( \alpha ) \\\
&\lambda_2 \sim \text{Exp}( \alpha )
\end{align}

$\alpha$ is called a *hyper-parameter* or *parent variable*. In literal terms, it is a parameter that influences other parameters. Our initial guess at $\alpha$ does not influence the model too strongly, so we have some flexibility in our choice.  A good rule of thumb is to set the exponential parameter equal to the inverse of the average of the count data. Since we're modeling $\lambda$ using an exponential distribution, we can use the expected value identity shown earlier to get:

$$\frac{1}{N}\sum_{i=0}^N \;C_i \approx E[\; \lambda \; |\; \alpha ] = \frac{1}{\alpha}$$

An alternative, and something I encourage the reader to try, would be to have two priors: one for each $\lambda_i$. Creating two exponential distributions with different $\alpha$ values reflects our prior belief that the rate changed at some point during the observations.

What about $\tau$? Because of the noisiness of the data, it's difficult to pick out a priori when $\tau$ might have occurred. Instead, we can assign a *uniform prior belief* to every possible day. This is equivalent to saying

\begin{align}
& \tau \sim \text{DiscreteUniform(1,70) }\\\\
& \Rightarrow P( \tau = k ) = \frac{1}{70}
\end{align}

So after all this, what does our overall prior distribution for the unknown variables look like? Frankly, *it doesn't matter*. What we should understand is that it's an ugly, complicated mess involving symbols only a mathematician could love. And things will only get uglier the more complicated our models become. Regardless, all we really care about is the posterior distribution.

We next turn to PyMC, a Python library for performing Bayesian analysis that is undaunted by the mathematical monster we have created.

Introducing our first hammer: PyMC
-----

PyMC is a Python library for programming Bayesian analysis [3]. It is a fast, well-maintained library. The only unfortunate part is that its documentation is lacking in certain areas, especially those that bridge the gap between beginner and hacker. One of this book's main goals is to solve that problem, and also to demonstrate why PyMC is so cool.

We will model the problem above using PyMC. This type of programming is called *probabilistic programming*, an unfortunate misnomer that invokes ideas of randomly-generated code and has likely confused and frightened users away from this field. The code is not random; it is probabilistic in the sense that we create probability models using programming variables as the model's components. Model components are first-class primitives within the PyMC framework.

B. Cronin [5] has a very motivating description of probabilistic programming:

>   Another way of thinking about this: unlike a traditional program, which only runs in the forward directions, a probabilistic program is run in both the forward and backward direction. It runs forward to compute the consequences of the assumptions it contains about the world (i.e., the model space it represents), but it also runs backward from the data to constrain the possible explanations. In practice, many probabilistic programming systems will cleverly interleave these forward and backward operations to efficiently home in on the best explanations.

Because of the confusion engendered by the term *probabilistic programming*, I'll refrain from using it. Instead, I'll simply say *programming*, since that's what it really is.

PyMC code is easy to read. The only novel thing should be the syntax, and I will interrupt the code to explain individual sections. Simply remember that we are representing the model's components ($\tau, \lambda_1, \lambda_2$ ) as variables:

```python
import pymc as pm

alpha = 1.0 / count_data.mean()  # Recall count_data is the
                               # variable that holds our txt counts
lambda_1 = pm.Exponential("lambda_1", alpha)
lambda_2 = pm.Exponential("lambda_2", alpha)

tau = pm.DiscreteUniform("tau", lower=0, upper=n_count_data)
```

In the code above, we create the PyMC variables corresponding to $\lambda_1$ and $\lambda_2$. We assign them to PyMC's *stochastic variables*, so-called because they are treated by the back end as random number generators. We can demonstrate this fact by calling their built-in `random()` methods.

```python
print("Random output:", tau.random(), tau.random(), tau.random())
```

    Random output: 64 5 12

```python
@pm.deterministic
def lambda_(tau=tau, lambda_1=lambda_1, lambda_2=lambda_2):
    out = np.zeros(n_count_data)
    out[:tau] = lambda_1  # lambda before tau is lambda1
    out[tau:] = lambda_2  # lambda after (and including) tau is lambda2
    return out
```

This code creates a new function `lambda_`, but really we can think of it as a random variable: the random variable $\lambda$ from above. Note that because `lambda_1`, `lambda_2` and `tau` are random, `lambda_` will be random. We are **not** fixing any variables yet.

`@pm.deterministic` is a decorator that tells PyMC this is a deterministic function. That is, if the arguments were deterministic (which they are not), the output would be deterministic as well. Deterministic functions will be covered in Chapter 2.

```python
observation = pm.Poisson("obs", lambda_, value=count_data, observed=True)

model = pm.Model([observation, lambda_1, lambda_2, tau])
```

The variable `observation` combines our data, `count_data`, with our proposed data-generation scheme, given by the variable `lambda_`, through the `value` keyword. We also set `observed = True` to tell PyMC that this should stay fixed in our analysis. Finally, PyMC wants us to collect all the variables of interest and create a `Model` instance out of them. This makes our life easier when we retrieve the results.

The code below will be explained in Chapter 3, but I show it here so you can see where our results come from. One can think of it as a *learning* step. The machinery being employed is called *Markov Chain Monte Carlo* (MCMC), which I also delay explaining until Chapter 3. This technique returns thousands of random variables from the posterior distributions of $\lambda_1, \lambda_2$ and $\tau$. We can plot a histogram of the random variables to see what the posterior distributions look like. Below, we collect the samples (called *traces* in the MCMC literature) into histograms.

```python
# Mysterious code to be explained in Chapter 3.
mcmc = pm.MCMC(model)
mcmc.sample(40000, 10000, 1)
```

     [-----------------100%-----------------] 40000 of 40000 complete in 6.5 sec

```python
lambda_1_samples = mcmc.trace('lambda_1')[:]
lambda_2_samples = mcmc.trace('lambda_2')[:]
tau_samples = mcmc.trace('tau')[:]
```

```python
figsize(12.5, 10)
# histogram of the samples:

ax = plt.subplot(311)
ax.set_autoscaley_on(False)

plt.hist(lambda_1_samples, histtype='stepfilled', bins=30, alpha=0.85,
         label="posterior of $\lambda_1$", color="#A60628", normed=True)
plt.legend(loc="upper left")
plt.title(r"""Posterior distributions of the variables
    $\lambda_1,\;\lambda_2,\;\tau$""")
plt.xlim([15, 30])
plt.xlabel("$\lambda_1$ value")

ax = plt.subplot(312)
ax.set_autoscaley_on(False)
plt.hist(lambda_2_samples, histtype='stepfilled', bins=30, alpha=0.85,
         label="posterior of $\lambda_2$", color="#7A68A6", normed=True)
plt.legend(loc="upper left")
plt.xlim([15, 30])
plt.xlabel("$\lambda_2$ value")

plt.subplot(313)
w = 1.0 / tau_samples.shape[0] * np.ones_like(tau_samples)
plt.hist(tau_samples, bins=n_count_data, alpha=1,
         label=r"posterior of $\tau$",
         color="#467821", weights=w, rwidth=2.)
plt.xticks(np.arange(n_count_data))

plt.legend(loc="upper left")
plt.ylim([0, .75])
plt.xlim([35, len(count_data) - 20])
plt.xlabel(r"$\tau$ (in days)")
plt.ylabel("probability");
```

### Interpretation

Recall that Bayesian methodology returns a *distribution*. Hence we now have distributions to describe the unknown $\lambda$s and $\tau$. What have we gained? Immediately, we can see the uncertainty in our estimates: the wider the distribution, the less certain our posterior belief should be. We can also see what the plausible values for the parameters are: $\lambda_1$ is around 18 and $\lambda_2$ is around 23. The posterior distributions of the two $\lambda$s are clearly distinct, indicating that it is indeed likely that there was a change in the user's text-message behaviour.

What other observations can you make? If you look at the original data again, do these results seem reasonable?

Notice also that the posterior distributions for the $\lambda$s do not look like exponential distributions, even though our priors for these variables were exponential. In fact, the posterior distributions are not really of any form that we recognize from the original model. But that's OK! This is one of the benefits of taking a computational point of view. If we had instead done this analysis using mathematical approaches, we would have been stuck with an analytically intractable (and messy) distribution. Our use of a computational approach makes us indifferent to mathematical tractability.

Our analysis also returned a distribution for $\tau$. Its posterior distribution looks a little different from the other two because it is a discrete random variable, so it doesn't assign probabilities to intervals. We can see that near day 45, there was a 50% chance that the user's behaviour changed. Had no change occurred, or had the change been gradual over time, the posterior distribution of $\tau$ would have been more spread out, reflecting that many days were plausible candidates for $\tau$. By contrast, in the actual results we see that only three or four days make any sense as potential transition points.

### Why would I want samples from the posterior, anyways?

We will deal with this question for the remainder of the book, and it is an understatement to say that it will lead us to some amazing results. For now, let's end this chapter with one more example.

We'll use the posterior samples to answer the following question: what is the expected number of texts at day $t, \; 0 \le t \le 70$ ? Recall that the expected value of a Poisson variable is equal to its parameter $\lambda$. Therefore, the question is equivalent to *what is the expected value of $\lambda$ at time $t$*?

In the code below, let $i$ index samples from the posterior distributions. Given a day $t$, we average over all possible $\lambda_i$ for that day $t$, using $\lambda_i = \lambda_{1,i}$ if $t \lt \tau_i$ (that is, if the behaviour change has not yet occurred), else we use $\lambda_i = \lambda_{2,i}$.

```python
figsize(12.5, 5)
# tau_samples, lambda_1_samples, lambda_2_samples contain
# N samples from the corresponding posterior distribution
N = tau_samples.shape[0]
expected_texts_per_day = np.zeros(n_count_data)
for day in range(0, n_count_data):
    # ix is a bool index of all tau samples corresponding to
    # the switchpoint occurring prior to value of 'day'
    ix = day < tau_samples
    # Each posterior sample corresponds to a value for tau.
    # for each day, that value of tau indicates whether we're "before"
    # (in the lambda1 "regime") or
    #  "after" (in the lambda2 "regime") the switchpoint.
    # by taking the posterior sample of lambda1/2 accordingly, we can average
    # over all samples to get an expected value for lambda on that day.
    # As explained, the "message count" random variable is Poisson distributed,
    # and therefore lambda (the poisson parameter) is the expected value of
    # "message count".
    expected_texts_per_day[day] = (lambda_1_samples[ix].sum()
                                   + lambda_2_samples[~ix].sum()) / N

plt.plot(range(n_count_data), expected_texts_per_day, lw=4, color="#E24A33",
         label="expected number of text-messages received")
plt.xlim(0, n_count_data)
plt.xlabel("Day")
plt.ylabel("Expected # text-messages")
plt.title("Expected number of text-messages received")
plt.ylim(0, 60)
plt.bar(np.arange(len(count_data)), count_data, color="#348ABD", alpha=0.65,
        label="observed texts per day")

plt.legend(loc="upper left");
```

Our analysis shows strong support for believing the user's behavior did change ($\lambda_1$ would have been close in value to $\lambda_2$ had this not been true), and that the change was sudden rather than gradual (as demonstrated by $\tau$'s strongly peaked posterior distribution). We can speculate what might have caused this: a cheaper text-message rate, a recent weather-to-text subscription, or perhaps a new relationship. (In fact, the 45th day corresponds to Christmas, and I moved away to Toronto the next month, leaving a girlfriend behind.)

##### Exercises

1\.  Using `lambda_1_samples` and `lambda_2_samples`, what is the mean of the posterior distributions of $\lambda_1$ and $\lambda_2$?

```python
# type your code here.
```

2\.  What is the expected percentage increase in text-message rates? `hint:` compute the mean of `lambda_1_samples/lambda_2_samples`. Note that this quantity is very different from `lambda_1_samples.mean()/lambda_2_samples.mean()`.

```python
# type your code here.
```

3\. What is the mean of $\lambda_1$ **given** that we know $\tau$ is less than 45.  That is, suppose we have been given new information that the change in behaviour occurred prior to day 45. What is the expected value of $\lambda_1$ now? (You do not need to redo the PyMC part. Just consider all instances where `tau_samples < 45`.)

```python
# type your code here.
```

### References

-  [1] Gelman, Andrew. N.p.. Web. 22 Jan 2013. [N is never large enough](http://andrewgelman.com/2005/07/31/n_is_never_larg/).
-  [2] Norvig, Peter. 2009. [The Unreasonable Effectiveness of Data](http://static.googleusercontent.com/media/research.google.com/en//pubs/archive/35179.pdf).
- [3] Patil, A., D. Huard and C.J. Fonnesbeck. 2010.
PyMC: Bayesian Stochastic Modelling in Python. Journal of Statistical
Software, 35(4), pp. 1-81.
- [4] Jimmy Lin and Alek Kolcz. Large-Scale Machine Learning at Twitter. Proceedings of the 2012 ACM SIGMOD International Conference on Management of Data (SIGMOD 2012), pages 793-804, May 2012, Scottsdale, Arizona.
- [5] Cronin, Beau. "Why Probabilistic Programming Matters." 24 Mar 2013. Google, Online Posting to Google . Web. 24 Mar. 2013. <https://plus.google.com/u/0/107971134877020469960/posts/KpeRdJKR6Z1>.

```python
from IPython.core.display import HTML

def css_styling():
    styles = open("../styles/custom.css", "r").read()
    return HTML(styles)
css_styling()
```

<style>
    @font-face {
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        font-family: "Computer Modern";
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    }
    div.cell{
        width:800px;
        margin-left:16% !important;
        margin-right:auto;
    }
    h1 {
        font-family: Helvetica, serif;
    }
    h4{
        margin-top:12px;
        margin-bottom: 3px;
       }
    div.text_cell_render{
        font-family: Computer Modern, "Helvetica Neue", Arial, Helvetica, Geneva, sans-serif;
        line-height: 145%;
        font-size: 130%;
        width:800px;
        margin-left:auto;
        margin-right:auto;
    }
    .CodeMirror{
            font-family: "Source Code Pro", source-code-pro,Consolas, monospace;
    }
    .prompt{
        display: None;
    }
    .text_cell_render h5 {
        font-weight: 300;
        font-size: 22pt;
        color: #4057A1;
        font-style: italic;
        margin-bottom: .5em;
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    .warning{
        color: rgb( 240, 20, 20 )
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</style>

```python

```