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The logic in this file largely borrows from Qwen2.5-Math codebase at https://github.com/QwenLM/Qwen2.5-Math:
"""
# flake8: noqa
import re
import regex
from latex2sympy2_extended import latex2sympy
from math import isclose
from sympy import N, simplify
from sympy.parsing.latex import parse_latex
from sympy.parsing.sympy_parser import parse_expr
from word2number import w2n
def convert_word_number(text: str) -> str:
try:
text = str(w2n.word_to_num(text))
except Exception:
pass
return text
def _fix_fracs(string):
substrs = string.split('\\frac')
new_str = substrs[0]
if len(substrs) > 1:
substrs = substrs[1:]
for substr in substrs:
new_str += '\\frac'
if len(substr) > 0 and substr[0] == '{':
new_str += substr
else:
try:
assert len(substr) >= 2
except Exception:
return string
a = substr[0]
b = substr[1]
if b != '{':
if len(substr) > 2:
post_substr = substr[2:]
new_str += '{' + a + '}{' + b + '}' + post_substr
else:
new_str += '{' + a + '}{' + b + '}'
else:
if len(substr) > 2:
post_substr = substr[2:]
new_str += '{' + a + '}' + b + post_substr
else:
new_str += '{' + a + '}' + b
string = new_str
return string
def _fix_a_slash_b(string):
if len(string.split('/')) != 2:
return string
a = string.split('/')[0]
b = string.split('/')[1]
try:
if 'sqrt' not in a:
a = int(a)
if 'sqrt' not in b:
b = int(b)
assert string == '{}/{}'.format(a, b)
new_string = '\\frac{' + str(a) + '}{' + str(b) + '}'
return new_string
except Exception:
return string
def _fix_sqrt(string):
_string = re.sub(r'\\sqrt(\w+)', r'\\sqrt{\1}', string)
return _string
def strip_answer_string(string):
string = str(string).strip()
# linebreaks
string = string.replace('\n', '')
# right "."
string = string.rstrip('.')
# remove inverse spaces
# replace \\ with \
string = string.replace('\\!', '')
# string = string.replace("\\ ", "")
# string = string.replace("\\\\", "\\")
# matrix
string = re.sub(r'\\begin\{array\}\{.*?\}', r'\\begin{pmatrix}', string)
string = re.sub(r'\\end\{array\}', r'\\end{pmatrix}', string)
string = string.replace('bmatrix', 'pmatrix')
# replace tfrac and dfrac with frac
string = string.replace('tfrac', 'frac')
string = string.replace('dfrac', 'frac')
string = (string.replace('\\neq', '\\ne').replace('\\leq', '\\le').replace('\\geq', '\\ge'))
# remove \left and \right
string = string.replace('\\left', '')
string = string.replace('\\right', '')
string = string.replace('\\{', '{')
string = string.replace('\\}', '}')
# Function to replace number words with corresponding digits
def replace_match(match):
word = match.group(1).lower()
if convert_word_number(word) == word:
return match.group(0)
else:
return convert_word_number(word)
string = re.sub(r'\\text\{([a-zA-Z]+)\}', replace_match, string)
# Before removing unit, check if the unit is squared (for surface area)
string = re.sub(r'(cm|inches)\}\^2', r'\1}', string)
# Remove unit: miles, dollars if after is not none
_string = re.sub(r'\\text{.*?}$', '', string).strip()
if _string != '' and _string != string:
# print("Warning: unit not removed: '{}' -> '{}'".format(string, _string))
string = _string
# Remove circ (degrees)
string = string.replace('^{\\circ}', '')
string = string.replace('^\\circ', '')
# remove dollar signs
string = string.replace('\\$', '')
string = string.replace('$', '')
string = string.replace('\\(', '').replace('\\)', '')
# convert word number to digit
string = convert_word_number(string)
# replace "\\text{...}" to "..."
string = re.sub(r'\\text\{(.*?)\}', r'\1', string)
for key in ['x=', 'y=', 'z=', 'x\\in', 'y\\in', 'z\\in', 'x\\to', 'y\\to', 'z\\to']:
string = string.replace(key, '')
string = string.replace('\\emptyset', r'{}')
string = string.replace('(-\\infty,\\infty)', '\\mathbb{R}')
# remove percentage
string = string.replace('\\%', '')
string = string.replace('\%', '')
string = string.replace('%', '')
# " 0." equivalent to " ." and "{0." equivalent to "{." Alternatively, add "0" if "." is the start of the string
string = string.replace(' .', ' 0.')
string = string.replace('{.', '{0.')
# cdot
# string = string.replace("\\cdot", "")
if (
string.startswith('{') and string.endswith('}') and string.isalnum()
or string.startswith('(') and string.endswith(')') and string.isalnum()
or string.startswith('[') and string.endswith(']') and string.isalnum()
):
string = string[1:-1]
# inf
string = string.replace('infinity', '\\infty')
if '\\infty' not in string:
string = string.replace('inf', '\\infty')
string = string.replace('+\\inity', '\\infty')
# and
string = string.replace('and', '')
string = string.replace('\\mathbf', '')
# use regex to remove \mbox{...}
string = re.sub(r'\\mbox{.*?}', '', string)
# quote
string.replace("'", '')
string.replace('"', '')
# i, j
if 'j' in string and 'i' not in string:
string = string.replace('j', 'i')
# replace a.000b where b is not number or b is end, with ab, use regex
string = re.sub(r'(\d+)\.0*([^\d])', r'\1\2', string)
string = re.sub(r'(\d+)\.0*$', r'\1', string)
# if empty, return empty string
if len(string) == 0:
return string
if string[0] == '.':
string = '0' + string
# to consider: get rid of e.g. "k = " or "q = " at beginning
if len(string.split('=')) == 2:
if len(string.split('=')[0]) <= 2:
string = string.split('=')[1]
string = _fix_sqrt(string)
string = string.replace(' ', '')
# \frac1b or \frac12 --> \frac{1}{b} and \frac{1}{2}, etc. Even works with \frac1{72} (but not \frac{72}1). Also does a/b --> \\frac{a}{b}
string = _fix_fracs(string)
# NOTE: X/Y changed to \frac{X}{Y} in dataset, but in simple cases fix in case the model output is X/Y
string = _fix_a_slash_b(string)
# Remove unnecessary '\' before integers
string = re.sub(r'\\(?=\-?\d+(\\|\)|,|\]|$))', '', string)
# Remove grade level (e.g., 12th grade) and just maintain the integer
string = re.sub(r'thgrade$', '', string)
# Normalize thousands-formatted numbers (e.g., 70,000 or -1,234,567.89) by removing commas
# This must run before the "list of integers" sorting to avoid misclassifying numbers with thousand separators.
if re.fullmatch(r'\s*-?\d{1,3}(?:,\d{3})+(?:\.\d+)?\s*', string):
string = string.replace(',', '')
# If the answer is a list of integers (without parenthesis), sort them
if re.fullmatch(r'(\s*-?\d+\s*,)*\s*-?\d+\s*', string):
# Split the string into a list of integers
try:
integer_list = list(map(int, string.split(',')))
except Exception:
integer_list = list(map(int, '-1,-1'.split(',')))
# Sort the list in ascending order
sorted_list = sorted(integer_list)
# Join the sorted list back into a comma-separated string
string = ','.join(map(str, sorted_list))
return string
def extract_answer(pred_str, use_last_number=True):
pred_str = pred_str.replace('\u043a\u0438', '')
if 'final answer is $' in pred_str and '$. I hope' in pred_str:
# minerva_math
tmp = pred_str.split('final answer is $', 1)[1]
pred = tmp.split('$. I hope', 1)[0].strip()
elif 'boxed' in pred_str:
ans = pred_str.split('boxed')[-1]
if len(ans) == 0:
return ''
elif ans[0] == '{':
stack = 1
a = ''
for c in ans[1:]:
if c == '{':
stack += 1
a += c
elif c == '}':
stack -= 1
if stack == 0:
break
a += c
else:
a += c
else:
a = ans.split('$')[0].strip()
pred = a
elif 'he answer is' in pred_str:
pred = pred_str.split('he answer is')[-1].strip()
elif 'final answer is' in pred_str:
pred = pred_str.split('final answer is')[-1].strip()
elif '答案是' in pred_str:
# Handle Chinese few-shot multiple choice problem answer extraction
pred = pred_str.split('答案是')[1].strip().split('\n\n')[0].strip()
elif 'ANSWER:' in pred_str:
pred = pred_str.split('ANSWER:')[-1].strip()
else: # use the last number
if use_last_number:
pattern = '-?\d*\.?\d+'
pred = re.findall(pattern, pred_str.replace(',', ''))
if len(pred) >= 1:
pred = pred[-1]
else:
pred = ''
else:
pred = ''
# multiple line
# pred = pred.split("\n")[0]
pred = re.sub(r'\n\s*', '', pred)
if pred != '' and pred[0] == ':':
pred = pred[1:]
if pred != '' and pred[-1] == '.':
pred = pred[:-1]
if pred != '' and pred[-1] == '/':
pred = pred[:-1]
pred = strip_answer_string(pred)
return pred
def choice_answer_clean(pred: str):
pred = pred.strip('\n').rstrip('.').rstrip('/').strip(' ').lstrip(':')
# Clean the answer based on the dataset
tmp = re.findall(r'\b(A|B|C|D|E)\b', pred.upper())
if tmp:
pred = tmp
else:
pred = [pred.strip().strip('.')]
pred = pred[-1]
# Remove the period at the end, again!
pred = pred.rstrip('.').rstrip('/')
return pred
def parse_digits(num):
num = regex.sub(',', '', str(num))
try:
return float(num)
except Exception:
if num.endswith('%'):
num = num[:-1]
if num.endswith('\\'):
num = num[:-1]
try:
return float(num) / 100
except Exception:
pass
return None
def is_digit(num):
# paired with parse_digits
return parse_digits(num) is not None
def str_to_pmatrix(input_str):
input_str = input_str.strip()
matrix_str = re.findall(r'\{.*,.*\}', input_str)
pmatrix_list = []
for m in matrix_str:
m = m.strip('{}')
pmatrix = r'\begin{pmatrix}' + m.replace(',', '\\') + r'\end{pmatrix}'
pmatrix_list.append(pmatrix)
return ', '.join(pmatrix_list)
def math_equal(
prediction,
reference,
include_percentage: bool = True,
is_close: bool = True,
timeout: bool = False,
) -> bool:
"""
Exact match of math if and only if:
1. numerical equal: both can convert to float and are equal
2. symbolic equal: both can convert to sympy expression and are equal
"""
if prediction is None or reference is None:
return False
if str(prediction.strip().lower()) == str(reference.strip().lower()):
return True
if (reference in ['A', 'B', 'C', 'D', 'E'] and choice_answer_clean(prediction) == reference):
return True
try: # 1. numerical equal
if is_digit(prediction) and is_digit(reference):
prediction = parse_digits(prediction)
reference = parse_digits(reference)
# number questions
if include_percentage:
gt_result = [reference / 100, reference, reference * 100]
else:
gt_result = [reference]
for item in gt_result:
try:
if is_close:
if numeric_equal(prediction, item):
return True
else:
if item == prediction:
return True
except Exception:
continue
return False
except Exception:
pass
if not prediction and prediction not in [0, False]:
return False
# 2. symbolic equal
reference = str(reference).strip()
prediction = str(prediction).strip()
## pmatrix (amps)
if 'pmatrix' in prediction and 'pmatrix' not in reference:
reference = str_to_pmatrix(reference)
## deal with [], (), {}
pred_str, ref_str = prediction, reference
if (prediction.startswith('[') and prediction.endswith(']') and not reference.startswith('(')
) or (prediction.startswith('(') and prediction.endswith(')') and not reference.startswith('[')):
pred_str = pred_str.strip('[]()')
ref_str = ref_str.strip('[]()')
for s in ['{', '}', '(', ')']:
ref_str = ref_str.replace(s, '')
pred_str = pred_str.replace(s, '')
if pred_str.lower() == ref_str.lower():
return True
## [a, b] vs. [c, d], return a==c and b==d
if (
regex.match(r'(\(|\[).+(\)|\])', prediction) is not None
and regex.match(r'(\(|\[).+(\)|\])', reference) is not None
):
pred_parts = prediction[1:-1].split(',')
ref_parts = reference[1:-1].split(',')
if len(pred_parts) == len(ref_parts):
if all([
math_equal(pred_parts[i], ref_parts[i], include_percentage, is_close) for i in range(len(pred_parts))
]):
return True
if ((prediction.startswith('\\begin{pmatrix}') or prediction.startswith('\\begin{bmatrix}'))
and (prediction.endswith('\\end{pmatrix}') or prediction.endswith('\\end{bmatrix}'))
and (reference.startswith('\\begin{pmatrix}') or reference.startswith('\\begin{bmatrix}'))
and (reference.endswith('\\end{pmatrix}') or reference.endswith('\\end{bmatrix}'))):
pred_lines = [
line.strip()
for line in prediction[len('\\begin{pmatrix}'):-len('\\end{pmatrix}')].split('\\\\')
if line.strip()
]
ref_lines = [
line.strip()
for line in reference[len('\\begin{pmatrix}'):-len('\\end{pmatrix}')].split('\\\\')
if line.strip()
]
matched = True
if len(pred_lines) == len(ref_lines):
for pred_line, ref_line in zip(pred_lines, ref_lines):
pred_parts = pred_line.split('&')
ref_parts = ref_line.split('&')
if len(pred_parts) == len(ref_parts):
if not all([
math_equal(
pred_parts[i],
ref_parts[i],
include_percentage,
is_close,
) for i in range(len(pred_parts))
]):
matched = False
break
else:
matched = False
if not matched:
break
else:
matched = False
if matched:
return True
if prediction.count('=') == 1 and reference.count('=') == 1:
pred = prediction.split('=')
pred = f'{pred[0].strip()} - ({pred[1].strip()})'
ref = reference.split('=')
ref = f'{ref[0].strip()} - ({ref[1].strip()})'
if symbolic_equal(pred, ref) or symbolic_equal(f'-({pred})', ref):
return True
elif (prediction.count('=') == 1 and len(prediction.split('=')[0].strip()) <= 2 and '=' not in reference):
if math_equal(prediction.split('=')[1], reference, include_percentage, is_close):
return True
elif (reference.count('=') == 1 and len(reference.split('=')[0].strip()) <= 2 and '=' not in prediction):
if math_equal(prediction, reference.split('=')[1], include_percentage, is_close):
return True
if symbolic_equal(prediction, reference):
return True
return False
def numeric_equal(prediction: float, reference: float):
return isclose(reference, prediction, rel_tol=1e-4)
def symbolic_equal(a, b):
def _parse(s):
for f in [parse_latex, parse_expr, latex2sympy]:
try:
return f(s.replace('\\\\', '\\'))
except Exception:
try:
return f(s)
except Exception:
pass
return s
a = _parse(a)
b = _parse(b)
# direct equal
try:
if str(a) == str(b) or a == b:
return True
except Exception:
pass
# simplify equal
try:
if a.equals(b) or simplify(a - b) == 0:
return True
except Exception:
pass
# equation equal
try:
if (abs(a.lhs - a.rhs)).equals(abs(b.lhs - b.rhs)):
return True
except Exception:
pass
try:
if numeric_equal(float(N(a)), float(N(b))):
return True
except Exception:
pass
# matrix
try:
# if a and b are matrix
if a.shape == b.shape:
_a = a.applyfunc(lambda x: round(x, 3))
_b = b.applyfunc(lambda x: round(x, 3))
if _a.equals(_b):
return True
except Exception:
pass
return False
def process_math500(prediction, ref_answer):
return float(math_equal(prediction, ref_answer))
def main():
prediction = r'''Here's a step-by-step solution:
1. **Identify the given information from the diagram and problem statement:**
* Triangle DEF is a right-angled triangle, with the right angle at E (indicated by the right-angle mark).
* The length of side EF is 7 (from the label on the diagram and the coordinates E=(0,7), F=(0,0)).
* We are given $\sin D = 0.7$.
2. **Recall the definition of sine in a right-angled triangle:**
For an angle in a right-angled triangle, the sine is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse.
In triangle DEF, for angle D:
* The side opposite to angle D is EF.
* The hypotenuse is DF.
Therefore, $\sin D = \frac{EF}{DF}$.
3. **Use the given $\sin D$ and EF to find the length of the hypotenuse DF:**
We have $\sin D = 0.7$ and $EF = 7$.
Substitute these values into the sine formula:
$0.7 = \frac{7}{DF}$
To solve for DF, multiply both sides by DF and then divide by 0.7:
$DF = \frac{7}{0.7}$
$DF = \frac{7}{\frac{7}{10}}$
$DF = 7 \times \frac{10}{7}$
$DF = 10$.
4. **Use the Pythagorean theorem to find the length of DE:**
Since triangle DEF is a right-angled triangle, we can use the Pythagorean theorem, which states that the square of the hypotenuse (DF) is equal to the sum of the squares of the other two sides (DE and EF):
$DE^2 + EF^2 = DF^2$
Substitute the known values $EF = 7$ and $DF = 10$:
$DE^2 + 7^2 = 10^2$
$DE^2 + 49 = 100$
5. **Solve for DE:**
Subtract 49 from both sides:
$DE^2 = 100 - 49$
$DE^2 = 51$
Take the square root of both sides:
$DE = \sqrt{51}$
The final answer is $\boxed{\sqrt{51}}$.'''
print(process_math500(extract_answer(prediction), '\\sqrt{51}'))
def main_units():
print(math_equal('\n\\boxed{70,\\!000}\n', '70000'))
print(extract_answer('The answer is \\boxed{70,\\!000}'))
print(strip_answer_string(extract_answer('The answer is \\boxed{70,\\!000}')))
print(math_equal(extract_answer('The answer is \\boxed{70,\\!000}'), '70000'))
if __name__ == '__main__':
main()
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