DeepSeek-Math/evaluation/eval/ocwcourses_eval_utils.py
ZhihongShao 21cc5c6701 init
2024-02-06 10:27:40 +08:00

264 lines
7.9 KiB
Python

import re
import numpy as np
import sympy
from sympy.core.sympify import SympifyError
from sympy.parsing.latex import parse_latex
import signal
INVALID_ANSWER = "[invalidanswer]"
class timeout:
def __init__(self, seconds=1, error_message="Timeout"):
self.seconds = seconds
self.error_message = error_message
def handle_timeout(self, signum, frame):
raise TimeoutError(self.error_message)
def __enter__(self):
signal.signal(signal.SIGALRM, self.handle_timeout)
signal.alarm(self.seconds)
def __exit__(self, type, value, traceback):
signal.alarm(0)
def normalize_numeric(s):
if s is None:
return None
for unit in [
"eV",
" \\mathrm{~kg} \\cdot \\mathrm{m} / \\mathrm{s}",
" kg m/s",
"kg*m/s",
"kg",
"m/s",
"m / s",
"m s^{-1}",
"\\text{ m/s}",
" \\mathrm{m/s}",
" \\text{ m/s}",
"g/mole",
"g/mol",
"\\mathrm{~g}",
"\\mathrm{~g} / \\mathrm{mol}",
"W",
"erg/s",
"years",
"year",
"cm",
]:
s = s.replace(unit, "")
s = s.strip()
for maybe_unit in ["m", "s", "cm"]:
s = s.replace("\\mathrm{" + maybe_unit + "}", "")
s = s.replace("\\mathrm{~" + maybe_unit + "}", "")
s = s.strip()
s = s.strip("$")
try:
return float(eval(s))
except:
try:
expr = parse_latex(s)
if expr.is_number:
return float(expr)
return INVALID_ANSWER
except:
return INVALID_ANSWER
def numeric_equality(n1, n2, threshold=0.01):
if n1 is None or n2 is None:
return False
if np.isclose(n1, 0) or np.isclose(n2, 0) or np.isclose(n1 - n2, 0):
return np.abs(n1 - n2) < threshold * (n1 + n2) / 2
else:
return np.isclose(n1, n2)
def normalize_symbolic_equation(s):
if not isinstance(s, str):
return INVALID_ANSWER
if s.startswith("\\["):
s = s[2:]
if s.endswith("\\]"):
s = s[:-2]
s = s.replace("\\left(", "(")
s = s.replace("\\right)", ")")
s = s.replace("\\\\", "\\")
if s.startswith("$") or s.endswith("$"):
s = s.strip("$")
try:
maybe_expression = parse_latex(s)
if not isinstance(maybe_expression, sympy.core.relational.Equality):
# we have equation, not expression
return INVALID_ANSWER
else:
return maybe_expression
except:
return INVALID_ANSWER
class SymbolicMathMixin:
"""
Methods useful for parsing mathematical expressions from text and determining equivalence of expressions.
"""
SUBSTITUTIONS = [ # used for text normalize
("an ", ""),
("a ", ""),
(".$", "$"),
("\\$", ""),
(r"\ ", ""),
(" ", ""),
("mbox", "text"),
(",\\text{and}", ","),
("\\text{and}", ","),
("\\text{m}", "\\text{}"),
]
REMOVED_EXPRESSIONS = [ # used for text normalizer
"square",
"ways",
"integers",
"dollars",
"mph",
"inches",
"ft",
"hours",
"km",
"units",
"\\ldots",
"sue",
"points",
"feet",
"minutes",
"digits",
"cents",
"degrees",
"cm",
"gm",
"pounds",
"meters",
"meals",
"edges",
"students",
"childrentickets",
"multiples",
"\\text{s}",
"\\text{.}",
"\\text{\ns}",
"\\text{}^2",
"\\text{}^3",
"\\text{\n}",
"\\text{}",
r"\mathrm{th}",
r"^\circ",
r"^{\circ}",
r"\;",
r",\!",
"{,}",
'"',
"\\dots",
]
def normalize_tex(self, final_answer: str) -> str:
"""
Normalizes a string representing a mathematical expression.
Used as a preprocessing step before parsing methods.
Copied character for character from appendix D of Lewkowycz et al. (2022)
"""
final_answer = final_answer.split("=")[-1]
for before, after in self.SUBSTITUTIONS:
final_answer = final_answer.replace(before, after)
for expr in self.REMOVED_EXPRESSIONS:
final_answer = final_answer.replace(expr, "")
# Extract answer that is in LaTeX math, is bold,
# is surrounded by a box, etc.
final_answer = re.sub(r"(.*?)(\$)(.*?)(\$)(.*)", "$\\3$", final_answer)
final_answer = re.sub(r"(\\text\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\textbf\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\overline\{)(.*?)(\})", "\\2", final_answer)
final_answer = re.sub(r"(\\boxed\{)(.*)(\})", "\\2", final_answer)
# Normalize shorthand TeX:
# \fracab -> \frac{a}{b}
# \frac{abc}{bef} -> \frac{abc}{bef}
# \fracabc -> \frac{a}{b}c
# \sqrta -> \sqrt{a}
# \sqrtab -> sqrt{a}b
final_answer = re.sub(r"(frac)([^{])(.)", "frac{\\2}{\\3}", final_answer)
final_answer = re.sub(r"(sqrt)([^{])", "sqrt{\\2}", final_answer)
final_answer = final_answer.replace("$", "")
# Normalize 100,000 -> 100000
if final_answer.replace(",", "").isdigit():
final_answer = final_answer.replace(",", "")
return final_answer
def parse_tex(self, text: str, time_limit: int = 5) -> sympy.Basic:
"""
Wrapper around `sympy.parse_text` that outputs a SymPy expression.
Typically, you want to apply `normalize_text` as a preprocessing step.
"""
try:
with timeout(seconds=time_limit):
parsed = parse_latex(text)
except (
# general error handling: there is a long tail of possible sympy/other
# errors we would like to catch
Exception
) as e:
print(f"failed to parse {text} with exception {e}")
return None
return parsed
def is_exp_equiv(self, x1: sympy.Basic, x2: sympy.Basic, time_limit=5) -> bool:
"""
Determines whether two sympy expressions are equal.
"""
try:
with timeout(seconds=time_limit):
try:
diff = x1 - x2
except (SympifyError, ValueError, TypeError) as e:
print(
f"Couldn't subtract {x1} and {x2} with exception {e}"
)
return False
try:
if sympy.simplify(diff) == 0:
return True
else:
return False
except (SympifyError, ValueError, TypeError) as e:
print(f"Failed to simplify {x1}-{x2} with {e}")
return False
except TimeoutError as e:
print(f"Timed out comparing {x1} and {x2}")
return False
except Exception as e:
print(f"failed on unrecognized exception {e}")
return False
def is_tex_equiv(self, x1: str, x2: str, time_limit=5) -> bool:
"""
Determines whether two (ideally normalized using `normalize_text`) TeX expressions are equal.
Does so by first checking for string exact-match, then falls back on sympy-equivalence,
following the (Lewkowycz et al. 2022) methodology.
"""
if x1 == x2:
# don't resort to sympy if we have full string match, post-normalization
return True
else:
return False
parsed_x2 = self.parse_tex(x2)
if not parsed_x2:
# if our reference fails to parse into a Sympy object,
# we forgo parsing + checking our generated answer.
return False
return self.is_exp_equiv(self.parse_tex(x1), parsed_x2, time_limit=time_limit)