lm_eval/tasks/leaderboard/math/utils.py (95 lines of code) (raw):
from typing import Dict, List
import datasets
try:
from math_verify import LatexExtractionConfig, parse, verify
except ModuleNotFoundError:
raise ModuleNotFoundError(
"`math-verify` is required for generating translation task prompt templates. \
please install math-verify via pip install lm-eval[math] or pip install -e .[math]",
)
INVALID_ANSWER = "[invalidanswer]"
# taken from
# https://github.com/wellecks/lm-evaluation-harness/blob/master/lm_eval/tasks/minerva_math.py
def doc_to_text(doc: dict) -> str:
return "Problem:" + "\n" + doc["problem"] + "\n\n" + "Solution:"
def process_docs(dataset: datasets.Dataset) -> datasets.Dataset:
def _process_doc(doc: dict) -> dict:
out_doc = {
"problem": doc["problem"],
"solution": doc["solution"],
"answer": remove_boxed(last_boxed_only_string(doc["solution"])),
}
if getattr(doc, "few_shot", None) is not None:
out_doc["few_shot"] = True
return out_doc
return dataset.map(_process_doc)
def list_fewshot_samples() -> list[dict]:
return [
{
"problem": "Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}",
"solution": "The expressions inside each square root must be non-negative. Therefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\\boxed{[2,5)}$.\nFinal Answer: The final answer is $[2,5)$. I hope it is correct.",
"few_shot": "1",
},
{
"problem": "If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$",
"solution": "We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\nFinal Answer: The final answer is $24$. I hope it is correct.",
"few_shot": "1",
},
{
"problem": "Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?",
"solution": "If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$:\n\\begin{align*}\n30n&=480\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align*}\nFinal Answer: The final answer is $16$. I hope it is correct.",
"few_shot": "1",
},
{
"problem": "If the system of equations\n\n\\begin{align*}\n6x-4y&=a,\\\n6y-9x &=b.\n\\end{align*}has a solution $(x, y)$ where $x$ and $y$ are both nonzero,\nfind $\\frac{a}{b},$ assuming $b$ is nonzero.",
"solution": "If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\nFinal Answer: The final answer is $-\\frac{2}{3}$. I hope it is correct.",
"few_shot": "1",
},
]
def process_results(doc: dict, results: List[str]) -> Dict[str, int]:
candidates = results[0]
parsed_candidate = parse(candidates)
parsed_answer = parse(doc["solution"], extraction_config=[LatexExtractionConfig()])
if verify(parsed_answer, parsed_candidate):
retval = 1
else:
retval = 0
output = {
"exact_match": retval,
}
return output
def last_boxed_only_string(string: str) -> str:
idx = string.rfind("\\boxed")
if "\\boxed " in string:
return "\\boxed " + string.split("\\boxed ")[-1].split("$")[0]
if idx < 0:
idx = string.rfind("\\fbox")
if idx < 0:
return INVALID_ANSWER
i = idx
right_brace_idx = None
num_left_braces_open = 0
while i < len(string):
if string[i] == "{":
num_left_braces_open += 1
if string[i] == "}":
num_left_braces_open -= 1
if num_left_braces_open == 0:
right_brace_idx = i
break
i += 1
if right_brace_idx is None:
retval = INVALID_ANSWER
else:
retval = string[idx : right_brace_idx + 1]
return retval
def remove_boxed(s: str) -> str:
try:
if "\\boxed " in s:
left = "\\boxed "
assert s[: len(left)] == left
return s[len(left) :]
left = "\\boxed{"
assert s[: len(left)] == left
assert s[-1] == "}"
return s[len(left) : -1]
except AssertionError:
return INVALID_ANSWER