[MATH] Too many values to unpack (expected 2)

[rkinas]

I am not able to run simple test from lighteval (math) tasks - I am getting "Too many values to unpack (expected 2)"

accelerate launch --multi_gpu --num_processes=4 run_evals_accelerate.py --model_args "pretrained=openchat/openchat-3.5-0106" --tasks "lighteval|math:algebra|0|0" --override_batch_size 1 --output_dir="./evals/"

[clefourrier]

Hi, could you provide your full logs?

[lewtun]

I can also reproduce the error with commit bbe3b5f8eff55c63dee8a28ebedc7a3a46556598

Edit: same issue on main with commit 6a48e4e7723611edda9f51d0fcb9c4d20b79b19d

Command to reproduce:

accelerate launch --multi_gpu --num_processes=8 run_evals_accelerate.py \
    --model_args "pretrained=Qwen/Qwen1.5-0.5B-Chat" \
    --use_chat_template \
    --tasks "lighteval|math:algebra|0|0" \
    --override_batch_size 1 \
    --output_dir "./scratch/evals"

Curiously, limiting the number of samples works:

accelerate launch --multi_gpu --num_processes=8 run_evals_accelerate.py \
    --model_args "pretrained=Qwen/Qwen1.5-0.5B-Chat" \
    --use_chat_template \
    --tasks "lighteval|math:algebra|0|0" \
    --override_batch_size 1 \
    --output_dir "./scratch/evals" \
    --max_samples 10

Logs attached.

logs.txt

cc @clefourrier @NathanHB for viz

[edbeeching]

Some notes on this.

It looks like the math_normalizer trys to parse a gold answer text:

The fraction $\\frac{1}{2y+1}$ fails to be defined only if the denominator is zero. This occurs when $y$ is the solution of the equation $$2y+1=0,$$ which is $y=-\\frac 12$. Thus the domain of $k(y)$ is $$\\boxed{\\left(-\\infty,-\\frac 12\\right)\\cup \\left(-\\frac 12,\\infty\\right)}.$$

This then fails in fix_fracs, which tries to parse the following:

(-\\infty,-\\frac12)\\cup(-\\frac12,\\infty)

and ends up failing when it tries to unpack:

a, b = substr # substr=12)\\cup(-

The parsing logic may need to be made more robust:
Here is an example code from a private repo that may help with this:

import re


def last_boxed_only_string(string):
    """
    Extracts the last LaTeX boxed or framed expression from a string.
    Args:
        string (str): The input string containing LaTeX expressions.
    Returns:
        str or None: The last boxed or framed expression, if found;
        otherwise, None.
    """

    idx = string.rfind("\\boxed")
    if idx < 0:
        idx = string.rfind("\\fbox")
        if idx < 0:
            return None

    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 = None
    else:
        retval = string[idx : right_brace_idx + 1]

    return retval


def remove_boxed(s):
    """
    Removes the LaTeX boxed command, returning the content inside the braces.
    Args:
        s (str): The string containing a LaTeX boxed expression.
    Returns:
        str or None: The content inside the boxed command, if valid;
        otherwise, None.
    """

    left = "\\boxed{"
    try:
        assert s[: len(left)] == left
        assert s[-1] == "}"
        length = len(left)
        return s[length:-1]
    except Exception:
        return None


def extract_boxed_answer(pred_str, strip_double_curly_brace=False):
    """
    Extracts the answer from a LaTeX boxed expression within
    a prediction string.
    Args:
        pred_str (str): The string containing one or more LaTeX
        boxed expressions.
        strip_double_curly_brace (bool): If True, removes an additional
        layer of braces.
    Returns:
        str or None: The extracted answer, if any; otherwise, None.
    """

    boxed_str = last_boxed_only_string(pred_str)
    if boxed_str is None:
        return None
    answer = remove_boxed(boxed_str)
    if answer is None:
        return None
    if strip_double_curly_brace:
        match = re.match("^\{(.*)\}$", answer)  # noqa: W605
        if match:
            answer = match.group(1)
    return answer


def normalize_final_answer(final_answer: str) -> str:
    """
    Normalizes a final answer string by removing or replacing various LaTeX
    and text elements.
    Args:
        final_answer (str): The answer string to normalize.
    Returns:
        str: The normalized answer string.
    """

    match = re.search(r"(.*?)Problem:", final_answer, flags=re.S)
    if match:
        final_answer = match.group(1)  # 返回匹配的第一部分，即"Problem"之前的所有文本
    """Normalize a final answer to a quantitative reasoning question."""
    # final_answer = final_answer.split('=')[-1]
    SUBSTITUTIONS = [
        ("an ", ""),
        ("a ", ""),
        (".$", "$"),
        ("\\$", ""),
        (r"\ ", ""),
        (" ", ""),
        ("mbox", "text"),
        (",\\text{and}", ","),
        ("\\text{and}", ","),
        ("\\text{m}", "\\text{}"),
        ("\\le", "<"),
    ]
    REMOVED_EXPRESSIONS = [
        "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",
        "\n",
        "\r",
        "\f",
    ]
    for before, after in SUBSTITUTIONS:
        final_answer = final_answer.replace(before, after)
    for expr in 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"(\\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)
    assert "\n" not in final_answer
    assert "\r" not in final_answer
    assert "\f" not in final_answer
    if len(re.findall(r"finalansweris(.*)", final_answer)) > 0:
        final_answer = re.findall(r"finalansweris(.*)", final_answer)[-1]

    if len(re.findall(r"answer?is:?(.*)", final_answer)) > 0:
        final_answer = re.findall(r"answer?is:?(.*)", final_answer)[-1]

    if len(re.findall(r"oxed\{(.*?)\}", final_answer)) > 0:
        final_answer = re.findall(r"oxed\{(.*?)\}", final_answer)[-1]

    if len(re.findall(r"\$(.*?)\$", final_answer)) > 0:
        final_answer = re.findall(r"\$(.*?)\$", final_answer)[-1]
    final_answer = final_answer.strip()
    if "rac" in final_answer and "\\frac" not in final_answer:
        final_answer = final_answer.replace("rac", "\\frac")

    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("$", "")

    if final_answer.replace(",", "").isdigit():
        final_answer = final_answer.replace(",", "")

    return final_answer