Merge branch 'main' into judgerbench

.github/workflows/lint.yml

@@ -17,7 +17,7 @@ jobs:
           python-version: '3.10'
       - name: Install pre-commit hook
         run: |
-          pip install pre-commit mmengine
+          pip install pre-commit==3.8.0 mmengine
           pre-commit install
       - name: Linting
         run: pre-commit run --all-files

configs/datasets/gaokao_math/README.md

@@ -0,0 +1,108 @@
+# GaoKao MATH Answer Evaluation Dataset
+A dataset for testing the performance of the model in the GaoKao MATH Answer Extraction task.
+Now support the following format of GAOKAO math questions:
+1. '单选题'：Single choice question
+2. '多选题'：Multiple choice question
+3. '填空题'：Fill in the blank question, can be multiple blanks
+4. '解答题'：Answer question, can be multiple answers
+
+Sample data:
+```json
+[
+    {
+        "id": "3b270bc4-570a-4d77-b122-a2fc372f7d6a",
+        "question": "过椭圆${x^2\\over {16}} +{ y^2 \\over {4}}=1$ %内一点$M(2,1)$ %引一条弦，使该弦被点$M$ %平分，则这条弦所在直线的方程为（ ）．\nA. $x+2y-4=0$ %\nB. $x-2y-4=0$ %\nC. $x+2y+4=0$ %\nD. $x-2y+4=0$ %\n\n",
+        "response": "本题主要考查直线与圆锥曲线．设所求直线与椭圆的一个交点为$A(x,y)$ %，由于中点$M(2,1)$ %，所以另一个交点$B$ %为$(4-x,2-y)$ %．因为$A$ %，$B$ %两点都在椭圆上，所以$x^2+4y^2=16$ %，$(4-x)^2+4(2-y)^2=16$ %，两式相减，整理可得$x+2y-4=0$ %．由于过$A$ %，$B$ %两点的直线只有一条，所以这条弦所在直线的方程为$x+2y-4=0$ %．故本题正确答案为A．\n答案是：A",
+        "extract_answer": "A",
+        "question_type": "单选题"
+    },
+    {
+        "id": "d60e42d7-30ee-44f9-a94d-aff6a8127750",
+        "question": "若函数$f(x)$ 具有下列性质：1.定义域为$(-1,1)$ ；2.对于任意的$x,y\\in(-1,1)$ ，都有$f(x)+f(y)=f\\left({\\dfrac{x+y}{1+xy}}\\right)$ ；3.当$-1< x< 0$ 时，$f(x)>0$ ，则称函数$f(x)$ 为$δ$ 的函数$.$ 若函数$f(x)$ 为$δ$ 的函数，则以下结论正确的是$(\\quad)$\nA. $\nB. x)$ 为奇函数\nC. $\nD. x)$ 为偶函数\nE. $\nF. x)$ 为单调递减函数\nG. $\nH. x)$ 为单调递增函数\n\n",
+        "response": "函数$f(x)$ 为$δ$ 的函数，令$x=y=0$ ，则$f(0)+f(0)=f(0)$ ，即$f(0)=0$ ，令$y=-x$ ，则$f(x)+f(-x)=f\\left(\\dfrac{x-x}{1-{x}^{2}}\\right)=f(0)=0$ ，则$f(-x)=-f(x)$ ，即函数$f(x)$ 是奇函数，设$-1< x< y< 1$ ，则$f(x)-f(y)=f(x)+f(-y)=f\\left(\\dfrac{x-y}{1-xy}\\right)$ ，$∵-1< x< y< 1$ ，$∴-1< \\dfrac{x-y}{1-xy}< 0$ ，则$f\\left(\\dfrac{x-y}{1-xy}\\right)>0$ ，即$f(x)-f(y)>0$ ，则$f(x)>f(y)$ ，即$f(x)$ 在$(-1,1)$ 上是减函数.故选$AC.$ 本题考查函数的奇偶性和单调性的判断，注意运用定义法，考查运算能力和推理能力，属于中档题.可令$x=y=0$ ，求得$f(0)=0$ ，再令$y=-x$ 可得$f(-x)=-f(x)$ ，可得$f(x)$ 的奇偶性；再令$-1< x< y< 1$ ，运用单调性的定义，结合其偶性的定义可得其单调性．\n答案是：A; C",
+        "extract_answer": "A, C",
+        "question_type": "多选题"
+    },
+    {
+        "id": "31b3f702-e60c-4a20-9a40-73bd72b92d1e",
+        "question": "请完成以下题目(1)曲线$$y=-5\\text{e}^{x}+3$$在点$$(0,-2)$$处的切线方程为___.(2)若曲线$$f(x)=x \\sin x+1$$在$$x=\\dfrac{ \\pi }{2}$$处的切线与直线$$ax+2y+1=0$$相互垂直,则实数$$a=$$___.\n\n",
+        "response": "(1)由$$y=-5\\text{e}^{x}+3$$,得$$y'=-5\\text{e}^{x}$$,所以切线的斜率$$k=y'|_{x=0}=-5$$,所以切线方程为$$y+2=-5(x-0)$$,即$$5x+y+2=0$$.(2)因为$$f'(x)= \\sin x+x \\cos x$$,所以$$f'\\left(\\dfrac{ \\pi }{2}\\right)= \\sin \\dfrac{ \\pi }{2}+\\dfrac{ \\pi }{2}\\cdot \\cos \\dfrac{ \\pi }{2}=1$$.又直线$$ax+2y+1=0$$的斜率为$$-\\dfrac{a}{2}$$,所以根据题意得$$1\\times \\left(-\\dfrac{a}{2}\\right)=-1$$,解得$$a=2$$.\n答案是：(1)$$5x+y+2=0$$ (2)$$2$$",
+        "extract_answer": "['(1)$$5x+y+2=0$$ (2)$$2$$']",
+        "question_type": "填空题"
+    },
+    {
+        "id": "16878941-1772-4290-bc61-00b193d5cf70",
+        "question": "已知函数$f\\left( x \\right)=\\left| 2x-1 \\right|$.（1）若不等式$f\\left( x+\\frac{1}{2} \\right)\\ge 2m+1\\left( m > 0 \\right)$的解集为$\\left( -\\infty ,-2 \\right]\\bigcup \\left[ 2,+\\infty \\right)$，求实数$m$的值；（2）若不等式$f\\left( x \\right)\\le {{2}^{y}}+\\frac{a}{{{2}^{y}}}+\\left| 2x+3 \\right|$对任意的实数$x,y\\in R$恒成立，求实数$a$的最小值.\n\n",
+        "response": "（1）直接写出不等式，解含有绝对值的函数不等式即可；（2）这是恒成立求参的问题,根据绝对值三角不等式得到左侧函数的最值，再结合均值不等式得最值.（1）由条件得$\\left| 2x \\right|\\le 2m+1$得$-m-\\frac{1}{2}\\le x\\le m+\\frac{1}{2}$，所以$m=\\frac{3}{2}$.（2）原不等式等价于$\\left| 2x-1 \\right|-\\left| 2x+3 \\right|\\le {{2}^{y}}+\\frac{a}{{{2}^{y}}}$，而$\\left| 2x-1 \\right|-\\left| 2x+3 \\right|\\le \\left| \\left( 2x-1 \\right)-\\left( 2x+3 \\right) \\right|=4$，所以${{2}^{y}}+\\frac{a}{{{2}^{y}}}\\ge 4$，则$a\\ge {{\\left[ {{2}^{y}}\\left( 4-{{2}^{y}} \\right) \\right]}_{\\text{max}}}=4$，当且仅当$y=1$时取得.\n答案是：(1) $m=\\frac{3}{2}$；(2) 最小值为$a=4$.",
+        "extract_answer": [
+            "(1) $m=\\frac{3}{2}$；(2) 最小值为$a=4$."
+        ],
+        "question_type": "解答题"
+    }
+]
+```
+## How to use
+
+### 1. Prepare the dataset
+```bash
+cd opencompass
+cp -rf /cpfs01/shared/public/liuhongwei/data/gaokao_math_dataset/gaokao_math ./data
+```
+📢：If you want to evaluate your own gaokao math data, replace the `test_v2.jsonl` with your own data, but follow the format above.
+
+### 2. Set the evaluation model
+
+open `opencompass.datasets.gaokao_math.gaokao_math_gen_9b076f` and set the model name and api url for evaluation, multiple urls are supported for acceleration.
+
+```python
+...
+
+gaokao_math_eval_cfg = dict(
+    evaluator=dict(type=GaoKaoMATHEvaluator, model_name='EVALUATE_MODEL_NAME', url=['http://0.0.0.0:23333/v1', 'http://...']))
+
+...
+
+```
+We recommand `Qwen2.5-72B-Instruct` model for evaluation.
+
+
+### 3. Set Extractor model and run the evaluation
+
+```python
+from mmengine.config import read_base
+from opencompass.models import HuggingFacewithChatTemplate
+
+
+with read_base():
+    from opencompass.datasets.gaokao_math.gaokao_math_gen_9b076f import gaokao_math_datasets
+
+
+trained_qwen2_1_5b_model = [ # trained extractor model
+    dict(
+        type=HuggingFacewithChatTemplate,
+        abbr='gaokao_math_extractor_1_5b_v02',
+        path='/cpfs01/shared/public/liuhongwei/models/gaokao_math_trained/gaokao_math_extractor_1_5b_v02',
+        max_out_len=1024,
+        batch_size=8,
+        run_cfg=dict(num_gpus=1),
+    )
+]
+
+datasets = sum([v for k, v in locals().items() if k.endswith("_datasets")], [])
+models = sum([v for k, v in locals().items() if k.endswith("_model")], [])
+
+...
+```
+
+### 4. Run the evaluation
+
+```bash
+python run.py eval.py --dump-eval-details # eval and dump the evaluation details to `results` folder
+```
+
+
+### 5. Evaluation results
+
+| Evaluator / Extractor | Qwen2.5-72B-Instruct | gaokao_math_extractor_1.5b_v0.2 |
+|-----------------------|-----------------------|----------------------------------|
+| Qwen2.5-72B-Instruct (ACC) | 95.85 | 95.2 |

configs/datasets/gaokao_math/gaokao_math_gen_f5fd28.py

@@ -0,0 +1,48 @@
+from opencompass.openicl.icl_prompt_template import PromptTemplate
+from opencompass.openicl.icl_retriever import ZeroRetriever
+from opencompass.openicl.icl_inferencer import GenInferencer
+from opencompass.datasets import GaoKaoMATHDataset, GaoKaoMATHEvaluator
+
+
+MATH_CN_PROMPT="""
+你是一个数学阅卷专家，任务是从给定的回答句子中提取精确的关键答案。你必须只提供提取的关键答案，不包括任何额外的文字。
+—
+我将为你提供一个问题、回答句子和问题类型。回答句子是对所提供问题的回应。利用提供的信息，你必须准确而精确地确定并从回答句子中提取预期的关键答案。请不要对问题发表主观看法。
+
+对于单选题，答案应该是选项字母，例如 "A"；
+对于多选题，答案应该是一个选项字母的列表，例如 ["A"] 或 ["A", "B", "C"]；
+对于填空题，答案应该是一个填入空白处的答案列表，列表的数量应该与问题中的空白数量相同，例如 ["$$\\frac{{1}}{{2}}$$"] 或 ["$$\\frac{{1}}{{2}}$$", "2"]。
+对于问答题，类似填空题，为每个小问抽出相应答案，例如 ["$$\\frac{{1}}{{2}}$$"] 或 ["$$\\frac{{1}}{{2}}$$", "2"]。
+
+如果回答句子提供了多个不同的答案，请仔细判断后面提供的答案是否是对前面答案的修正或修改。如果是这样，提取这个修正或修改后的答案作为最终答案。相反，如果回答句子在多个答案之间波动而没有明确的最终答案，你应该输出 [No valid answer]。
+—
+问题类型: {question_type}
+原始问题: {question}
+回答: {response}
+提取的关键答案:
+"""
+
+gaokao_math_reader_cfg = dict(input_columns=['question', 'response', 'question_type'], output_column='extract_answer')
+
+
+gaokao_math_infer_cfg = dict(
+    prompt_template=dict(
+        type=PromptTemplate,
+        template=dict(round=[
+            dict(role='HUMAN', prompt=MATH_CN_PROMPT),
+        ])),
+    retriever=dict(type=ZeroRetriever),
+    inferencer=dict(type=GenInferencer, max_out_len=512))
+
+gaokao_math_eval_cfg = dict(
+    evaluator=dict(type=GaoKaoMATHEvaluator, model_name='Qwen/Qwen2.5-72B-Instruct', url=['http://22.8.73.119:23333/v1', 'http://22.8.4.97:23333/v1', 'http://22.8.22.254:23333/v1', 'http://22.8.17.14:23333/v1']))
+
+gaokao_math_datasets = [
+    dict(
+        type=GaoKaoMATHDataset,
+        abbr='GaoKaoMATH',
+        path='./data/gaokao_math/test_2k.json',
+        reader_cfg=gaokao_math_reader_cfg,
+        infer_cfg=gaokao_math_infer_cfg,
+        eval_cfg=gaokao_math_eval_cfg)
+]

configs/datasets/math/math_0shot_llm_judge_gen_393424.py

@@ -0,0 +1,78 @@
+from opencompass.openicl.icl_prompt_template import PromptTemplate
+from opencompass.openicl.icl_retriever import ZeroRetriever
+from opencompass.openicl.icl_inferencer import GenInferencer
+from opencompass.datasets import MATHDataset, MATHEvaluator, math_postprocess_v2, GaoKaoMATHEvaluator
+from opencompass.utils.model_postprocessors import naive_model_postprocess, xfinder_postprocess
+from opencompass.utils.postprocessors.naive import MATH_NAVIE_PROMPT_TEMPLATE
+
+# ----------------------------- Eval Parameters -----------------------------
+## Postprocess function
+post_func = 're' # 're', 'xfinder_model', 'naive_model'
+
+## Evalute function
+eval_func = 'naive_model' # 're', 'naive_model'
+
+## Model api url
+xfinder_url = 'http://0.0.0.0:23333/v1' # for 'xFinder-qwen1505' if post_func is 'xfinder_model'
+naive_model_name = 'Qwen/Qwen2.5-72B-Instruct' # replace with your model name
+naive_model_url = ['http://22.8.6.22:23333/v1', 'http://22.8.67.84:23333/v1', 'http://22.8.72.81:23333/v1', 'http://22.9.42.143:23333/v1'] # Multi-apis for accerlation
+
+# ----------------------------- Detailed Config -----------------------------
+
+math_reader_cfg = dict(input_columns=['problem'], output_column='solution')
+
+math_infer_cfg = dict(
+    prompt_template=dict(
+        type=PromptTemplate,
+        template=dict(
+            round=[
+                dict(role='HUMAN', prompt='{problem}\nPlease reason step by step, and put your final answer within \\boxed{}.'),
+            ]
+        ),
+    ),
+    retriever=dict(type=ZeroRetriever),
+    inferencer=dict(type=GenInferencer, max_out_len=1024),
+)
+
+if post_func == 're':
+    pred_postprocessor = dict(type=math_postprocess_v2)
+elif post_func == 'xfinder_model':
+    pred_postprocessor = dict(
+        type=xfinder_postprocess,
+        question_type='math',
+        model_name='xFinder-qwen1505',
+        num_processes=128,
+        api_url=xfinder_url,
+    )
+elif post_func == 'naive_model':
+    pred_postprocessor = dict(
+        type=naive_model_postprocess,
+        custom_instruction=MATH_NAVIE_PROMPT_TEMPLATE,
+        model_name=naive_model_name,
+        num_processes=64,
+        api_url=naive_model_url,
+    )
+
+if eval_func == 're':
+    evaluator = dict(type=MATHEvaluator, version='v2')
+elif eval_func == 'naive_model':
+    evaluator = dict(
+        type=GaoKaoMATHEvaluator,
+        model_name=naive_model_name,
+        url=naive_model_url,
+    )
+
+math_eval_cfg = dict(
+    evaluator=evaluator, pred_postprocessor=pred_postprocessor,
+)
+
+math_datasets = [
+    dict(
+        type=MATHDataset,
+        abbr='math',
+        path='opencompass/math',
+        reader_cfg=math_reader_cfg,
+        infer_cfg=math_infer_cfg,
+        eval_cfg=math_eval_cfg,
+    )
+]

configs/datasets/flames/flames_gen_1a58bb.py → configs/datasets/subjective/flames/flames_gen_1a58bb.py

@@ -58,5 +58,6 @@
             name=_name,
             reader_cfg=subjective_reader_cfg,
             infer_cfg=subjective_infer_cfg,
-            eval_cfg=subjective_eval_cfg
+            eval_cfg=subjective_eval_cfg,
+            mode='singlescore',
         ))