Open DietDietDiet opened 2 months ago
Hi, @DietDietDiet , CMMMU 应该是尚未在 VLMEvalKit 中支持,作者没有提过 PR
@kennymckormick 这个官方可以支持一下吗
Hi, @DietDietDiet We have added this benchmark to our TODO list, but cannot guarantee a time at this moment. If you urgently need this benchmark, you can follow the MMMU example to create a PR and add this benchmark (we will help with merging very fast).
支持CMMMU评测: 大概步骤为加载数据集、拼接并构建调用LLM的指令、最麻烦的就是各种对结果的归一化。当前当前只能对【dev 、val】这两个数据进行评估了,因为test文件未开源答案。具体数据地址分别为:
第一步:数据获取: dev: https://github.com/CMMMU-Benchmark/CMMMU/tree/main/cmmmu-data-dev val : https://github.com/CMMMU-Benchmark/CMMMU/tree/main/cmmmu-data-val
第二步:拼接指令预测结果
PROMPT = {
"task_instructions": [
"请回答以下多项选择题,并选出正确选项。这些题目可能包括单选和多选题型。如果所提供的信息不足以确定一个明确的答案,那么请根据可用的数据和你的判断来选择最可能正确的选项。",
"请回答以下判断题,并根据题目描述和所给的信息来判断问题中陈述的对错。如果信息不完整或不足以作出绝对判断,请运用你的逻辑推理和现有信息来做出最可能的判断。",
"请回答以下填空题,并根据题目的要求和所提供的信息来给出最恰当的答案。如果信息不足以确切回答,那么请依据现有的数据和你的推理能力来填写最合理的答案。",
],
"multi_choice_example_format": ["问题:{}\n选项:\n{}\n正确答案:\n"],
"T/F_example_format": ["问题:{}\n正确答案:\n"],
"short_ans_example_format": ["问题:{}\n正确答案:\n"],
}
def construct_prompt(sample):
question = sample["question"]
task_instructions = PROMPT["task_instructions"]
if sample["type"] == "选择":
formatted_options = ""
start_chr = "A"
for i in range(1, 5):
formatted_options += f"({start_chr}) {sample[f'option{i}']}\n"
start_chr = chr(ord(start_chr) + 1)
current_example_template = PROMPT["multi_choice_example_format"][0]
current_example = current_example_template.format(question, formatted_options)
final_input_prompt = task_instructions[0] + "\n\n" + current_example
elif sample["type"] == "判断":
current_example_template = PROMPT["T/F_example_format"][0]
current_example = current_example_template.format(question)
final_input_prompt = task_instructions[1] + "\n\n" + current_example
else: # For fill in the blanks questions.
current_example_template = PROMPT["short_ans_example_format"][0]
current_example = current_example_template.format(question)
final_input_prompt = task_instructions[2] + "\n\n" + current_example
for i,img in enumerate(sample["img_list"]):
final_input_prompt = final_input_prompt.replace(img, f"图片 {i}")
return final_input_prompt
以上代码中函数construct_prompt(sample)需要传的sample为读取的每一行内容。
第三步: 获取到LLM返回的纯文本的结果后,需要通过如下进行归一化,这个的确比较麻烦。
import json
import os
import random
import re
from collections import Counter, defaultdict
from loguru import logger as eval_logger
def load_cmmmu(cmmmu_dir):
"""
读取Json文件
:param table_vqa_dir:
:param split:
:return:
"""
dataset = [json.loads(value.strip()) for value in
open(os.path.join(cmmmu_dir)).readlines()]
if "answer" not in dataset[0]:
return None
else:
return dataset
def cmmmu_process_results(doc, results):
pred = results[0]
if doc["type"] == "选择":
index2ans, all_choices = get_multi_choice_info([doc[f"option{i}"] for i in range(1, 5)])
parsed_pred = get_multi_choice_prediction(pred, all_choices, index2ans)
elif doc["type"] == "判断":
parsed_pred = get_TF_prediction(pred)
else:
parsed_pred = get_fill_blank_prediction(pred, doc["answer"])
return {"cmmmu_acc": {"id": doc["id"], "subdomain": doc["subcategory"], "question_type": doc["type"], "answer": doc["answer"], "parsed_pred": parsed_pred}}
def get_multi_choice_info(options):
start_chr = "A"
all_choices = []
index2ans = {}
for i, option in enumerate(options):
index2ans[chr(ord(start_chr) + i)] = option
all_choices.append(chr(ord(start_chr) + i))
return index2ans, all_choices
def get_multi_choice_prediction(response, all_choices, index2ans):
for char in [",", ".", "!", "?", ";", ":", "'"]:
response = response.strip(char)
response = " " + response + " " # add space to avoid partial match
candidates = []
for choice in all_choices: # (A) (B) (C) (D)
# Add the choice to candidates each time it appears in the response
candidates.extend([choice for _ in range(response.count(f"({choice})"))])
if len(candidates) == 0:
for choice in all_choices: # A B C D
# Similarly, add the choice for each occurrence
candidates.extend([choice for _ in range(response.count(f"{choice}"))])
if len(candidates) == 0 and len(response.split()) >= 1:
for index, ans in index2ans.items():
# Add index for each occurrence of ans in response
candidates.extend([index for _ in range(response.count(ans))])
# if all above doesn't get candidates, check if the content is larger than 5 tokens and try to parse the example
if len(candidates) == 0 and len(response.split()) >= 1:
for index, ans in index2ans.items():
if ans in response:
candidates.append(index)
index_ans = False # it's content ans.
if len(candidates) == 0: # still not get answer, randomly choose one.
return random.choice(all_choices)
# return ''
else:
# Count the occurrence of each candidate
candidate_counts = Counter(candidates)
# Select the most frequent candidates
max_count = max(candidate_counts.values())
most_frequent_candidates = [c for c in all_choices if candidate_counts.get(c, 0) == max_count]
# Combine the most frequent candidates in ABCD order
return "".join(most_frequent_candidates)
def get_fill_blank_prediction(response, answer):
"""get the prediction from the generated response,
return a list of predicted strings or numbers"""
def get_key_subresponses(response):
key_responses = []
response = response.strip("。").strip()
sub_responses = re.split(r"。|\n", response)
indicators_of_keys = ["是", "为", "所以", "等于", "方案", "选择", "正确答案", "因此", "最后", "答案", "结果"]
key_responses = []
for index, resp in enumerate(sub_responses):
# if last one, accept it's an equation (the entire response can be just one sentence with equation)
if index == len(sub_responses) - 1:
indicators_of_keys.extend(["="])
shortest_key_response = None # the shortest response that may contain the answer (tail part of the response)
for indicator in indicators_of_keys:
if indicator in resp:
if not shortest_key_response:
shortest_key_response = resp.split(indicator)[-1].strip()
else:
if len(resp.split(indicator)[-1].strip()) < len(shortest_key_response):
shortest_key_response = resp.split(indicator)[-1].strip()
if shortest_key_response:
# and it's not trivial
if shortest_key_response.strip() not in [":", ",", ".", "!", "?", ";", ":", "'"]:
key_responses.append(shortest_key_response)
if len(key_responses) == 0: # did not found any
return [response]
return key_responses
key_responses = get_key_subresponses(response)
pred_list = key_responses.copy() # keep the original string response
for resp in key_responses:
pred_list.extend(extract_numbers(resp))
tmp_pred_list = []
for i in range(len(pred_list)):
tmp_pred_list.extend(normalize_str(pred_list[i], answer))
pred_list = tmp_pred_list
# remove duplicates
pred_list = list(set(pred_list))
return pred_list
def extract_numbers(string):
# Pattern for numbers with Chinese commas
pattern_commas = r"-?\d{1,3}(?:,\d{3})+"
# Pattern for scientific notation
pattern_scientific = r"-?\d+(?:\.\d+)?[eE][+-]?\d+"
# Pattern for simple numbers without Chinese commas
pattern_simple = r"-?(?:\d+\.\d+|\.\d+|\d+)(?![eE][+-]?\d+)(?!,\d)"
# Extract numbers with Chinese commas
numbers_with_commas = re.findall(pattern_commas, string)
# Extract numbers in scientific notation
numbers_scientific = re.findall(pattern_scientific, string)
# Extract simple numbers without Chinese commas
numbers_simple = re.findall(pattern_simple, string)
# Combine all extracted numbers
all_numbers = numbers_with_commas + numbers_scientific + numbers_simple
return all_numbers
def normalize_str(string, answer):
# check if characters in the string
# if number, numerize it.
if string == None:
return [string]
string = string.strip()
is_number = check_is_number(string)
if is_number:
string = string.replace(",", "")
string = float(string)
# leave 2 decimal
string = round(string, 2)
return [string]
else: # it's likely to be a string
if len(string) > len(answer) + 20 or count_letters(string) > count_letters(answer) + 2:
return []
return [string]
def check_is_number(string):
try:
float(string.replace(",", ""))
return True
except ValueError:
# check if there's comma inside
return False
def get_TF_prediction(response):
"""get the prediction from the generated response,
return a list of predicted strings or numbers"""
def get_key_subresponses(response):
key_responses = []
response = response.strip("。").strip()
sub_responses = re.split(r"。|\n", response)
indicators_of_keys = ["是", "为", "所以", "判断", "陈述", "说法", "表达", "答案", "结果"]
key_responses = []
for index, resp in enumerate(sub_responses):
shortest_key_response = None # the shortest response that may contain the answer (tail part of the response)
for indicator in indicators_of_keys:
if indicator in resp:
if not shortest_key_response:
shortest_key_response = resp.split(indicator)[-1].strip()
else:
if len(resp.split(indicator)[-1].strip()) < len(shortest_key_response):
shortest_key_response = resp.split(indicator)[-1].strip()
if shortest_key_response:
# and it's not trivial
if shortest_key_response.strip() not in [":", ",", ".", "!", "?", ";", ":", "'"]:
key_responses.append(shortest_key_response)
if len(key_responses) == 0: # did not found any
return [response]
return key_responses
key_responses = get_key_subresponses(response)
pred_list = key_responses.copy() # keep the original string response
# remove duplicates
pred_list = list(set(pred_list))
return pred_list
def count_letters(string):
return sum(c.isalpha() and "a" <= c <= "z" or "A" <= c <= "Z" for c in string)
这里:res=cmmmu_process_results(doc, results),doc为读取原始文件的每一行,results为["LLM返回的纯文本结果"]
第四步:然后就是量化
import json
import os
import random
import re
from collections import Counter, defaultdict
from loguru import logger as eval_logger
from normalization_cmmmu import normalize_str
DOMAIN_CAT2SUB_CAT = {
"艺术与设计": ["艺术", "艺术理论", "设计", "音乐"],
"商业": ["会计", "经济", "金融", "管理", "营销"],
"科学": ["生物", "化学", "地理", "数学", "物理"],
"健康与医学": ["基础医学", "临床医学", "诊断学与实验室医学", "制药", "公共卫生"],
"人文社会科学": ["历史", "文献学", "社会学", "心理学"],
"技术与工程": ["农业", "建筑学", "计算机科学", "电子学", "能源和电力", "材料", "机械工程"],
}
def calculate_ins_level_acc(results):
correct_sum = 0
entries_sum = 0
for cat_results in results.values():
correct_sum += cat_results["correct_num"]
entries_sum += cat_results["entries_num"]
if entries_sum == 0:
return 0
return correct_sum / entries_sum
def cmmmu_aggregate_results(results):
evaluation_result = {}
subset_to_eval_samples = defaultdict(list)
for result in results:
subset_to_eval_samples[result["subdomain"]].append(result)
for subset, sub_eval_samples in subset_to_eval_samples.items():
metric_dict = eval_cmmmu(sub_eval_samples)
evaluation_result[subset] = metric_dict
printable_results = {}
for domain, in_domain_cats in DOMAIN_CAT2SUB_CAT.items():
in_domain_cat_results = {}
for cat_name in in_domain_cats:
if cat_name in evaluation_result.keys():
in_domain_cat_results[cat_name] = evaluation_result[cat_name]
else:
pass
in_domain_ins_acc = calculate_ins_level_acc(in_domain_cat_results)
in_domain_data_num = sum([cat_results["entries_num"] for cat_results in in_domain_cat_results.values()])
printable_results["Overall-" + domain] = {
"num": int(in_domain_data_num),
"acc": round(in_domain_ins_acc, 3),
}
# add sub category
for cat_name, cat_results in in_domain_cat_results.items():
printable_results[cat_name] = {
"num": int(cat_results["entries_num"]),
"acc": round(cat_results["acc"], 3),
}
all_ins_acc = calculate_ins_level_acc(evaluation_result)
printable_results["Overall"] = {
"num": sum([cat_results["entries_num"] for cat_results in evaluation_result.values()]),
"acc": round(all_ins_acc, 3),
}
print(printable_results)
return printable_results["Overall"]["acc"]
def eval_cmmmu(entries):
correct_cnt = 0
for entry in entries:
parsed_pred = entry.get("parsed_pred", "")
correct = False
if entry.get("question_type") == "选择":
if parsed_pred == entry["answer"]:
correct_cnt += 1
correct = True
elif entry.get("question_type") == "填空":
norm_answers = normalize_str(entry["answer"], entry["answer"])
for pred in parsed_pred:
# already normalized
if isinstance(pred, str): # if it's a string, then find if ans in the pred_i
for norm_ans in norm_answers:
# only see if the string answer in the string pred
# print(norm_ans, pred)
if isinstance(norm_ans, str) and norm_ans in pred:
if not correct:
correct_cnt += 1
correct = True
break
else: # it's a number
if pred in norm_answers:
if not correct:
correct_cnt += 1
correct = True
break
else:
positive_keywords = ["正确", "对", "准确", "肯定", "对的"]
negative_keywords = ["不对", "错误", "不正确", "不准确", "不合适", "否定", "错的", "错"]
ambiguous_keywords = ["对错", "是否正确", "否正确", "或者", "是否", "正确性", "对不"]
def judge_similarity(pred_list, positive_keywords, negative_keywords):
positive_count = 0
negative_count = 0
for pred in pred_list:
if any(pos_word in pred for pos_word in positive_keywords):
positive_count += 1
elif any(neg_word in pred for neg_word in negative_keywords):
negative_count += 1
if positive_count > negative_count:
return "对"
elif negative_count > positive_count:
return "错"
else:
return random.choice(["对", "错"])
answer = entry["answer"]
parsed_pred = [word for word in parsed_pred if not any(ambiguous in word for ambiguous in ambiguous_keywords)]
result = judge_similarity(parsed_pred, positive_keywords, negative_keywords)
if result == answer:
correct_cnt += 1
correct = True
if correct:
entry["judge"] = "正确"
else:
entry["judge"] = "错误"
if len(entries) == 0:
print("entries_num == 0, please check your file")
results_count = {"correct_num": 0, "entries_num": 0, "acc": 0}
else:
results_count = {"correct_num": correct_cnt, "entries_num": len(entries), "acc": correct_cnt / len(entries)}
return results_count
if __name__ == '__main__':
tmp1={'id': 11999, 'subdomain': '营销', 'question_type': '选择', 'answer': 'B', 'parsed_pred': 'B'}
tmp2={'id': 9262, 'subdomain': '营销', 'question_type': '选择', 'answer': 'B', 'parsed_pred': 'B'}
results=list()
results.append(tmp1)
results.append(tmp2)
res=cmmmu_aggregate_results(results)
print(res)
hi,cmmmu官网上提到vlmevalkit已经支持了CMMMU,但支持的配置集中暂未列出改配置,请问下是什么情况呀
我已经在这里实现了哈,你可以参考这个。
hi,cmmmu官网上提到vlmevalkit已经支持了CMMMU,但支持的配置集中暂未列出改配置,请问下是什么情况呀