如何实现函数检测Counter中元素占比是否相近？

直接用以下函数实现需求，核心思路是通过对比占比的最大值与最小值的绝对差，结合math.isclose的abs_tol参数判断所有元素占比是否相近：

import math
from collections import Counter

def are_labels_balanced(counter, abs_tol=2):
    total = sum(counter.values())
    if total == 0:
        return True  # 空数据集默认平衡，可根据实际需求调整
    percents = [count / total * 100 for count in counter.values()]
    if len(percents) <= 1:
        return True
    max_p = max(percents)
    min_p = min(percents)
    return math.isclose(max_p, min_p, abs_tol=abs_tol)

测试示例 #

# 原示例：占比60%、20%、20%，不相近
c1 = Counter(["Dog", "Cat", "Dog", "Horse", "Dog"])
print(are_labels_balanced(c1))  # 输出: False

# 平衡示例：三类各占约33.33%
c2 = Counter(["Dog","Cat","Horse","Dog","Cat","Horse"])
print(are_labels_balanced(c2))  # 输出: True

# 两类各占50%，完全平衡
c3 = Counter(["A","A","B","B"])
print(are_labels_balanced(c3))  # 输出: True

# 两类占比60%、40%，差值超2%
c4 = Counter(["A","A","A","B","B"])
print(are_labels_balanced(c4))  # 输出: False

逻辑说明 #

1. 先计算总样本数，避免除以0的异常情况
2. 生成每个元素的占比百分比列表
3. 若只有1类标签或无样本，直接判定为平衡
4. 通过对比占比的最大值和最小值，利用math.isclose判断两者的绝对差是否在允许的公差范围内——只要最大最小占比满足条件，所有元素的占比必然彼此相近

内容的提问来源于stack exchange，提问作者Anna