如何在Pandas中按条件为整个分组标记标签？

需求说明 #

当working_group分组内存在任意一行满足size >=2且process为pending时，整个分组的state列都标记为not_done；其余分组标记为something_else。

解决方案 #

原代码仅给满足条件的单行打标签，未扩展到整个分组。以下是两种高效实现方式：

方法1：先筛选目标分组再批量标记（性能更优）

先提取所有符合条件的working_group，再通过匹配分组给整组打标签：

import numpy as np
import pandas as pd

df = pd.DataFrame({"id": [1, 2, 3, 4, 5, 6, 7, 8],
                        "process": ["pending", "finished", "finished", "finished", "finished", "finished", "finished", "pending"],
                        "working_group": ["a", "a", "c", "d", "d", "f", "g", "g"],
                        "size": [2, 2, 1, 2, 2, 1, 2, 2]})

# 1. 提取所有满足条件的working_group（去重）
pending_groups = df[(df['size'] >= 2) & (df['process'] == 'pending')]['working_group'].unique()

# 2. 给整个分组标记标签
df['state'] = np.where(df['working_group'].isin(pending_groups), 'not_done', 'something_else')

方法2：分组后用transform判断（逻辑更直观）

通过groupby结合transform，直接在组内判断是否存在满足条件的行，返回同长度的结果列：

import numpy as np
import pandas as pd

df = pd.DataFrame({"id": [1, 2, 3, 4, 5, 6, 7, 8],
                        "process": ["pending", "finished", "finished", "finished", "finished", "finished", "finished", "pending"],
                        "working_group": ["a", "a", "c", "d", "d", "f", "g", "g"],
                        "size": [2, 2, 1, 2, 2, 1, 2, 2]})

# 按working_group分组，判断组内是否有满足条件的行，返回对应标签
df['state'] = df.groupby('working_group').transform(
    lambda group: 'not_done' if ((group['size'] >= 2) & (group['process'] == 'pending')).any() else 'something_else'
)['process']  # 取任意列均可，transform返回与原DataFrame同长度的Series

运行结果 #

执行上述任意一种方法后，得到的DataFrame如下：

id   process working_group  size     state
0   1   pending             a     2  not_done
1   2  finished             a     2  not_done
2   3  finished             c     1  something_else
3   4  finished             d     2  something_else
4   5  finished             d     2  something_else
5   6  finished             f     1  something_else
6   7  finished             g     2  not_done
7   8   pending             g     2  not_done

内容的提问来源于stack exchange，提问作者Googme