如何使用与pandas.groupby方法相反的方法将熊猫排捆在一起

IIUC,你可以组成两个连续的小组 cumcount :

# start new group on identical values
g1 = df['court'].eq(df['court'].shift()).cumsum()
# in the previous groups, restart if the value was already seen
g2 = df['court'].groupby([df['court'], g1]).cumcount()

df['group'] = df.groupby([g1, g2]).ngroup()

# variant
# df['group'] = (g1.diff().ne(0)|g2.diff().ne(0)).cumsum().sub(1)

输出

   match_id  court  group
0  50311513      1      0
1  50313011      2      0
2  50313009      2      1
3  50317691      1      1
4  50315247      2      2
5  50318597      1      2
6  50318877      1      3
7  50318983      1      4
8  50318831      1      5
9  50318595      1      6

中间体:

   match_id  court  g1  g2  group
0  50311513      1   0   0      0
1  50313011      2   0   0      0
2  50313009      2   1   0      1
3  50317691      1   1   0      1
4  50315247      2   1   1      2
5  50318597      1   1   1      2
6  50318877      1   2   0      3
7  50318983      1   3   0      4
8  50318831      1   4   0      5
9  50318595      1   5   0      6

一般化

如果您需要处理更多的组,您可以使用一个自定义(迭代)函数,该函数带有一个集合来跟踪已经看到的项目。如果过去看到的项目集已经包含当前项目,则启动一个新组:

def grouper(s):
    S = set()
    group = []
    n = 0
    for val in s:
        if val in S:
            n += 1
            S = {val}
        else:
            S.add(val)
        group.append(n)
    return group

df['group'] = grouper(df['court'])

实例

    match_id  court  group
0   50311513      1      0
1   50313011      2      0
2   50313012      3      0
3   50313009      2      1
4   50317691      1      1
5   50315247      2      2
6   50315248      3      2
7   50318597      1      2
8   50318877      1      3
9   50318983      1      4
10  50319873      3      4
11  50318831      1      5
12  50318595      1      6