Python 如何将数据帧分割成块，并在每次索引重新开始时水平合并

我有一个数据帧，实际上是垂直连接的多个数据帧。我想横向合并它们，但按索引值拆分时遇到问题。我希望在索引为“finish”的地方开始一个新的块，我希望避免手动执行此操作，因为实际的数据帧大约有20个部分，每个部分的长度都不相同

这是原始数据帧

        12_boys  12_girls
finish                   
1        Team A    Team A
2        Team B    Team B
3        Team C    Team C
4        Team D    Team D
5        Team E    Team E
finish  14_boys  14_girls
1        Team A    Team A
2        Team B    Team B
3        Team C    Team C
4        Team D    Team D
finish  16_boys  16_girls
1        Team A    Team A
2        Team B    Team B
3        Team C    Team C
4        Team D    Team D

这是我想要的

       12_boys 14_boys 16_boys 12_girls 14_girls 16_girls
finish                                                   
1       Team A  Team A  Team A   Team A   Team A   Team A
2       Team B  Team B  Team B   Team B   Team B   Team B
3       Team C  Team C  Team C   Team C   Team C   Team C
4       Team D  Team D  Team D   Team D   Team D   Team D
5       Team E     NaN     NaN   Team E      NaN      NaN

我能想到的最接近的方法是手动拆分和合并，但这不会转换列名

data1 = data.iloc[0:6]
data2 = data.iloc[6:10]
data3 = data.iloc[11:15]
data_merge = pd.merge(data1, data2, on='finish', how='outer')
data_merge = pd.merge(data_merge, data3, on='finish', how='outer')

Output:

    12_boys_x   12_girls_x  12_boys_y   12_girls_y  12_boys     12_girls
finish                      
1   Team A  Team A  Team A  Team A  Team A  Team A
2   Team B  Team B  Team B  Team B  Team B  Team B
3   Team C  Team C  Team C  Team C  Team C  Team C
4   Team D  Team D  Team D  Team D  Team D  Team D
5   Team E  Team E  NaN     NaN     NaN     NaN
finish  14_boys     14_girls    NaN     NaN     NaN     NaN

---

我们可以首先通过检查索引更改为

finish

的位置来识别组。然后，我们使用

GroupBy

和

pd.concat

将每组紧靠在一起：

grps = (df.index == 'finish').cumsum()

dfs = []
for grp, dfg in df.groupby(grps):
    if grp != 0:
        dfg.columns = dfg.iloc[0].values
        dfs.append(dfg)
    else: dfs.append(dfg)

df_new = pd.concat(dfs, axis=1, sort=False).iloc[:-1]

  12_boys 12_girls 14_boys 14_girls 16_boys 16_girls
1  Team A   Team A  Team A   Team A  Team A   Team A
2  Team B   Team B  Team B   Team B  Team B   Team B
3  Team C   Team C  Team C   Team C  Team C   Team C
4  Team D   Team D  Team D   Team D  Team D   Team D
5  Team E   Team E     NaN      NaN     NaN      NaN