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Pandas 数据帧重置多索引的索引计数

[英]Pandas dataframe reset index count of multiindex

constructed a dataframe by concatenating several dataframes with the keys [a,b,c] as Index通过将几个数据帧与键 [a,b,c] 作为索引连接起来构建一个数据帧

+-------+----------+----------+
| Index | IndexPos | SomeData |
+-------+----------+----------+
| a     |        1 | some1     |
|       |        2 | some2     |
|       |        3 | some3     |
| b     |        1 | some1     |
|       |        2 | some2     |
|       |        3 | some3     |
| c     |        1 | some1     |
|       |        2 | some2     |
|       |        3 | some3     |
+-------+----------+----------+

and now want slice it down to the last 2 elements like:现在想把它切成最后 2 个元素,如:

df.groupby(df.index.levels[0].name).tail(2)

After that I want to recount the remaining elements IndexPos to get this:之后,我想重新计算剩余的元素 IndexPos 以得到这个:

+-------+----------+----------+
| Index | IndexPos | SomeData |
+-------+----------+----------+
| a     |        1 | some2     |
|       |        2 | some3     |
| b     |        1 | some2     |
|       |        2 | some3     |
| c     |        1 | some2     |
|       |        2 | some3     |
+-------+----------+----------+

Is there a way to do this, or do I have to slice it before concatenating?有没有办法做到这一点,或者我必须在连接之前将其切片?

First groupby on level=0 and get the last two rows from each group using tail , then using groupby + cumcount on sliced dataframe create a sequential counter for each group and set it as new index at level=1 :首先在level=0groupby并使用tail从每个组中获取最后两行,然后在切片数据帧上使用groupby + cumcount为每个组创建一个顺序计数器并将其设置为level=1新索引:

d = df.groupby(level=0).tail(2)
d = d.droplevel(1).set_index(d.groupby(level=0).cumcount().add(1), append=True)

Or using factorize in place of groupby + cumcount inspired by @anky's solution:或者使用factorize代替groupby + cumcount灵感来自@anky 的解决方案:

d = df.groupby(level=0).tail(2)
d = d.droplevel(1).set_index(d.index.get_level_values(1).factorize()[0] + 1, append=True)

Result:结果:

print(d)

        SomeData
Index           
a     1    some2
      2    some3
b     1    some2
      2    some3
c     1    some2
      2    some3

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