Pandas - 按行交错/压缩两个DataFrame

Question

假设我有两个数据帧:

>> df1

   0  1  2
0  a  b  c
1  d  e  f

>> df2

   0  1  2
0  A  B  C
1  D  E  F

如何交错行？ 即得到这个:

>> interleaved_df

   0  1  2
0  a  b  c
1  A  B  C
2  d  e  f
3  D  E  F

(注意我的真实DF具有相同的列,但行数不同).

---

我试过的

灵感来自这个问题(非常相似,但在列上提问):

import pandas as pd
from itertools import chain, zip_longest

df1 = pd.DataFrame([['a','b','c'], ['d','e','f']])  
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']])

concat_df = pd.concat([df1,df2])

new_index = chain.from_iterable(zip_longest(df1.index, df2.index))
# new_index now holds the interleaved row indices

interleaved_df = concat_df.reindex(new_index)

ValueError: cannot reindex from a duplicate axis

最后一次调用失败,因为df1和df2有一些相同的索引值(我的真实DF也是如此).

有任何想法吗？

Answer 1

您可以在连接后对索引进行排序,然后重置索引即

import pandas as pd

df1 = pd.DataFrame([['a','b','c'], ['d','e','f']])  
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']])

concat_df = pd.concat([df1,df2]).sort_index().reset_index(drop=True)

输出:

   0  1  2
0  a  b  c
1  A  B  C
2  d  e  f
3  D  E  F

编辑(OmerB):无论索引值如何,都要保持顺序.

import pandas as pd
df1 = pd.DataFrame([['a','b','c'], ['d','e','f']]).reset_index()  
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']]).reset_index()

concat_df = pd.concat([df1,df2]).sort_index().set_index('index')

Answer 2

用 toolz.interleave

In [1024]: from toolz import interleave

In [1025]: pd.DataFrame(interleave([df1.values, df2.values]))
Out[1025]:
   0  1  2
0  a  b  c
1  A  B  C
2  d  e  f
3  D  E  F