Moritz :
Splitting columns by patterns is easy:
import pandas as pd
_df = pd.DataFrame([['1 / 2 / 3', '4 / 5 / 6'], ['7 / 8 / 9', '10 / 11 / 12']])
_df.apply(lambda x: x.str.split(' / '))
0 1
0 [1, 2, 3] [4, 5, 6]
1 [7, 8, 9] [10, 11, 12]
But how can I create a dataframe using expand=True
as multiindex? I do not know where I could pass the index.
_df.apply(lambda x: x.str.split(' / ', expand=True))
ValueError: If using all scalar values, you must pass an index
Expected output (names of columns are not important, can be arbitrary):
A B
a b c a b c
0 1 2 3 4 5 6
1 7 8 9 10 11 12
anky_91 :
here is one way using df.stack
and unstack
with a little help using swaplevel
:
s=_df.stack().str.split(' / ')
out = (pd.DataFrame(s.tolist(),index=s.index).unstack()
.swaplevel(axis=1).sort_index(axis=1))
0 1
0 1 2 0 1 2
0 1 2 3 4 5 6
1 7 8 9 10 11 12
To match the specific output, we can use:
from string import ascii_lowercase
out.rename(columns=dict(enumerate(ascii_lowercase)))
a b
a b c a b c
0 1 2 3 4 5 6
1 7 8 9 10 11 12
Or better yet :)
from string import ascii_lowercase, ascii_uppercase
out.rename(columns=dict(enumerate(ascii_uppercase)), level=0, inplace=True)
out.rename(columns=dict(enumerate(ascii_lowercase)), level=1, inplace=True)
print(out)
A B
a b c a b c
0 1 2 3 4 5 6
1 7 8 9 10 11 12
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