DataFrame.loc multiple columns replace

Python 3.6.8 pandas==0.25.3

#! -*- coding:utf-8 -*-
import pandas as pd
df = pd.DataFrame([
    {'a':'a1','b':'b1','c':'c1','d':'d1','e':'e1'},
    {'a':'a2','b':'b2','c':'c2','d':'d2','e':'e2'}
    ])
import copy
df2 = copy.deepcopy(df)

 

print(df)
#     a   b   c   d   e
# 0  a1  b1  c1  d1  e1
# 1  a2  b2  c2  d2  e2


df.loc[df['a']=='a2', ['c']] = df['e']
print(df)
#     a   b   c   d   e
# 0  a1  b1  c1  d1  e1
# 1  a2  b2  e2  d2  e2

# loc muti columns replace has some problem
df.loc[df['a']=='a2', ['b','c']] = df[['d','e']]
print(df)
#     a    b    c   d   e
# 0  a1   b1   c1  d1  e1
# 1  a2  NaN  NaN  d2  e2

[phofl]

Yep I would agree with counter-intuitive. Just thought this is a nice feature, if it is not a bug:)

[Liam3851]

You’re using label-based indexing using .loc. Pandas therefore does not infer that you want to replace column b with column d and column c with column e-- that would be positional logic.

Either you can use .iloc for this use case, or else rename the columns before setting:

In [6]: df.loc[df['a']=='a2', ['b','c']] = df[['d','e']].rename(columns={'d':'b', 'e':'c'})

In [7]: df
Out[7]:
    a   b   c   d   e
0  a1  b1  c1  d1  e1
1  a2  d2  e2  d2  e2

Alternatively you can also use the setting using .to_numpy, see the warning box at https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#basics. In this case you need to do the alignment on the right hand side yourself:

In [9]: df.loc[df['a']=='a2', ['b','c']] = df.loc[df['a'] == 'a2', ['d','e']].to_numpy()

In [10]: df
Out[10]:
    a   b   c   d   e
0  a1  b1  c1  d1  e1
1  a2  d2  e2  d2  e2