Pandas replacing values on specific columns

Question

I am aware of these two similar questions:

Pandas replace values

Pandas: Replacing column values in dataframe

I used a different approach for substituting values from which I think it should be the cleanest one. But it does not work. I know how to work around it, but I would like to understand why it does not work:

In [108]: df=pd.DataFrame([[1, 2, 8],[3, 4, 8], [5, 1, 8]], columns=['A', 'B', 'C']) 

In [109]: df
Out[109]: 
   A  B  C
0  1  2  8
1  3  4  8
2  5  1  8

In [110]: df.loc[:, ['A', 'B']].replace([1, 3, 2], [3, 6, 7], inplace=True)

In [111]: df
Out[111]: 
   A  B  C
0  1  2  8
1  3  4  8
2  5  1  8

In [112]: df.loc[:, 'A'].replace([1, 3, 2], [3, 6, 7], inplace=True)

In [113]: df
Out[113]: 
   A  B  C
0  3  2  8
1  6  4  8
2  5  1  8

If I slice only one column In [112] it works different to slicing several columns In [110]. As I understand the .loc method it returns a view and not a copy. In my logic this means that making an inplace change on the slice should change the whole DataFrame. This is what happens at line In [110].

Answer 1

Here is the answer by one of the developers: https://github.com/pydata/pandas/issues/11984

This should ideally show a SettingWithCopyWarning, but I think this is quite difficult to detect.

You should NEVER do this type of chained inplace setting. It is simply bad practice.

idiomatic is:

In [7]: df[['A','B']] = df[['A','B']].replace([1, 3, 2], [3, 6, 7])

In [8]: df
Out[8]: 
   A  B  C
0  3  7  8
1  6  4  8
2  5  3  8

(you can do with df.loc[:,['A','B']] as well, but more clear as above.