根据条件为列名添加后缀

Question

i want to add the prefix '_nan' to columns that are all nan. I have the following code that prints what I want but does not reassign the columns in the actual dataframe and I am not sure why. Does anyone have any ideas why this is happening? Thanks in advance

df = pd.DataFrame({ 'a':[1, 0, 0, 0],
                    'b':[np.nan, np.nan, np.nan, np.nan],
                    'c':[np.nan, np.nan, np.nan, np.nan]})

a = df.loc[:,df.isna().all()].columns

df[[*a]] = df[[*a]].add_suffix('_nan')

Answer 1

You can use list comprehension:

df.columns = [x + '_nan' if df[x].isna().all() else x for x in df.columns]

Output:

    a   b_nan   c_nan
0   1     NaN     NaN
1   0     NaN     NaN
2   0     NaN     NaN
3   0     NaN     NaN

Answer 2

why this is happening?

After some experiments I found that when you assign pandas.DataFrame slice to slice of pandas.DataFrame then pandas apparently does bother solely about order of columns (given in list ), not their names, consider following example:

import pandas as pd
df_1 = pd.DataFrame({'a':[1,2,3],'b':[4,5,6],'c':[7,8,9]})
df_2 = pd.DataFrame({'x':[10,20,30],'y':[400,500,600]})
df_1[['a','b']] = df_2[['x','y']]
print(df_1)

output

    a    b  c
0  10  400  7
1  20  500  8
2  30  600  9

whilst

...
df_1[['a','b']] = df_2[['y','x']]
print(df_1)

produce

     a   b  c
0  400  10  7
1  500  20  8
2  600  30  9