why numpy place is not replacing empty strings

Question

Hello i have an dataframe as shown below

daf = pd.DataFrame({'A':[10,np.nan,20,np.nan,30]})
daf['B'] = ''

the above code has created a data frame with column B having empty strings

     A    B
0   10.0    
1   NaN 
2   20.0    
3   NaN 
4   30.0

the problem here is i need to replace column with all empty strings,(note here entire column should be empty) with values provided with numpy place last argument here it is 1

so i used following code

np.place(daf.to_numpy(),((daf[['A','B']] == '').all() & (daf[['A','B']] == '')).to_numpy(),[1])

which did nothing it gave same output

    A    B
0   10.0    
1   NaN 
2   20.0    
3   NaN 
4   30.0

but when i assign daf['B'] = np.nan the code seems to work fine by checking if entire column is null, then replace it with 1

here is the data frame

      A      B
0   10.0    NaN
1   NaN     NaN
2   20.0    NaN
3   NaN     NaN
4   30.0    NaN

    

replace where those nan with 1 where the entire column is nan

np.place(daf.to_numpy(),(daf[['A','B']].isnull() & daf[['A','B']].isnull().all()).to_numpy(),[1])

which gave correct output

    A        B
0   10.0    1.0
1   NaN     1.0
2   20.0    1.0
3   NaN     1.0
4   30.0    1.0

can some one tell me how to work with empty strings replacing , and give a reason why its not working with empty string as input

Answer 1

If I'm understanding your question correctly, you're wanting to replace a column with empty strings with a column of 1s. This can be done with pandas.replace()

daf.replace('', 1.0)

      A    B
0  10.0  1.0
1   NaN  1.0
2  20.0  1.0
3   NaN  1.0
4  30.0  1.0

This function also works with regex if you want to be more granular with the replacement.