Replace column values of sliced rows with None in Pandas Dataframe

Question

I have the following df:

colA colB colC
12    33  66
13    35  67
14    44  77 
15    55  79
18    56  81

I would like to replace the values of colB and colC with None starting from index 2 all the way to the end of df. The expected output is:

colA colB   colC
12    33    66
13    35    67
14    None  None 
15    None  None
18    None  None

Answer 1

Use DataFrame.loc with any index and columns names in list:

df.loc[df.index[2:], ['colB','colC']] = None

If there is default RangeIndex use 2: :

df.loc[2:, ['colB','colC']] = None
print (df)
   colA  colB  colC
0    12  33.0  66.0
1    13  35.0  67.0
2    14   NaN   NaN
3    15   NaN   NaN
4    18   NaN   NaN

Because numeric values are None s converted to NaN s.

If need integers with missing values useInt64 :

df[['colB','colC']] = df[['colB','colC']].astype('Int64')
print (df)

   colA  colB  colC
0    12    33    66
1    13    35    67
2    14  <NA>  <NA>
3    15  <NA>  <NA>
4    18  <NA>  <NA>

Answer 2

You can do something like this -

df.loc[2:, "colB":] = None

Basically using the loc method to select the rows starting from index 2 and the columns colB and colC, and then assign the value None to them. This will replace the values of colB and colC with None starting from index 2.

Answer 3

Apart from pandas.DataFrame.loc ( that jezrael's mentions ), one can use pandas.DataFrame.iloc as follows

df.iloc[2:, 1:] = None

[Out]:

   colA  colB  colC
0    12  33.0  66.0
1    13  35.0  67.0
2    14   NaN   NaN
3    15   NaN   NaN
4    18   NaN   NaN

Note that colB and colC are floats, because NaN is a float. If one doesn't want those columns to be float64 , one approach would be to use pandas.Int64Dtype as follows

df[['colB', 'colC']] = df[['colB', 'colC']].astype(pd.Int64Dtype())
  
[Out]:

   colA  colB  colC
0    12    33    66
1    13    35    67
2    14  <NA>  <NA>
3    15  <NA>  <NA>
4    18  <NA>  <NA>