Replace column values of sliced rows with None in Pandas Dataframe-CodePudding

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

CodePudding user response：

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 Nones converted to NaNs.

If need integers with missing values use Int64:

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>

CodePudding user response：

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.

CodePudding user response：

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>