如果缺少另一列，则替换一列中的值

Question

DATA

I have a dataframe called data looks like following:

Name              ID
JAMES             252
STEPHEN           578
JOY               nan
ROGELIO           473
FACS              nan
CLIFFORD          793

data['Name'] is a column of strings, and data['ID'] has numeric values.

GOAL

I want to replace data['Name'] with missing value NaN whenever data['ID'] is missing, ie nan.

The result would be:

Name              ID
JAMES             252
STEPHEN           578
NaN               nan
ROGELIO           473
NaN               nan
CLIFFORD          793

I have searched online but similar answers are all about using fillna() which is not what I want. Do you have any suggestions on how to do this?

Answer 1

You can use .loc function to find all the index's where df['ID'] is null and set df['NAME'] as np.nan there

import numpy as np

df.loc[df['ID'].isnull() , 'NAME'] = np.nan

Answer 2

How about this method?

import pandas as pd
import numpy as np
a = {'Name':['JAMES','STEPHEN','JOY','ROGELIO','FACS','CLIFFORD'],'ID':[252,578,np.nan,473,np.nan,793]}
df = pd.DataFrame(a)

df.loc[df['ID'].isnull() , 'Name'] = np.nan
print(df)

Output:

       Name     ID
0     JAMES  252.0
1   STEPHEN  578.0
2       NaN    NaN
3   ROGELIO  473.0
4       NaN    NaN
5  CLIFFORD  793.0

If you wish to drop the NaN values, add the following:

df = df.dropna(how='any')
print(df)

Output:

       Name     ID
0     JAMES  252.0
1   STEPHEN  578.0
3   ROGELIO  473.0
5  CLIFFORD  793.0

Edit: I did the other way around, now it's correct.

Answer 3

pandas.DataFrame.mask is perfect for this :

df.mask(df['ID'].isnull())

Output:

       Name     ID
0     JAMES  252.0
1   STEPHEN  578.0
2       NaN    NaN
3   ROGELIO  473.0
4       NaN    NaN
5  CLIFFORD  793.0