如何根据 Pandas 中另一列的值填充一列的缺失值？

Question

I have a dataframe looks like below:

import numpy as np
import pandas as pd
d = {'col1': [np.nan, 19, 32, np.nan, 54, 67], 'col2': [0, 1, 0, 1, 1, 1]}
df = pd.DataFrame(d)

I want to fill the missing values in "col1" based on the values of "col2". To be specific: I want to fill the missing values in "col1" with 0 if "col2" is 0, else leave the "col1" as it is. In this case, my output should look like:

d_updated = {'col1': [0, 19, 32, np.nan, 54, 67], 'col2': [0, 1, 0, 1, 1, 1]}
df_updated = pd.DataFrame(d_updated)

To have the above output, I try to get the index which "col2" have values equal to 0 and use fillna():

ix = list(df[df["col2"] == 0].index)
df["col2"].loc[ix].fillna(0, inplace = True)

However, my approach doesn't work and I don't know why. Thanks ahead.

Answer 1

Try, using loc with boolean indexing:

df.loc[(df['col1'].isna()) & (df['col2'] == 0), 'col1'] = df['col2']

Output:

   col1  col2
0   0.0     0
1  19.0     1
2  32.0     0
3   NaN     1
4  54.0     1
5  67.0     1

Answer 2

m=(df.col2==0 )&(df.col1.isna())#boolean select using loc

Then any of the following can do

df.loc[m,'col1']=df.loc[m,'col1'].fillna(0, inplace=True)

or
df.loc[m,'col1'] = df.loc[m,'col1'].replace('nan', np.nan).fillna(0)