Pandas Lambda Function with Nan Support

Within pandas 0.24.2, I use

df.apply(lambda x: x['col_name'] if x[col1] is np.nan else expressions_another, axis=1)

because pd.isnull() doesn't work.

in my work,I found the following phenomenon,

No running results:

df['prop'] = df.apply(lambda x: (x['buynumpday'] / x['cnumpday']) if pd.isnull(x['cnumpday']) else np.nan, axis=1)

Results exist:

df['prop'] = df.apply(lambda x: (x['buynumpday'] / x['cnumpday']) if x['cnumpday'] is not np.nan else np.nan, axis=1)

So far, I still don't know the deeper reason, but I have these experiences, for object, use [is np.nan()] or pd.isna(). For a float, use np.isnan() or pd.isna().

You need to use np.nan()

#import numpy as np
df2=df.apply(lambda x: 2 if np.isnan(x['Col1']) else 1, axis=1)   

df2
Out[1307]: 
0    1
1    1
2    1
3    2
dtype: int64

You need pandas.isnull for check if scalar is NaN:

df = pd.DataFrame({ 'Col1' : [1,2,3,np.NaN],
                 'Col2' : [8,9,7,10]})  

df2 = df.apply(lambda x: x['Col2'] if pd.isnull(x['Col1']) else x['Col1'], axis=1)

print (df)
   Col1  Col2
0   1.0     8
1   2.0     9
2   3.0     7
3   NaN    10

print (df2)
0     1.0
1     2.0
2     3.0
3    10.0
dtype: float64

But better is use Series.combine_first:

df['Col1'] = df['Col1'].combine_first(df['Col2'])

print (df)
   Col1  Col2
0   1.0     8
1   2.0     9
2   3.0     7
3  10.0    10

Another solution with Series.update:

df['Col1'].update(df['Col2'])
print (df)
   Col1  Col2
0   8.0     8
1   9.0     9
2   7.0     7
3  10.0    10

Assuming that you do have a second column, that is:

df = pd.DataFrame({ 'Col1' : [1,2,3,np.NaN], 'Col2': [1,2,3,4]})

The correct solution to this problem would be:

df['Col1'].fillna(df['Col2'], inplace=True)