Keep NaN values using apply and lambda in python

Question

I am working with python and I have a list of value in column A including some NaN values. I need to code them as 1 and 0 depending if they are above or below the median (0.4). I've tried the code below which is working fairly well

df["A_median"] = (df["A"].apply(lambda count: 0 if count<median_a else 1))

But it is giving me the NaN values as 1 instead of keeping them.

  A   A_median
36.6  1
NaN   1
NaN   1
0.1   0

The final table instead should be as follow

  A   A_median
36.6  1
NaN   NaN
NaN   NaN
7     0

Any idea on how I can fix this?

Thanks

Quang Hoang · Accepted Answer · 2020-06-05 20:00:11Z

3

Is it:

median_a = 0.4
df['A_median'] = (df['A'] > median_a).where(df.A.notna())

Output:

      A  A_median
0  36.6       1.0
1   NaN       NaN
2   NaN       NaN
3   0.1       0.0

answered Jun 5, 2020 at 20:00

Quang Hoang

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