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I want to fill the NaN values on my dataframe on column c with the mean for only rows who has as category B, and ignore the others.

print (df) Category b c 0 A 1 5.0 1 C 1 NaN 2 A 1 4.0 3 B 2 NaN 4 A 2 1.0 5 B 2 Nan 6 C 1 3.0 7 C 1 2.0 8 B 1 NaN

So what I'm doing for the moment is :

df.c = df.c.fillna(df.c.mean())

But it fill all the NaN values, while I want only to fill the 3rd, 5th and the 8th rows who had category value equal to B.

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2 Answers 2

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Combine fillna with slicing assignment

df.loc[df.Category.eq('B'), 'c'] = (df.loc[df.Category.eq('B'), 'c']. fillna(df.c.mean())) Out[736]: Category b c 0 A 1 5.0 1 C 1 NaN 2 A 1 4.0 3 B 2 3.0 4 A 2 1.0 5 B 2 3.0 6 C 1 3.0 7 C 1 2.0 8 B 1 3.0

Or a direct assignment with 2 masks

df.loc[df.Category.eq('B') & df.c.isna(), 'c'] = df.c.mean() Out[745]: Category b c 0 A 1 5.0 1 C 1 NaN 2 A 1 4.0 3 B 2 3.0 4 A 2 1.0 5 B 2 3.0 6 C 1 3.0 7 C 1 2.0 8 B 1 3.0
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2 Comments

Thank you @Andy for the answer that was perfect, beside I'd like to know which one is horizontally scalable for more data.
@saul: You are welcome. I think the 2nd is more scalable and clean :)
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This would be the answer for your question: 

df.c = df.apply( lambda row: row['c'].fillna(df.c.mean()) if row['Category']=='B' else row['c'] ,axis=1)
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