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Pandas conditional swap

I have a pandas dataframe as shown below:

Col1 Col2 Col3 0 A 7 NaN 1 B 16 NaN 1 B 16 15

What I want to do is exchange Col2 for Col3, where Col3 is NaN . Based on other posts and answers on SO, I have this code:

 df[['Col2', 'Col3']] = df[['Col3', 'Col2']].where(df[['Col3']].isnull())

But this does not seem to work properly and gives me the following:

  Col1 Col2 Col3 0 A NaN NaN 1 B NaN NaN 1 B NaN NaN

Is there something I am missing here?

Update: My desired result is as follows:

  Col1 Col2 Col3 0 A NaN 7 1 B NaN 16 1 B 16 15

thanks

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2 answers

You can use loc to exchange:

 df.loc[df['Col3'].isnull(), ['Col2', 'Col3']] = df.loc[df['Col3'].isnull(), ['Col3', 'Col2']].values

Note that .values is required to ensure that the .values is done correctly, otherwise Pandas will try to align the names of indexes and columns, and no exchange will occur.

You can also just reassign each line separately if you think the code is cleaner:

 null_idx = df['Col3'].isnull() df.loc[null_idx, 'Col3'] = df['Col2'] df.loc[null_idx, 'Col2'] = np.nan

Result:

  Col1 Col2 Col3 0 A NaN 7.0 1 B NaN 16.0 2 B 16.0 15.0
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Try the following: (faster)

 df["Col3"], df["Col2"] = np.where(df['Col3'].isnull(), [df["Col2"], df["Col3"]], [df["Col3"], df["Col2"] ]) df Col1 Col2 Col3 0 A NaN 7.0 1 B NaN 16.0 1 B 16.0 15.0 %timeit df.loc[df['Col3'].isnull(), ['Col2', 'Col3']] = df.loc[df['Col3'].isnull(), ['Col3', 'Col2']].values 100 loops, best of 3: 2.68 ms per loop %timeit df["Col3"], df["Col2"] = np.where(df['Col3'].isnull(), [df["Col2"], df["Col3"]], [df["Col3"], df["Col2"] ]) 1000 loops, best of 3: 592 µs per loop
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