pandas equivalent of np.where

PythonPandasNumpyWhere Clause

Python Problem Overview


np.where has the semantics of a vectorized if/else (similar to Apache Spark's when/otherwise DataFrame method). I know that I can use np.where on pandas.Series, but pandas often defines its own API to use instead of raw numpy functions, which is usually more convenient with pd.Series/pd.DataFrame.

Sure enough, I found pandas.DataFrame.where. However, at first glance, it has completely different semantics. I could not find a way to rewrite the most basic example of np.where using pandas where:

# df is pd.DataFrame
# how to write this using df.where?
df['C'] = np.where((df['A']<0) | (df['B']>0), df['A']+df['B'], df['A']/df['B'])

Am I missing something obvious? Or is pandas' where intended for a completely different use case, despite same name as np.where?

Python Solutions


Solution 1 - Python

Try:

(df['A'] + df['B']).where((df['A'] < 0) | (df['B'] > 0), df['A'] / df['B'])

The difference between the numpy where and DataFrame where is that the default values are supplied by the DataFrame that the where method is being called on (docs).

I.e.

np.where(m, A, B)

is roughly equivalent to

A.where(m, B)

If you wanted a similar call signature using pandas, you could take advantage of the way method calls work in Python:

pd.DataFrame.where(cond=(df['A'] < 0) | (df['B'] > 0), self=df['A'] + df['B'], other=df['A'] / df['B'])

or without kwargs (Note: that the positional order of arguments is different from the numpy where argument order):

pd.DataFrame.where(df['A'] + df['B'], (df['A'] < 0) | (df['B'] > 0), df['A'] / df['B'])

Solution 2 - Python

I prefer using pandas' mask over where since it is less counterintuitive (at least for me).

(df['A']/df['B']).mask(df['A']<0) | (df['B']>0), df['A']+df['B'])

Here, column A and B are added where the condition holds, otherwise their ratio stays untouched.

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