pandas equivalent of np.where

Solution 1:

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'])