Python pandas apply function if a column value is not NULL

Question

I have a dataframe (in Python 2.7, pandas 0.15.0):

df=        A    B               C 0    NaN   11             NaN 1    two  NaN  ['foo', 'bar'] 2  three   33             NaN 

I want to apply a simple function for rows that does not contain NULL values in a specific column. My function is as simple as possible:

def my_func(row):     print row 

And my apply code is the following:

df[['A','B']].apply(lambda x: my_func(x) if(pd.notnull(x[0])) else x, axis = 1) 

It works perfectly. If I want to check column 'B' for NULL values the pd.notnull() works perfectly as well. But if I select column 'C' that contains list objects:

df[['A','C']].apply(lambda x: my_func(x) if(pd.notnull(x[1])) else x, axis = 1) 

then I get the following error message: ValueError: ('The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()', u'occurred at index 1')

Does anybody know why pd.notnull() works only for integer and string columns but not for 'list columns'?

And is there a nicer way to check for NULL values in column 'C' instead of this:

df[['A','C']].apply(lambda x: my_func(x) if(str(x[1]) != 'nan') else x, axis = 1) 

Thank you!

Answer 1

The problem is that pd.notnull(['foo', 'bar']) operates elementwise and returns array([ True, True], dtype=bool). Your if condition trys to convert that to a boolean, and that's when you get the exception.

To fix it, you could simply wrap the isnull statement with np.all:

df[['A','C']].apply(lambda x: my_func(x) if(np.all(pd.notnull(x[1]))) else x, axis = 1) 

Now you'll see that np.all(pd.notnull(['foo', 'bar'])) is indeed True.

Answer 2

I had a column contained lists and NaNs. So, the next one worked for me.

df.C.map(lambda x: my_func(x) if type(x) == list else x)