Use apply lambda with if else conditional without computing the condition twice

Question

Is there a way to apply the lambda statement without having to compute the x.split(' ')[0] twice? I know it can be done using a function i.e. .apply(lambda x: pre_dir(x) and take care of the logic there, but wondering if it can be done in a one-liner.

address.insert(6, 'PRE_DIR', address['STREETNAME'].apply(lambda x: x.split(' ')[0] if x.split(' ')[0] in ['N', 'S', 'E', 'W'] else ''))

user7864386 · Accepted Answer · 2022-04-19 17:56:15Z

1

You could use where instead of apply. In other words, replace

address['STREETNAME'].apply(lambda x: x.split(' ')[0] if x.split(' ')[0] in ['N', 'S', 'E', 'W'] else '')

by

address['STREETNAME'].str.split(' ').str[0].where(lambda x: x.isin(['N', 'S', 'E', 'W']), '')

Example: For the following DataFrame

address = pd.DataFrame({'STREETNAME':['North Ave', '1st St', 'W 34th St']})

the above code produces the following column:

    0     
    1     
    2    W

answered Apr 19, 2022 at 17:56

user7864386

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mozway · Accepted Answer · 2022-04-19 18:03:15Z

0

with python ≥ 3.8 you can use a assignment expression:

lambda x: y if (y:=x.split(' ')[0]) in ['N', 'S', 'E', 'W'] else '')

That said, IIUC, you could use a regex here:

address['STREETNAME'].str.extract('^([NSEW]) ').fillna('')

answered Apr 19, 2022 at 18:03

mozway

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2 Comments

Infinity Cliff Over a year ago

this would not work for my situation as not all road names start with NSEW, but any road name could contain those letters

mozway Over a year ago

@InfinityCliff you mean the second approach? The first one should give the exact same result as the other answer, just avoiding to compute the split twice unnecessarily ;)

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