Python - are there other ways to apply a function and filter in a list comprehension?
list-comprehension, python
Solution
A generator expression:
cleaned = [i for i in (word.strip() for word in words) if i]
Using `filter()` and `map()`:
cleaned = filter(None, map(str.strip, words))
The latter produces a generator in Python 3; apply `list()` to it or combine `map()` with a list comprehension:
cleaned = [i for i in map(str.strip, words) if i]
Problem
this has been irking me for years. given I have a list of words : ``` words = [ 'one', 'two', 'three', '', ' four', 'five ', 'six', \ 'seven', 'eight ', ' nine', 'ten', ''] ``` even though it's super lightweight, I still feel weird writing this list comprehension: ``` cleaned = [ i.strip() for i in words if i.strip() ] ``` i don't like applying strip() twice. it just seems silly. it's slightly/negligibly faster like this: ``` _words = [ w.strip() for w in words ] cleaned = [ w for w in _words if w ] ``` which is also the same as ``` cleaned = [ i for i in [ w.strip() for w in words ] if i ] ``` I'm wondering if there are other ways to write this. I was largely interested in a nested loops form of list comprehensions ( see Idiom for flattening a shallow nested list: how does it work? ) , but I couldn't figure anything out. update I put benchmark up on github, outlining my original 3 approaches, and ones shared below. - https://gist.github.com/jvanasco/8793879 The fastest is @Martijn Pieters `filter()`; converting the inner list to a generator expression is a negligible hit to speed, but should be better for memory management (according to python's docs ). All the speed differences involved are , expectedly, negligible and not worth sharing.