Split by comma and strip whitespace in Python
Use list comprehension -- simpler, and just as easy to read as a
my_string = "blah, lots , of , spaces, here "result = [x.strip() for x in my_string.split(',')]# result is ["blah", "lots", "of", "spaces", "here"]
I came to add:
but saw it had already been mentioned by Jason Orendorff in a comment.
Reading Glenn Maynard's comment in the same answer suggesting list comprehensions over map I started to wonder why. I assumed he meant for performance reasons, but of course he might have meant for stylistic reasons, or something else (Glenn?).
So a quick (possibly flawed?) test on my box applying the three methods in a loop revealed:
[word.strip() for word in string.split(',')]$ time ./list_comprehension.py real 0m22.876smap(lambda s: s.strip(), string.split(','))$ time ./map_with_lambda.py real 0m25.736smap(str.strip, string.split(','))$ time ./map_with_str.strip.py real 0m19.428s
map(str.strip, string.split(',')) the winner, although it seems they are all in the same ballpark.
Certainly though map (with or without a lambda) should not necessarily be ruled out for performance reasons, and for me it is at least as clear as a list comprehension.
Python 2.6.5 on Ubuntu 10.04
Split using a regular expression. Note I made the case more general with leading spaces. The list comprehension is to remove the null strings at the front and back.
import re string = " blah, lots , of , spaces, here "pattern = re.compile("^\s+|\s*,\s*|\s+$")print([x for x in pattern.split(string) if x])['blah', 'lots', 'of', 'spaces', 'here']
This works even if
^\s+ doesn't match:
"foo, bar "print([x for x in pattern.split(string) if x])['foo', 'bar']>>>string =
Here's why you need ^\s+:
compile("\s*,\s*|\s+$")print([x for x in pattern.split(string) if x])[' blah', 'lots', 'of', 'spaces', 'here']pattern = re.
See the leading spaces in blah?
Clarification: above uses the Python 3 interpreter, but results are the same in Python 2.