Remove trailing newline from the elements of a string list
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Track title: Darkness Approaches Looping
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Chapters
00:00 Remove Trailing Newline From The Elements Of A String List
00:28 Accepted Answer Score 297
00:53 Answer 2 Score 130
01:03 Answer 3 Score 70
01:18 Answer 4 Score 8
01:30 Answer 5 Score 3
02:28 Thank you
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Full question
https://stackoverflow.com/questions/7984...
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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
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Tags
#python #list #strip
#avk47
    Rise to the top 3% as a developer or hire one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------
Music by Eric Matyas
https://www.soundimage.org
Track title: Darkness Approaches Looping
--
Chapters
00:00 Remove Trailing Newline From The Elements Of A String List
00:28 Accepted Answer Score 297
00:53 Answer 2 Score 130
01:03 Answer 3 Score 70
01:18 Answer 4 Score 8
01:30 Answer 5 Score 3
02:28 Thank you
--
Full question
https://stackoverflow.com/questions/7984...
--
Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...
--
Tags
#python #list #strip
#avk47
ACCEPTED ANSWER
Score 297
You can either use a list comprehension
my_list = ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']
stripped = [s.strip() for s in my_list]
or alternatively use map():
stripped = list(map(str.strip, my_list))
In Python 2, map() directly returned a list, so you didn't need the call to list. In Python 3, the list comprehension is more concise and generally considered more idiomatic.
ANSWER 2
Score 130
list comprehension?
[x.strip() for x in lst]
ANSWER 3
Score 70
You can use lists comprehensions:
strip_list = [item.strip() for item in lines]
Or the map function:
# with a lambda
strip_list = map(lambda it: it.strip(), lines)
# without a lambda
strip_list = map(str.strip, lines)
ANSWER 4
Score 8
This can be done using list comprehensions as defined in PEP 202
[w.strip() for w in  ['this\n', 'is\n', 'a\n', 'list\n', 'of\n', 'words\n']]