本文翻譯自：Split by comma and strip whitespace in Python
I have some python code that splits on comma, but doesn't strip the whitespace: 我有一些在逗號處分割的python代碼，但沒有去除空格：
>>> string = "blah, lots , of , spaces, here "
>>> mylist = string.split(',')
>>> print mylist
['blah', ' lots ', ' of ', ' spaces', ' here ']
I would rather end up with whitespace removed like this: 我寧願這樣刪除空格：
['blah', 'lots', 'of', 'spaces', 'here']
I am aware that I could loop through the list and strip() each item but, as this is Python, I'm guessing there's a quicker, easier and more elegant way of doing it. 我知道我可以遍歷list和strip（）每個項目，但是，因爲這是Python，所以我猜有一種更快，更輕鬆和更優雅的方法。
#1樓
參考：https://stackoom.com/question/H59g/用逗號分割並在Python中去除空格
#2樓
Split using a regular expression. 使用正則表達式拆分。 Note I made the case more general with leading spaces. 注意我用前導空格使情況更一般。 The list comprehension is to remove the null strings at the front and back. 列表理解是刪除前面和後面的空字符串。
>>> import re
>>> string = " blah, lots , of , spaces, here "
>>> pattern = re.compile("^\s+|\s*,\s*|\s+$")
>>> print([x for x in pattern.split(string) if x])
['blah', 'lots', 'of', 'spaces', 'here']
This works even if ^\\s+ doesn't match: 即使^\\s+不匹配也可以：
>>> string = "foo, bar "
>>> print([x for x in pattern.split(string) if x])
['foo', 'bar']
>>>
Here's why you need ^\\s+: 這就是您需要^ \\ s +的原因：
>>> pattern = re.compile("\s*,\s*|\s+$")
>>> print([x for x in pattern.split(string) if x])
[' blah', 'lots', 'of', 'spaces', 'here']
See the leading spaces in blah? 看到等等的主要空間嗎？
Clarification: above uses the Python 3 interpreter, but results are the same in Python 2. 說明：上面使用的是Python 3解釋器，但結果與Python 2相同。
#3樓
I came to add: 我來補充：
map(str.strip, string.split(','))
but saw it had already been mentioned by Jason Orendorff in a comment . 但是看到Jason Orendorff在評論中已經提到了它。
Reading Glenn Maynard's comment in the same answer suggesting list comprehensions over map I started to wonder why. 在同一個答案中讀到格倫·梅納德（Glenn Maynard）的評論，這暗示了人們對地圖的理解，我開始懷疑爲什麼。 I assumed he meant for performance reasons, but of course he might have meant for stylistic reasons, or something else (Glenn?). 我以爲他是出於性能方面的考慮，但是當然他可能是出於風格方面的原因，或者其他原因（Glenn？）。
So a quick (possibly flawed?) test on my box applying the three methods in a loop revealed: 因此，在我的盒子上快速地（可能有缺陷？）應用了以下三種方法的測試：
[word.strip() for word in string.split(',')]
$ time ./list_comprehension.py
real 0m22.876s
map(lambda s: s.strip(), string.split(','))
$ time ./map_with_lambda.py
real 0m25.736s
map(str.strip, string.split(','))
$ time ./map_with_str.strip.py
real 0m19.428s
making map(str.strip, string.split(',')) the winner, although it seems they are all in the same ballpark. 使map(str.strip, string.split(','))成爲贏家，儘管看起來他們都在同一個球場。
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. 當然，出於性能原因，不一定要排除map（有或沒有lambda），對我來說，它至少與列表理解一樣清晰。
Edit: 編輯：
Python 2.6.5 on Ubuntu 10.04 Ubuntu 10.04上的Python 2.6.5
#4樓
s = 'bla, buu, jii'
sp = []
sp = s.split(',')
for st in sp:
print st
#5樓
re (as in regular expressions) allows splitting on multiple characters at once: re （如正則表達式中）允許一次拆分多個字符：
$ string = "blah, lots , of , spaces, here "
$ re.split(', ',string)
['blah', 'lots ', ' of ', ' spaces', 'here ']
This doesn't work well for your example string, but works nicely for a comma-space separated list. 這對於您的示例字符串而言效果不佳，但對於逗號分隔的列表則效果很好。 For your example string, you can combine the re.split power to split on regex patterns to get a "split-on-this-or-that" effect. 對於您的示例字符串，您可以結合使用re.split功能來分割正則表達式模式，以獲得“此或該分割”效果。
$ re.split('[, ]',string)
['blah',
'',
'lots',
'',
'',
'',
'',
'of',
'',
'',
'',
'spaces',
'',
'here',
'']
Unfortunately, that's ugly, but a filter will do the trick: 不幸的是，這很醜陋，但是filter可以解決問題：
$ filter(None, re.split('[, ]',string))
['blah', 'lots', 'of', 'spaces', 'here']
Voila! 瞧！
#6樓
import re
result=[x for x in re.split(',| ',your_string) if x!='']
this works fine for me. 這對我來說很好。