python拆分而不创建空格

Question

I understand why it is important to create blanks using split thanks to this question , but sometimes it is necessary not to grab them.

lets say you parsed some css and got the following strings:

s1 = 'background-color:#000;color:#fff;border:1px #ccc dotted;'
s2 = 'color:#000;background-color:#fff;border:1px #333 dotted'

both are valid css even though there is a semicolon lacking at the end of the string. when splitting the strings, you get the following:

>>> s1.split(';')
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted', '']
>>> s2.split(';')
['color:#000', 'background-color:#fff', 'border:1px #333 dotted']

that extra semicolon creates a blank item in the list. now if I want to manipulate further I would need to test the beginning and end of each list, and remove them if they are blank, which is not that bad, but seems avoidable.

question:

is there a method that is essentially the same as split but does not include trailing blank items? or is there simply a way to remove those just like a string has strip to remove the trailing whitespace

Answer 1

Simply remove the items with the None filter:

filter(None, s1.split(';'))

Demo:

>>> s1 = 'background-color:#000;color:#fff;border:1px #ccc dotted;'
>>> filter(None, s1.split(';'))
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted']

Calling filter() with None removes all 'empty' or numeric 0 items; anything that would evaluate to false in a boolean context.

filter(None, ....) eats list comprehensions for breakfast:

>>> import timeit
>>> timeit.timeit('filter(None, a)', "a = [1, 2, 3, None, 4, 'five', ''] * 100")
9.410392045974731
>>> timeit.timeit('[i for i in a if i]', "a = [1, 2, 3, None, 4, 'five', ''] * 100")
44.9318630695343

Answer 2

You can use a list comprehension to filter out the empty strings, as an empty string is considered False :

>>> s1 = 'background-color:#000;color:#fff;border:1px #ccc dotted;'
>>> [i for i in s1.split(';') if i]
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted']

Alternatively, you can rstrip() the semicolon first:

>>> s1.rstrip(';').split(';')
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted']

Answer 3

Apply str.strip to the string before doing the split :

>>> s1 = 'background-color:#000;color:#fff;border:1px #ccc dotted;'
...     
>>> s1.strip(';').split(';')
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted']

Works for both leading and trailing ';' :

>>> s1 = ';background-color:#000;color:#fff;border:1px #ccc dotted;'
>>> s1.strip(';').split(';')
['background-color:#000', 'color:#fff', 'border:1px #ccc dotted']

I am not sure why you would want to avoid this as a strip before split is going to be faster than both LC and filter :

>>> s1 = ';background-color:#000;color:#fff;border:1px #ccc dotted;'*1000
>>> %timeit filter(None, s1.split(';'))
1000 loops, best of 3: 638 us per loop
>>> %timeit s1.strip(';').split(';')
1000 loops, best of 3: 570 us per loop
>>> %timeit [i for i in s1.split(';') if i]
100 loops, best of 3: 931 us per loop