有没有办法在 Python 中按第 n 个分隔符分割字符串？
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Is there a way to split a string by every nth separator in Python?
提问by Gnuffo1
For example, if I had the following string:
例如，如果我有以下字符串：
"this-is-a-string"
“这是一个字符串”
Could I split it by every 2nd "-" rather than every "-" so that it returns two values ("this-is" and "a-string") rather than returning four?
我可以按每第二个“-”而不是每个“-”分割它，以便它返回两个值（“this-is”和“a-string”）而不是返回四个？
回答by Gumbo
Here's another solution:
这是另一个解决方案：
span = 2
words = "this-is-a-string".split("-")
print ["-".join(words[i:i+span]) for i in range(0, len(words), span)]
回答by John La Rooy
>>> s="a-b-c-d-e-f-g-h-i-j-k-l" # use zip(*[i]*n)
>>> i=iter(s.split('-')) # for the nth case
>>> map("-".join,zip(i,i))
['a-b', 'c-d', 'e-f', 'g-h', 'i-j', 'k-l']
>>> i=iter(s.split('-'))
>>> map("-".join,zip(*[i]*3))
['a-b-c', 'd-e-f', 'g-h-i', 'j-k-l']
>>> i=iter(s.split('-'))
>>> map("-".join,zip(*[i]*4))
['a-b-c-d', 'e-f-g-h', 'i-j-k-l']
Sometimes itertools.izip is faster as you can see in the results
有时 itertools.izip 更快，正如您在结果中看到的
>>> from itertools import izip
>>> s="a-b-c-d-e-f-g-h-i-j-k-l"
>>> i=iter(s.split("-"))
>>> ["-".join(x) for x in izip(i,i)]
['a-b', 'c-d', 'e-f', 'g-h', 'i-j', 'k-l']
Here is a version that sort ofworks with an odd number of parts depending what output you desire in that case. You might prefer to trim the '-'off the end of the last element with .rstrip('-')for example.
这里是一个版本，那种具有奇数个部分的工作取决于你在这种情况下愿望输出。例如，您可能更喜欢修剪'-'最后一个元素的末尾.rstrip('-')。
>>> from itertools import izip_longest
>>> s="a-b-c-d-e-f-g-h-i-j-k-l-m"
>>> i=iter(s.split('-'))
>>> map("-".join,izip_longest(i,i,fillvalue=""))
['a-b', 'c-d', 'e-f', 'g-h', 'i-j', 'k-l', 'm-']
Here are some timings
这里有一些时间
$ python -m timeit -s 'import re;r=re.compile("[^-]+-[^-]+");s="a-b-c-d-e-f-g-h-i-j-k-l"' 'r.findall(s)'
100000 loops, best of 3: 4.31 usec per loop
$ python -m timeit -s 'from itertools import izip;s="a-b-c-d-e-f-g-h-i-j-k-l"' 'i=iter(s.split("-"));["-".join(x) for x in izip(i,i)]'
100000 loops, best of 3: 5.41 usec per loop
$ python -m timeit -s 's="a-b-c-d-e-f-g-h-i-j-k-l"' 'i=iter(s.split("-"));["-".join(x) for x in zip(i,i)]'
100000 loops, best of 3: 7.3 usec per loop
$ python -m timeit -s 's="a-b-c-d-e-f-g-h-i-j-k-l"' 't=s.split("-");["-".join(t[i:i+2]) for i in range(0, len(t), 2)]'
100000 loops, best of 3: 7.49 usec per loop
$ python -m timeit -s 's="a-b-c-d-e-f-g-h-i-j-k-l"' '["-".join([x,y]) for x,y in zip(s.split("-")[::2], s.split("-")[1::2])]'
100000 loops, best of 3: 9.51 usec per loop
回答by recursive
Regular expressions handle this easily:
正则表达式很容易处理这个：
import re
s = "aaaa-aa-bbbb-bb-c-ccccc-d-ddddd"
print re.findall("[^-]+-[^-]+", s)
Output:
输出：
['aaaa-aa', 'bbbb-bb', 'c-ccccc', 'd-ddddd']
Update for Nick D:
尼克 D 的更新：
n = 3
print re.findall("-".join(["[^-]+"] * n), s)
Output:
输出：
['aaaa-aa-bbbb', 'bb-c-ccccc']
回答by EmFi
EDIT:The original code I posted didn't work. This version does:
编辑：我发布的原始代码不起作用。这个版本做了：
I don't think you can split on every other one, but you could split on every - and join every pair.
我不认为你可以分开每一个，但你可以分开每一个 - 并加入每一对。
chunks = []
content = "this-is-a-string"
split_string = content.split('-')
for i in range(0, len(split_string) - 1,2) :
if i < len(split_string) - 1:
chunks.append("-".join([split_string[i], split_string[i+1]]))
else:
chunks.append(split_string[i])
回答by elzapp
I think several of the already given solutions are good enough, but just for fun, I did this version:
我认为已经给出的几个解决方案已经足够好了，但只是为了好玩，我做了这个版本：
def twosplit(s,sep):
first=s.find(sep)
if first>=0:
second=s.find(sep,first+1)
if second>=0:
return [s[0:second]] + twosplit(s[second+1:],sep)
else:
return [s]
else:
return [s]
print twosplit("this-is-a-string","-")
回答by SpliFF
l = 'this-is-a-string'.split()
nl = []
ss = ""
c = 0
for s in l:
c += 1
if c%2 == 0:
ss = s
else:
ss = "%s-%s"%(ss,s)
nl.insert(ss)
print nl
