Is there a way to split a string by every nth separator in Python?

Question

For example, if I had the following string:

"this-is-a-string"

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?

Solution 1

Here’s another solution:

span = 2
words = "this-is-a-string".split("-")
print ["-".join(words[i:i+span]) for i in range(0, len(words), span)]

Solution 2

>>> 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

>>> 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 of works 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.

>>> 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

$ 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

Solution 3

Regular expressions handle this easily:

import re
s = "aaaa-aa-bbbb-bb-c-ccccc-d-ddddd"
print re.findall("[^-]+-[^-]+", s)

Output:

['aaaa-aa', 'bbbb-bb', 'c-ccccc', 'd-ddddd']

Update for Nick D:

n = 3
print re.findall("-".join(["[^-]+"] * n), s)

Output:

['aaaa-aa-bbbb', 'bb-c-ccccc']

Solution 4

EDIT: The original code I posted didn't work. This version does:

I don't think you can split on every other one, but you could split on every - and join every pair.

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])

View more solutions

Comments

[Mamey]

sorry, i misread your question first time and rewrote it, n was a leftover from previous. Now it gives a list of strings.

[Jed Smith]

Probably the most elegant solution which is still readable, the rest are stretching it.

[Gumbo]

… and only for an even number of words.

[B Bulfin]

Nick: Not so. See my update. Gumbo: Also not so. Just a simple change to the regex will handle that case as well if it is desired.

[Nick Dandoulakis]

@recursive, ok but I don't see the d-ddddd in the output ;-)

[hasen]

sorry, have to -1, too complicated, uses regex,

[Gumbo]

You’re using the wrong code for my proposal. I’m operating on the words an not the string. python -m timeit -s 's="a-b-c-d-e-f-g-h-i-j-k-l".split("-")' '["-".join(s[i:i+2]) for i in range(0, len(s), 2)]'

[RedGlyph]

Nicely done, but fails for an odd number of elements. It shouldn't be too hard to overcome though.

[John La Rooy]

@Gumbo, sorry, I fixed it to match your comment, I've just moved the split() out of the setup clause and used t as a temporary variable

[Bruno Feroleto]

This is very complicated (long to read/decipher), compared to many of the other solutions proposed here…

[Gnuffo1]

This seems the simplest for working for a variable length between seperations.

[Mamey]

rubbish. it's actually much easier to decipher. length is largely irrelevant and it could be shortened by making it less readable. It should have good performance since the loop only has a simple test condition to deal with. Also it has the most flexibility for handling other processing inside the loop. Also the winning answer will crash on a string with an odd number of hyphens. Iter and list ops might be pythonic but that doesn't necessarily make them 'better'.

[Patrick Artner]

Your code has many errors: # TypeError: insert() takes exactly 2 arguments (1 given) can be fixed by using append instead -> returns ['-this-is-a-string']. While it now runs, the result is false. Fixing the split character to '-': returns ['-this', 'is-a'] which is better but still wrong. By moving the c += 1 line after the `else-clause you can fix that as well. After those fixes the solution is fine