Pythonic way to split a line into groups of four words?

Question

Suppose I have string that look like the following, of varying length, but with the number of "words" always equal to multiple of 4.

9c 75 5a 62 32 3b 3a fe 40 14 46 1c 6e d5 24 de
c6 11 17 cc 3d d7 99 f4 a1 3f 7f 4c

I would like to chop them into strings like 9c 75 5a 62 and 32 3b 3a fe

I could use a regex to match the exact format, but I wonder if there is a more straightforward way to do it, because regex seems like overkill for what should be a simple problem.

Thijs van Dien · Accepted Answer · 2013-06-06T18:50:07.643

A staight-forward way of doing this is as follows:

wordlist = words.split()
for i in xrange(0, len(wordlist), 4):
    print ' '.join(wordlist[i:i+4])

If for some reason you can't make a list of all words (e.g. an infinite stream), you could do this:

from itertools import groupby, izip
words = (''.join(g) for k, g in groupby(words, ' '.__ne__) if k)
for g in izip(*[iter(words)] * 4):
    print ' '.join(g)

Disclaimer: I didn't come up with this pattern; I found it in a similar topic a while back. It arguably relies on an implementation detail, but it would be quite a bit more ugly when done in a different way.

This assumes that there are no spaces? – merlin2011 Jun 06 '13 at 18:28 — merlin2011, Jun 06 '13 at 18:28

iruvar · Answer 2 · 2013-06-06T18:47:33.843

1

A slightly functional way of doing it based on itertools grouper recipe

 for x in grouper(words.split(), 4):
    print ' '.join(x)

edited Jun 06 '13 at 18:47

answered Jun 06 '13 at 18:39

iruvar

22,736
7
53
82

I do not see grouper in intertools. Do you mean groupby? Have you tried this? – dansalmo Jun 06 '13 at 18:46
@dansalmo, check the [recipe](http://docs.python.org/2/library/itertools.html#recipes) section – iruvar Jun 06 '13 at 18:46

dansalmo · Answer 3 · 2013-06-06T18:53:58.250

>>> words = '9c 75 5a 62 32 3b 3a fe 40 14 46 1c 6e d5 24 de'.split()
>>> [' '.join(words[i*4:i*4+4]) for i in range(len(words)/4)]
['9c 75 5a 62', '32 3b 3a fe', '40 14 46 1c', '6e d5 24 de']

or based on 1_CR's answer

from itertools import izip_longest

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

[' '.join(x) for x in grouper(words.split(), 4)]

score 0 · Answer 4 · answered Jun 06 '13 at 18:43

0

giantString= '9c 75 5a 62 32 3b 3a fe 40 14 46 1c 6e d5 24 de c6 11 17 cc 3d d7 99 f4 a1 3f 7f 4c'

splitGiant = giantString.split(' ')
stringHolders = []
for item in xrange(len(splitGiant)/4):
    stringHolders.append(splitGiant[item*4:item*4+4])

stringHolder2 = []

for item in stringHolders:
    stringHolder2.append(' '.join(item))

print stringHolder2

the most long convoluted way to do this possible.

answered Jun 06 '13 at 18:43

TehTris

3,139
1
21
33

1

why don't You use `xrange(0, len(splitGiant), 4)` this allows You to reduce calculation – oleg Jun 06 '13 at 18:48
because my brain is broken. – TehTris Jun 06 '13 at 18:59
Just a little hint: Python (PEP8) encourages `underscore_style` over `camelCase`. – Thijs van Dien Jun 06 '13 at 19:35
sorry @stretch_armstrong – TehTris Jun 06 '13 at 19:43

Pythonic way to split a line into groups of four words?

4 Answers4