How to retrieve groups of n items in list

Question

In Python, how do I print every triple (or groups of n) items in a list?

I have searched and found various solutions using the itertools module for handling pairs, but I have failed to adapt them for groups of three. I will refrain from including my attempts here as I'm uncertain if they are at all indicative of the correct way to go.

Example:

my_list = ["abra", "cada", "bra", "hum", "dee", "dum"]

I would like to print the first triplet in one line, and then the next triplet on the next line:

"Abra cada bra"
"Hum dee dum"

Edit: In order to expand generality, the question has been edited to cover "groups of n items" instead of just triples.

what do you mean, items 0, 1, 2, ; then 3, 4, 5; 6, 7, 8? Or something completely different? — Antti Haapala -- Слава Україні, Aug 29 '15 at 08:50
http://stackoverflow.com/questions/1915170/split-a-generator-iterable-every-n-items-in-python-splitevery — Antti Haapala -- Слава Україні, Aug 29 '15 at 08:54

score 2 · Answer 1 · answered Aug 29 '15 at 08:52

2

you could use straightforward for loop

i = 0
list = ["abra", "cada", "bra", "hum", "dee", "dum"]
while i < len(list):
  print(" ".join(list[i:i+3]))
  i += 3

answered Aug 29 '15 at 08:52

fl00r

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Keep in mind not to name a variable like a data type. So the list `list` should rather be named something like `l` instead of `list`. – albert Aug 29 '15 at 08:55

Stuart · Answer 2 · 2015-08-29T09:02:57.133

2

A list comprehension using indices is a fairly straightforward way of doing this.

print([my_list[i:i+3] for i in range(0, len(my_list), 3)])

or to print it as desired:

for i in range(0, len(my_list), 3):
    print(' '.join(list[i:i+3]))

Alternatively:

for t in zip(my_list[::3], my_list[1::3], my_list[2::3]):
    print(' '.join(t))

edited Aug 29 '15 at 09:02

answered Aug 29 '15 at 08:57

Stuart

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sureshvv · Answer 3 · 2015-08-29T09:16:38.873

2

Try:

l1 = iter(list)
print '\n'.join([ ' '.join(x) for x in zip(*[l1]*3) ])

edited Aug 29 '15 at 09:16

answered Aug 29 '15 at 09:04

sureshvv

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what happened to your slicing answer? – Padraic Cunningham Aug 29 '15 at 09:22
This is even better. 3 can be changed easily to n. After all he had "n" in the title of the question. – sureshvv Aug 29 '15 at 09:24

Padraic Cunningham · Accepted Answer · 2015-08-29T09:18:20.607

If you want an itertools answer:

from itertools import islice

my_list = ["abra", "cada", "bra", "hum", "dee", "dum"]
it = iter(my_list)
for sli in iter(lambda:list(islice(it, 3)), []):
    print("".join(sli))

Or using zip and iter:

my_list = ["abra", "cada", "bra", "hum", "dee", "dum"]
it = iter(my_list)
for sli in zip(it,it,it):
    print("".join(sli))

If you want to number each match and format:

n = 3
it = iter(my_list)
rn = map(str, range(1, n+1))
for sli in zip(it, it, it):
    print(" ".join(["{}: {}".format(a,b) for a,b in zip(rn, sli)]))

Output:

1: abra 2: cada 3: bra
1: hum 2: dee 3: dum

For grouping an arbitrary number of elements, there is the grouper recipe from itertools:

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)

for grp in (grouper(my_list,3)):
    print("".join(grp))

Thanks for including these alternatives! I have accepted this answer as it covers some situations I may have to deal with. But for future readers, I should mention that the `for loop` solutions below may just as well be the simplest solutions. — P A N, Aug 29 '15 at 09:26
No worries, you tagged itertools so I provided some itertools solutions, using iter and zip would still probably be better than the for loop and slicing though — Padraic Cunningham, Aug 29 '15 at 09:28

How to retrieve groups of n items in list

4 Answers4