splitting a list into n groups

[Peter Otten]

Eddie Corns wrote:

> Rajarshi Guha <rajarshi at presidency.com> writes:
> 
>>Hi,
>>  is there an efficient (pythonic) way in which I could split a list into
>>say 5 groups? By split I mean the the first x members would be one group,
>>the next x members another group and so on 5 times. (Obviously x =
>>lengthof list/5)
> 
> How about:
> 
> ------------------------------------------------------------
> def span (x, n):
>     return range(0, x, n)
> 
> def group (l, num):
>     n = (len(l)/num) + 1
>     return [l[s:s+n] for s in span (len(l), n)]
> 
> # test cases
> l1 = range(100)
> print group (l1, 5)
> print group (l1, 6)
> print group (l1, 4)
> ------------------------------------------------------------

More test cases :-)

for k in range(20):
    l1 = range(k)
    lol = group(l1, 5)
    print len(lol), lol

0 []
1 [[0]]
2 [[0], [1]]
3 [[0], [1], [2]]
4 [[0], [1], [2], [3]]
3 [[0, 1], [2, 3], [4]]
3 [[0, 1], [2, 3], [4, 5]]
4 [[0, 1], [2, 3], [4, 5], [6]]
4 [[0, 1], [2, 3], [4, 5], [6, 7]]
5 [[0, 1], [2, 3], [4, 5], [6, 7], [8]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10]]
4 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]]
5 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12]]
5 [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11], [12, 13]]
4 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14]]
4 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17]]
5 [[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15], [16, 17,
18]]

Peter