How to create all the permutations of a list based on consecutive indices as a whole list in python?

Question

This is my code (I use python 2.x):

import itertools

def grouper(input_list, n = 2):
    for i in xrange(len(input_list) - (n - 1)):
        yield input_list[i:i+n]

def list_of_lists(num):
    nums = list(map(int,str(num)))
    for first, second, third, fourth in grouper(nums, 4):
        x = first, second, third, fourth
        xx = list(x)
        print xx

This is my input:

a = 1232331
list_of_lists(a)

This is my output:

[1, 2, 3, 2]
[2, 3, 2, 3]
[3, 2, 3, 3]
[2, 3, 3, 1]

But I want the output to be:

[[1, 2, 3, 2], [2, 3, 2, 3], [3, 2, 3, 3], [2, 3, 3, 1]]

@Jean-FrançoisFabre at a wild stab - some form of "sliding window" — Jon Clements, Mar 25 '17 at 09:40

McGrady · Answer 1 · 2017-03-25T09:39:00.497

0

You can use zip method to get the combinations:

try:
    from itertools import izip as zip
except ImportError:
    pass


x = [1, 2, 3, 4, 5, 6]


def getCombin(n, l):
    if len(l) < n:
        return None
    else:
        return list(zip(*(x[i:] for i in range(n))))


print(getCombin(3,x))

Output:

[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6)]

If you're using Python2.x you can import izip:

from itertools import izip as zip

edited Mar 25 '17 at 09:39

answered Mar 25 '17 at 09:28

McGrady

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As stated above, this solution is quite fragile. Also, please don't make a habit of dumping "Try this:" and "Output:" with zero explanation, as such answers are generally unhelpful. – TigerhawkT3 Mar 25 '17 at 09:37
@TigerhawkT3 I update my answer – McGrady Mar 25 '17 at 09:41
@Jean-FrançoisFabre I fixed it. – McGrady Mar 25 '17 at 09:42

How to create all the permutations of a list based on consecutive indices as a whole list in python?

This is my code (I use python 2.x):

This is my input:

This is my output:

But I want the output to be:

1 Answers1