Getting all ordered sublists of ordered lists

Question

We are struggling with the following problem:
For example the following list, should be divided in m lists with the same order.

Lets say [9,7,6,5,2,10,8,4,3,1] with m=3 a few of the end results should be:

[[9],[7,6],[5,2,10,8,4,3,1]]
[[9,7],[6,5,2,10,8,4,3],[1]]
[[9,7,6,5],[2,10,8,4],[3,1]]

etc.

How can I achieve it?

See this solution [link](https://stackoverflow.com/questions/464864/how-to-get-all-possible-combinations-of-a-list-s-elements/24402296) — sahasrara62, Dec 18 '18 at 09:29
@prashantrana This solution doesn't take the order into account. — Stijn, Dec 18 '18 at 09:32
@Stijin that was hint . Create a nested list of size 3 and using this link put values in that . — sahasrara62, Dec 18 '18 at 09:35
What is the statement of your problem? I can read that you're going to declare it, but then you start with an example. — Gsk, Dec 18 '18 at 09:48
@Gsk We want to have all the possible permutations of a list, but with respect to the order. I thought the example would clarify. — Stijn, Dec 18 '18 at 09:55

L Martens · Accepted Answer · 2018-12-18T11:45:55.317

1

from itertools import combinations, permutations


perm=[]
index = [9,7,6,5,2,10,8,4,3,1]
perm.append(index)
M = 3
slicer = [x for x in combinations(range(1, len(index)), M - 1)]
slicer = [(0,) + x + (len(index),) for x in slicer]

result = [tuple(p[s[i]:s[i + 1]] for i in range(len(s) - 1)) for s in slicer for p in perm]

edited Dec 18 '18 at 11:45

answered Dec 18 '18 at 11:44

L Martens

26
2

That's very helpful! Thanks a lot – Stijn Dec 18 '18 at 11:45

Ravikiran · Answer 2 · 2018-12-18T12:26:37.977

0

Solution:

n= length of the list

k=m=3

def part(n, k):
 def _part(n, k, pre):
    if n <= 0:
        return []
    if k == 1:
        if n <= pre:
            return [[n]]
        return []
    ret = []
    for i in range(min(pre, n), 0, -1):
        ret += [[i] + sub for sub in _part(n-i, k-1, i)]
    return ret
 return _part(n, k, n)

import itertools
perm=[]
partitions=part(n, m)
for i in partitions:
    perm.append(list(itertools.permutations(i,m)))
perm1=list(itertools.chain.from_iterable(perm))  
y= [9,7,6,5,2,10,8,4,3]
new=[]
for i in perm1:
   k=0
   for j in i:
       new.append(list(y[k:k+j]))
       k=k+j
       if k==9:
         print(new)
         new=[]

edited Dec 18 '18 at 12:26

answered Dec 18 '18 at 10:10

Ravikiran

571
3
10

Thanks for this solution, but do you have any idea how to use itertools.combinations?So we can get all possible combinations out of it? Since I tried a lot with it but it doesn't take order into account. – Stijn Dec 18 '18 at 10:54
Hey , I have updated the code which will print all the possible outputs. – Ravikiran Dec 18 '18 at 12:27

Getting all ordered sublists of ordered lists

2 Answers2