How do I return a list of list of this code?

Question

I found this code for printing all the permutations from an input list, i want to return a nested list but have been unable to do so. The list returned is either empty or the nested list of has all the same permutations. I can also do with a little help in understanding the logic.

ref : https://stackoverflow.com/questions/104420/how-do-i-generate-all-permutations-of-a-list.[one of the solutions]. edit 1 : list assignment added.

l = []
def perm(a, k=0):
   if k == len(a):
      print(a)
      l.append(a)
   else:
      for i in range(k, len(a)):
         a[k], a[i] = a[i] ,a[k]
         perm(a, k+1)
         a[k], a[i] = a[i], a[k]

perm([1,2,3])
print(l)

Output for list l : [[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]

I am not allowed to use itertools.

Thanks,

What do you mean "has all the same permutations?" I see all permutations without repeat. Run it in console. — Mike, Sep 16 '22 at 03:19
You're modifying the list in place. So if you append them all to a list to return, they're all the same list. You should make a copy when you append to the result list. — Barmar, Sep 16 '22 at 03:22
Show your variant where all the same permutations are returned. Probably it is easy to fix this. — Michael Butscher, Sep 16 '22 at 03:24

Alexander · Accepted Answer · 2022-09-16T05:11:17.080

0

At each recursive step the code is altering the list in place. So what the code below does is make a copy of the list once it reaches the base case, and returns it. Then it collects the return values into an empty list on it's way back up the recursive call stack and outputs all lists combined.

def perm(a, k=0):
    if k == len(a):
        return [a[:]]
    lst = []
    for i in range(k, len(a)):
        a[k], a[i] = a[i] ,a[k]
        lst += perm(a, k+1)
        a[k], a[i] = a[i], a[k]
    return lst

print(perm([1,2,3]))

Output:

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

Here is a slightly improved version

def perm(a, k=[]):
    if not a: return [k]
    lst = []
    for i in range(len(a)):
        lst += perm(a[:i] + a[i+1:], k + [a[i]])
     return lst
print(perm([1,2,3]))

edited Sep 16 '22 at 05:11

answered Sep 16 '22 at 03:36

Alexander

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I was wondering if there is a simpler alternative than this. – boom Sep 16 '22 at 04:45
@boom what do you mean by simpler? – Alexander Sep 16 '22 at 04:46
Alternative to this code. I want to still get the permutations ,but can it be done with a different logic? – boom Sep 16 '22 at 04:48
@boom yes it can, but do you have something specific in mind? Are you having trouble understanding the logic in this example – Alexander Sep 16 '22 at 04:54

How do I return a list of list of this code?

1 Answers1