python - get all possible permutations with replacement without recursion

Question

In python, how can I write a function that get some number n and print all possible words(a-z) permutations(with replacement) when length=n.

Without using recursion and without using external functions like itertools.permutations, itertools.combinations etc..

For example: n=2 should print

aa ba ... za
ab bb ... zb
...
az bz ... zz

n=3

aaa baa ... zaa
aba aca ... aza
aab aac ... aaz
abb abc ... abz
acb acc ... acz
...
azz bzz ... zzz

See the code in the documentation https://docs.python.org/3/library/itertools.html#itertools.combinations — vaultah, Feb 27 '16 at 17:14
I think it is important that you say 'with replacement'... at least that is what you are doing in your examples — Juan Leni, Feb 27 '16 at 17:17
The correct link for combinations_with_replacement is here: https://docs.python.org/3/library/itertools.html#itertools.combinations_with_replacement — Juan Leni, Feb 27 '16 at 17:19
@mtk99 this examples doenst print the opposite (print "ab" but doesnt print "ba"). I can use "reverse" but I'm pretty sure there is better way — A.Sh, Feb 27 '16 at 17:25
[`itertools.permutations`](https://docs.python.org/3/library/itertools.html#itertools.permutations) — vaultah, Feb 27 '16 at 17:27
@vaultah it doesnt print replacements and opposites, as i wrote to mtk99. plus I dont want to get the iterable objects, I know I can fill it by myself in a tuple with a-z but I think maybe there is better way — A.Sh, Feb 27 '16 at 17:28
Those **are not** "combinations with replacement", since you want to generate for example `'az'` and `'za'`. By definition, combinations include elements only in the same order that they appeared in the input. The question title said "combinations", but the body says "permutations" - these are **not the same thing**. "Permutations with replacement" are equivalent to a Cartesian product of a sequence with itself. I have closed the question as a duplicate, and fixed the title. — Karl Knechtel, Mar 01 '23 at 21:18

Juan Leni · Accepted Answer · 2016-03-03T20:28:30.563

Basically you are counting. Here is an example. A tried to keep it simple so it is easy to follow:

def get_string(idxlist, item_list):
    return ''.join([item_list[i] for i in idxlist])

def increment_vector(idxlist, max_count):
    idxlist[0] += 1
    for i in xrange(len(idxlist)-1):
        if idxlist[i] < max_count:
            break
        idxlist[i]=0
        idxlist[i+1] += 1

def generate(n, item_list):
    max_count = len(item_list)

    idxlist = [0] * n
    while idxlist[-1] < max_count:
        print ( get_string( idxlist, item_list )),
        increment_vector(idxlist, max_count)
        if idxlist[0]==0:
            print 

item_list = map(chr, range(97, 123)) # letters from a-z

generate(3, item_list)

The only point where you really work with your items is in get_string.

EDIT: small adjustment so the output is formatted like in your question

python - get all possible permutations with replacement without recursion

1 Answers1