How to simplify this code about iteration combination in python

Question

So i have this code below that i created. This code will list every possible combination of a certain value, in this example is "a" "b" "c" "d". If i use this code, the result would be like this: a, aa, ab, ac, ad, aaa, aab, aac, aad, aba, abb, abc, etc. How to simplify this for loop code, so that i can input more values without createing more for loop?

n = "abcd"

for c in n:
    print(c)
    for c1 in n:
        print(c+c1)
        for c2 in n:
            print(c+c1+c2)
            for c3 in n:
                print(c+c1+c2+c3)

Possible duplicate of How to get all possible combinations of a list’s elements? — Zaraki Kenpachi
– Zaraki Kenpachi, Commented Jun 18, 2019 at 10:05
If you don't want to use itertools, you can go for recursion. I've already answered to a similar question here — MePsyDuck
– MePsyDuck, Commented Jun 18, 2019 at 10:05
Possible duplicate of all permutations of string without using itertools — MePsyDuck
– MePsyDuck, Commented Jun 18, 2019 at 10:07

Jkind9 · Accepted Answer · 2019-06-18 10:05:11Z

2

https://docs.python.org/2/library/itertools.html

Itertools permutations and combinations is what you're after.

itertools.permutations([1, 2, 3])

answered Jun 18, 2019 at 10:05

Jkind9

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ncica · Accepted Answer · 2019-06-18 10:20:57Z

2

from itertools import combinations
n = "abcd"
print ([''.join(l) for i in range(len(n)) for l in combinations(n, i+1)])

output:

['a', 'b', 'c', 'd', 'ab', 'ac', 'ad', 'bc', 'bd', 'cd', 'abc', 'abd', 'acd', 'bcd', 'abcd']

edit:

from itertools import combinations_with_replacement

n = "abcd"
comb = []

for i in range(1, len(n)+1):
    comb += list(combinations_with_replacement(n, i))
print([''.join(c) for c in comb])

output:

['a', 'b', 'c', 'd', 'aa', 'ab', 'ac', 'ad', 'bb', 'bc', 'bd', 'cc', 'cd', 'dd', 'aaa', 'aab', 'aac', 'aad', 'abb', 'abc', 'abd', 'acc', 'acd', 'add', 'bbb', 'bbc', 'bbd', 'bcc', 'bcd', 'bdd', 'ccc', 'ccd', 'cdd', 'ddd', 'aaaa', 'aaab', 'aaac', 'aaad', 'aabb', 'aabc', 'aabd', 'aacc', 'aacd', 'aadd', 'abbb', 'abbc', 'abbd', 'abcc', 'abcd', 'abdd', 'accc', 'accd', 'acdd', 'addd', 'bbbb', 'bbbc', 'bbbd', 'bbcc', 'bbcd', 'bbdd', 'bccc', 'bccd', 'bcdd', 'bddd', 'cccc', 'cccd', 'ccdd', 'cddd', 'dddd']

edited Jun 18, 2019 at 10:20

answered Jun 18, 2019 at 10:05

ncica

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5 Comments

frankegoesdown Over a year ago

But it has wrong output where aaaa, bbbb, cccc, dddd, etc.

Stradivari Over a year ago

I think OP is asking for permutations instead of combinations, as he expects 'aba' and 'aab'

Felnyr Over a year ago

Yeah, he is looking for permutations with repetition.

Ganaarya Over a year ago

Sorry, my bad...yeah i mean permutaions with repetition

Ganaarya Over a year ago

Almost there, but it seems combinations_with_replacement does not include results like "ba" "ca" "cb" etc

ncica · Accepted Answer · 2019-07-19 15:50:02Z

1

To get the same result of 340 elements you can use itertools.product:

product(n, repeat=1)
product(n, repeat=2)
...
product(n, repeat=4)

To print the result you can use the following loop:

from itertools import product
n = "abcd"

for i in range(1, 5):  
    for prod in product(n, repeat=i): 
        print(''.join(prod))

To get an extra level you can easily increase the 5 in range. Note: the order of printing is slightly different than in your code.

edited Jul 19, 2019 at 15:50

ncica

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answered Jun 18, 2019 at 10:20

ikkuh

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Comments

vurmux · Accepted Answer · 2019-06-18 10:07:29Z

You can use itertools module to solve your problem. You need combinations_with_replacement function with all possible lengths:

import itertools as it

n = "abcd"
result = []

for l in range(len(n)):
    result += list(it.combinations_with_replacement(n, l+1))
print(result)

[('a',), ('b',), ('c',), ('d',), ('a', 'a'), ('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'b'), ('b', 'c'), ('b', 'd'), ('c', 'c'), ('c', 'd'), ('d', 'd'), ('a', 'a', 'a'), ('a', 'a', 'b'), ('a', 'a', 'c'), ('a', 'a', 'd'), ('a', 'b', 'b'), ('a', 'b', 'c'), ('a', 'b', 'd'), ('a', 'c', 'c'), ('a', 'c', 'd'), ('a', 'd', 'd'), ('b', 'b', 'b'), ('b', 'b', 'c'), ('b', 'b', 'd'), ('b', 'c', 'c'), ('b', 'c', 'd'), ('b', 'd', 'd'), ('c', 'c', 'c'), ('c', 'c', 'd'), ('c', 'd', 'd'), ('d', 'd', 'd')]

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