Combinations without duplicates of all elements in Python

To clarify the picture, if i have a string:

'pac'


I would want to get the list of every permutation of it, in this example:

['p', 'a', 'c', 'pa', 'pc', 'pac']


Just like i would type any of it in some search engine like the one on Ebay and it would pop up "pac" as a suggestion.

Code bellow is what i achieved thus far, but it's obviously not working as it should. 'Names' is just a list with multiple names, for instance: ['pac', 'greg', 'witch']

letters = {}
for name in names:
temp =
letter =
temp2 =

for let in name:
let = let.lower()
temp.append(let)
letter.append(let)

for i in range(0, len(name)):
for j in range(1, len(name) - i):
print(i, j, end=' ')

to_add = letter[i] + temp[j]
print(to_add, temp2)

temp2.append(to_add)
letter.append(to_add)
temp = temp2
temp2 =

letters[name] = letter
return letters


If there is built-in function feel free to share it with me, but that's not the core of the problem.

Possible duplicate of stackoverflow.com/questions/8306654/…
– dmitryro
Jun 30 at 3:11

Just use permutations and list comprehension - your code contains a lot of extra work.
– dmitryro
Jun 30 at 3:11

Possible duplicate of Finding all possible permutations of a given string in python
– Matthew Story
Jun 30 at 3:12

What happened to 'ac'? And it looks like you want combinations, not permutations. The set of all combinations is called the power set, see the itertools docs for a recipe.
– PM 2Ring
Jun 30 at 3:27

1 Answer
1

User standard library itertools:

itertools

In [48]: T = 'pac'

In [49]: list(itertools.chain(*[itertools.combinations(T, i+1) for i in range(len(T))]))
Out[49]: [('p',), ('a',), ('c',), ('p', 'a'), ('p', 'c'), ('a', 'c'), ('p', 'a', 'c')]


It break down into:

1, itertools.combinations(['p', 'a', 'c'], i) to generate all the subsample of 'pac' with sample size i

itertools.combinations(['p', 'a', 'c'], i)

i

2, repeat i for i=1, 2, 3, and put the resultant three itertools.combination objects in a list

i

i=1, 2, 3

itertools.combination

3, unpack the list from #2, and use the those as the parameter to generate an itertools.chain object (as the name implies, it will chain its args together). See more on arguments unpack

itertools.chain

In a lot of real world use cases, especially when the total number of the elements are large, you don't actual want to make a list out of the itertools.chain object. The point of using itertools are often to achieve memory efficiency by avoiding having to putting all its members in memory.

itertools.chain

itertools

(if you don't want tuples just add a ''.join to get them back to strings)

tuple

''.join

I understand what is going on here, but I think you should generally explain what is going on with the line: list(itertools.chain(*[itertools.combinations(list(T), i+1) for i in range(len(T))])) so that way OP doesnt just copy+paste, but learns HOW it works.
– Fallenreaper
Jun 30 at 3:16

list(itertools.chain(*[itertools.combinations(list(T), i+1) for i in range(len(T))]))

Ah, sure, I should do that thanks @Fallenreaper !
– CT Zhu
Jun 30 at 3:18

You really don't need list(T) and your claim for memory efficiency is lost (for this example - but generally true) because you expand the arguments to itertools.chain(). You could maintain memory efficiency with itertools.chain.from_iterable(itertools.combination(T, i+1) for i in range(len(T))) - a generator expression vs an expanded list comprehension.
– AChampion
Jun 30 at 3:32

list(T)

itertools.chain()

itertools.chain.from_iterable(itertools.combination(T, i+1) for i in range(len(T)))

@AChampion, good points!
– CT Zhu
Jun 30 at 3:40

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