How do I generate all permutations of a list?

For Python 2.6 onwards:

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

This returns as a generator. Use list(permutations(xs)) to return as a list.

First, import itertools:

import itertools 

Permutation (order matters):

print(list(itertools.permutations([1,2,3,4], 2))) [(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)] 

Combination (order does NOT matter):

print(list(itertools.combinations('123', 2))) [('1', '2'), ('1', '3'), ('2', '3')] 

Cartesian product (with several iterables):

print(list(itertools.product([1,2,3], [4,5,6]))) [(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)] 

Cartesian product (with one iterable and itself):

print(list(itertools.product([1,2], repeat=3))) [(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)] 

def permutations(head, tail=''): if len(head) == 0: print(tail) else: for i in range(len(head)): permutations(head[:i] + head[i+1:], tail + head[i]) 

called as:

permutations('abc') 

#!/usr/bin/env python def perm(a, k=0): if k == len(a): print a else: for i in xrange(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]) 

Output:

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

As I'm swapping the content of the list it's required a mutable sequence type as input. E.g. perm(list("ball")) will work and perm("ball") won't because you can't change a string.

This Python implementation is inspired by the algorithm presented in the book Computer Algorithms by Horowitz, Sahni and Rajasekeran.

This solution implements a generator, to avoid holding all the permutations on memory:

def permutations (orig_list): if not isinstance(orig_list, list): orig_list = list(orig_list) yield orig_list if len(orig_list) == 1: return for n in sorted(orig_list): new_list = orig_list[:] pos = new_list.index(n) del(new_list[pos]) new_list.insert(0, n) for resto in permutations(new_list[1:]): if new_list[:1] + resto <> orig_list: yield new_list[:1] + resto 

Regular implementation (no yield - will do everything in memory):

def getPermutations(array): if len(array) == 1: return [array] permutations = [] for i in range(len(array)): # get all perm's of subarray w/o current item perms = getPermutations(array[:i] + array[i+1:]) for p in perms: permutations.append([array[i], *p]) return permutations 

Yield implementation:

def getPermutations(array): if len(array) == 1: yield array else: for i in range(len(array)): perms = getPermutations(array[:i] + array[i+1:]) for p in perms: yield [array[i], *p] 

The basic idea is to go over all the elements in the array for the 1st position, and then in 2nd position go over all the rest of the elements without the chosen element for the 1st, etc. You can do this with recursion, where the stop criteria is getting to an array of 1 element - in which case you return that array.

In a functional style

def addperm(x,l): return [ l[0:i] + [x] + l[i:] for i in range(len(l)+1) ] def perm(l): if len(l) == 0: return [[]] return [x for y in perm(l[1:]) for x in addperm(l[0],y) ] print perm([ i for i in range(3)]) 

The result:

[[0, 1, 2], [1, 0, 2], [1, 2, 0], [0, 2, 1], [2, 0, 1], [2, 1, 0]] 

The following code is an in-place permutation of a given list, implemented as a generator. Since it only returns references to the list, the list should not be modified outside the generator. The solution is non-recursive, so uses low memory. Work well also with multiple copies of elements in the input list.

def permute_in_place(a): a.sort() yield list(a) if len(a) <= 1: return first = 0 last = len(a) while 1: i = last - 1 while 1: i = i - 1 if a[i] < a[i+1]: j = last - 1 while not (a[i] < a[j]): j = j - 1 a[i], a[j] = a[j], a[i] # swap the values r = a[i+1:last] r.reverse() a[i+1:last] = r yield list(a) break if i == first: a.reverse() return if __name__ == '__main__': for n in range(5): for a in permute_in_place(range(1, n+1)): print a print for a in permute_in_place([0, 0, 1, 1, 1]): print a print 

A quite obvious way in my opinion might be also:

def permutList(l): if not l: return [[]] res = [] for e in l: temp = l[:] temp.remove(e) res.extend([[e] + r for r in permutList(temp)]) return res 

list2Perm = [1, 2.0, 'three'] listPerm = [[a, b, c] for a in list2Perm for b in list2Perm for c in list2Perm if ( a != b and b != c and a != c ) ] print listPerm 

Output:

[ [1, 2.0, 'three'], [1, 'three', 2.0], [2.0, 1, 'three'], [2.0, 'three', 1], ['three', 1, 2.0], ['three', 2.0, 1] ] 

I used an algorithm based on the factorial number system- For a list of length n, you can assemble each permutation item by item, selecting from the items left at each stage. You have n choices for the first item, n-1 for the second, and only one for the last, so you can use the digits of a number in the factorial number system as the indices. This way the numbers 0 through n!-1 correspond to all possible permutations in lexicographic order.

from math import factorial def permutations(l): permutations=[] length=len(l) for x in xrange(factorial(length)): available=list(l) newPermutation=[] for radix in xrange(length, 0, -1): placeValue=factorial(radix-1) index=x/placeValue newPermutation.append(available.pop(index)) x-=index*placeValue permutations.append(newPermutation) return permutations permutations(range(3)) 

output:

[[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]] 

This method is non-recursive, but it is slightly slower on my computer and xrange raises an error when n! is too large to be converted to a C long integer (n=13 for me). It was enough when I needed it, but it's no itertools.permutations by a long shot.

Note that this algorithm has an n factorial time complexity, where n is the length of the input list

Print the results on the run:

global result result = [] def permutation(li): if li == [] or li == None: return if len(li) == 1: result.append(li[0]) print result result.pop() return for i in range(0,len(li)): result.append(li[i]) permutation(li[:i] + li[i+1:]) result.pop() 

Example:

permutation([1,2,3]) 

Output:

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

One can indeed iterate over the first element of each permutation, as in tzwenn's answer. It is however more efficient to write this solution this way:

def all_perms(elements): if len(elements) <= 1: yield elements # Only permutation possible = no permutation else: # Iteration over the first element in the result permutation: for (index, first_elmt) in enumerate(elements): other_elmts = elements[:index]+elements[index+1:] for permutation in all_perms(other_elmts): yield [first_elmt] + permutation 

This solution is about 30 % faster, apparently thanks to the recursion ending at len(elements) <= 1 instead of 0. It is also much more memory-efficient, as it uses a generator function (through yield), like in Riccardo Reyes's solution.

This is inspired by the Haskell implementation using list comprehension:

def permutation(list): if len(list) == 0: return [[]] else: return [[x] + ys for x in list for ys in permutation(delete(list, x))] def delete(list, item): lc = list[:] lc.remove(item) return lc 

For performance, a numpy solution inspired by Knuth, (p22) :

from numpy import empty, uint8 from math import factorial def perms(n): f = 1 p = empty((2*n-1, factorial(n)), uint8) for i in range(n): p[i, :f] = i p[i+1:2*i+1, :f] = p[:i, :f] # constitution de blocs for j in range(i): p[:i+1, f*(j+1):f*(j+2)] = p[j+1:j+i+2, :f] # copie de blocs f = f*(i+1) return p[:n, :] 

Copying large blocs of memory saves time - it's 20x faster than list(itertools.permutations(range(n)) :

In [1]: %timeit -n10 list(permutations(range(10))) 10 loops, best of 3: 815 ms per loop In [2]: %timeit -n100 perms(10) 100 loops, best of 3: 40 ms per loop 

If you don't want to use the builtin methods such as:

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

you can implement permute function yourself

from collections.abc import Iterable def permute(iterable: Iterable[str]) -> set[str]: perms = set() if len(iterable) == 1: return {*iterable} for index, char in enumerate(iterable): perms.update([char + perm for perm in permute(iterable[:index] + iterable[index + 1:])]) return perms if __name__ == '__main__': print(permute('abc')) # {'bca', 'abc', 'cab', 'acb', 'cba', 'bac'} print(permute(['1', '2', '3'])) # {'123', '312', '132', '321', '213', '231'} 

Disclaimer: shameless plug by package author. :)

The trotter package is different from most implementations in that it generates pseudo lists that don't actually contain permutations but rather describe mappings between permutations and respective positions in an ordering, making it possible to work with very large 'lists' of permutations, as shown in this demo which performs pretty instantaneous operations and look-ups in a pseudo-list 'containing' all the permutations of the letters in the alphabet, without using more memory or processing than a typical web page.

In any case, to generate a list of permutations, we can do the following.

import trotter my_permutations = trotter.Permutations(3, [1, 2, 3]) print(my_permutations) for p in my_permutations: print(p) 

Output:

 A pseudo-list containing 6 3-permutations of [1, 2, 3]. [1, 2, 3] [1, 3, 2] [3, 1, 2] [3, 2, 1] [2, 3, 1] [2, 1, 3] 

The beauty of recursion:

>>> import copy >>> def perm(prefix,rest): ... for e in rest: ... new_rest=copy.copy(rest) ... new_prefix=copy.copy(prefix) ... new_prefix.append(e) ... new_rest.remove(e) ... if len(new_rest) == 0: ... print new_prefix + new_rest ... continue ... perm(new_prefix,new_rest) ... >>> perm([],['a','b','c','d']) ['a', 'b', 'c', 'd'] ['a', 'b', 'd', 'c'] ['a', 'c', 'b', 'd'] ['a', 'c', 'd', 'b'] ['a', 'd', 'b', 'c'] ['a', 'd', 'c', 'b'] ['b', 'a', 'c', 'd'] ['b', 'a', 'd', 'c'] ['b', 'c', 'a', 'd'] ['b', 'c', 'd', 'a'] ['b', 'd', 'a', 'c'] ['b', 'd', 'c', 'a'] ['c', 'a', 'b', 'd'] ['c', 'a', 'd', 'b'] ['c', 'b', 'a', 'd'] ['c', 'b', 'd', 'a'] ['c', 'd', 'a', 'b'] ['c', 'd', 'b', 'a'] ['d', 'a', 'b', 'c'] ['d', 'a', 'c', 'b'] ['d', 'b', 'a', 'c'] ['d', 'b', 'c', 'a'] ['d', 'c', 'a', 'b'] ['d', 'c', 'b', 'a'] 

ANOTHER APPROACH (without libs)

def permutation(input): if len(input) == 1: return input if isinstance(input, list) else [input] result = [] for i in range(len(input)): first = input[i] rest = input[:i] + input[i + 1:] rest_permutation = permutation(rest) for p in rest_permutation: result.append(first + p) return result 

Input can be a string or a list

print(permutation('abcd')) print(permutation(['a', 'b', 'c', 'd'])) 

from __future__ import print_function def perm(n): p = [] for i in range(0,n+1): p.append(i) while True: for i in range(1,n+1): print(p[i], end=' ') print("") i = n - 1 found = 0 while (not found and i>0): if p[i]<p[i+1]: found = 1 else: i = i - 1 k = n while p[i]>p[k]: k = k - 1 aux = p[i] p[i] = p[k] p[k] = aux for j in range(1,(n-i)/2+1): aux = p[i+j] p[i+j] = p[n-j+1] p[n-j+1] = aux if not found: break perm(5) 

Here is an algorithm that works on a list without creating new intermediate lists similar to Ber's solution at https://stackoverflow.com/a/108651/184528.

def permute(xs, low=0): if low + 1 >= len(xs): yield xs else: for p in permute(xs, low + 1): yield p for i in range(low + 1, len(xs)): xs[low], xs[i] = xs[i], xs[low] for p in permute(xs, low + 1): yield p xs[low], xs[i] = xs[i], xs[low] for p in permute([1, 2, 3, 4]): print p 

You can try the code out for yourself here: http://repl.it/J9v

This algorithm is the most effective one, it avoids of array passing and manipulation in recursive calls, works in Python 2, 3:

def permute(items): length = len(items) def inner(ix=[]): do_yield = len(ix) == length - 1 for i in range(0, length): if i in ix: #avoid duplicates continue if do_yield: yield tuple([items[y] for y in ix + [i]]) else: for p in inner(ix + [i]): yield p return inner() 

Usage:

for p in permute((1,2,3)): print(p) (1, 2, 3) (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 1, 2) (3, 2, 1) 

def pzip(c, seq): result = [] for item in seq: for i in range(len(item)+1): result.append(item[i:]+c+item[:i]) return result def perm(line): seq = [c for c in line] if len(seq) <=1 : return seq else: return pzip(seq[0], perm(seq[1:])) 

Generate all possible permutations

I'm using python3.4:

def calcperm(arr, size): result = set([()]) for dummy_idx in range(size): temp = set() for dummy_lst in result: for dummy_outcome in arr: if dummy_outcome not in dummy_lst: new_seq = list(dummy_lst) new_seq.append(dummy_outcome) temp.add(tuple(new_seq)) result = temp return result 

Test Cases:

lst = [1, 2, 3, 4] #lst = ["yellow", "magenta", "white", "blue"] seq = 2 final = calcperm(lst, seq) print(len(final)) print(final) 

I see a lot of iteration going on inside these recursive functions, not exactly pure recursion...

so for those of you who cannot abide by even a single loop, here's a gross, totally unnecessary fully recursive solution

def all_insert(x, e, i=0): return [x[0:i]+[e]+x[i:]] + all_insert(x,e,i+1) if i<len(x)+1 else [] def for_each(X, e): return all_insert(X[0], e) + for_each(X[1:],e) if X else [] def permute(x): return [x] if len(x) < 2 else for_each( permute(x[1:]) , x[0]) perms = permute([1,2,3]) 

To save you folks possible hours of searching and experimenting, here's the non-recursive permutaions solution in Python which also works with Numba (as of v. 0.41):

@numba.njit() def permutations(A, k): r = [[i for i in range(0)]] for i in range(k): r = [[a] + b for a in A for b in r if (a in b)==False] return r permutations([1,2,3],3) [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]] 

To give an impression about performance:

%timeit permutations(np.arange(5),5) 243 µs ± 11.1 µs per loop (mean ± std. dev. of 7 runs, 1 loop each) time: 406 ms %timeit list(itertools.permutations(np.arange(5),5)) 15.9 µs ± 8.61 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each) time: 12.9 s 

So use this version only if you have to call it from njitted function, otherwise prefer itertools implementation.

Anyway we could use sympy library , also support for multiset permutations

import sympy from sympy.utilities.iterables import multiset_permutations t = [1,2,3] p = list(multiset_permutations(t)) print(p) # [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]] 

Answer is highly inspired by Get all permutations of a numpy array

Another solution:

def permutation(flag, k =1 ): N = len(flag) for i in xrange(0, N): if flag[i] != 0: continue flag[i] = k if k == N: print flag permutation(flag, k+1) flag[i] = 0 permutation([0, 0, 0]) 

This is the asymptotically optimal way O(n*n!) of generating permutations after initial sorting.

There are n! permutations at most and hasNextPermutation(..) runs in O(n) time complexity

In 3 steps,

1. Find largest j such that a[j] can be increased
2. Increase a[j] by smallest feasible amount
3. Find lexicogrpahically least way to extend the new a[0..j]

''' Lexicographic permutation generation consider example array state of [1,5,6,4,3,2] for sorted [1,2,3,4,5,6] after 56432(treat as number) ->nothing larger than 6432(using 6,4,3,2) beginning with 5 so 6 is next larger and 2345(least using numbers other than 6) so [1, 6,2,3,4,5] ''' def hasNextPermutation(array, len): ' Base Condition ' if(len ==1): return False ''' Set j = last-2 and find first j such that a[j] < a[j+1] If no such j(j==-1) then we have visited all permutations after this step a[j+1]>=..>=a[len-1] and a[j]<a[j+1] a[j]=5 or j=1, 6>5>4>3>2 ''' j = len -2 while (j >= 0 and array[j] >= array[j + 1]): j= j-1 if(j==-1): return False # print(f"After step 2 for j {j} {array}") ''' decrease l (from n-1 to j) repeatedly until a[j]<a[l] Then swap a[j], a[l] a[l] is the smallest element > a[j] that can follow a[l]...a[j-1] in permutation before swap we have a[j+1]>=..>=a[l-1]>=a[l]>a[j]>=a[l+1]>=..>=a[len-1] after swap -> a[j+1]>=..>=a[l-1]>=a[j]>a[l]>=a[l+1]>=..>=a[len-1] a[l]=6 or l=2, j=1 just before swap [1, 5, 6, 4, 3, 2] after swap [1, 6, 5, 4, 3, 2] a[l]=5, a[j]=6 ''' l = len -1 while(array[j] >= array[l]): l = l-1 # print(f"After step 3 for l={l}, j={j} before swap {array}") array[j], array[l] = array[l], array[j] # print(f"After step 3 for l={l} j={j} after swap {array}") ''' Reverse a[j+1...len-1](both inclusive) after reversing [1, 6, 2, 3, 4, 5] ''' array[j+1:len] = reversed(array[j+1:len]) # print(f"After step 4 reversing {array}") return True array = [1,2,4,4,5] array.sort() len = len(array) count =1 print(array) ''' The algorithm visits every permutation in lexicographic order generating one by one ''' while(hasNextPermutation(array, len)): print(array) count = count +1 # The number of permutations will be n! if no duplicates are present, else less than that # [1,4,3,3,2] -> 5!/2!=60 print(f"Number of permutations: {count}")