Finding all possible permutations of a given strin

Question

I have a string. I want to generate all permutations from that string, by changing the order of characters in it. For example, say:

x='stack'

what I want is a list like this,

l=['stack','satck','sackt'.......]

Currently I am iterating on the list cast of the string, picking 2 letters randomly and transposing them to form a new string, and adding it to set cast of l. Based on the length of the string, I am calculating the number of permutations possible and continuing iterations till set size reaches the limit. There must be a better way to do this.

Answer 1

Stack overflow users have already posted some strong solutions but I wanted to show yet another solution. This one I find to be more intuitive

The idea is that for a given string: we can recurse by the algorithm (pseudo code):

permutations = char + permutations(string - char) for char in string

Hope it helps someone!

def permutations(string):
    """Create all permutations of a string with non-repeating characters
    """
    permutation_list = []
    if len(string) == 1:
        return [string]
    else:
        for char in string:
            [permutation_list.append(char + a) for a in permutations(string.replace(char, ""))]
    return permutation_list

Answer 2

why do you not simple do:

from itertools import permutations
perms = [''.join(p) for p in permutations(['s','t','a','c','k'])]
print perms
print len(perms)
print len(set(perms))

you get no duplicate as you can see :

 ['stack', 'stakc', 'stcak', 'stcka', 'stkac', 'stkca', 'satck', 'satkc', 
'sactk', 'sackt', 'saktc', 'sakct', 'sctak', 'sctka', 'scatk', 'scakt', 'sckta',
 'sckat', 'sktac', 'sktca', 'skatc', 'skact', 'skcta', 'skcat', 'tsack', 
'tsakc', 'tscak', 'tscka', 'tskac', 'tskca', 'tasck', 'taskc', 'tacsk', 'tacks', 
'taksc', 'takcs', 'tcsak', 'tcska', 'tcask', 'tcaks', 'tcksa', 'tckas', 'tksac', 
'tksca', 'tkasc', 'tkacs', 'tkcsa', 'tkcas', 'astck', 'astkc', 'asctk', 'asckt', 
'asktc', 'askct', 'atsck', 'atskc', 'atcsk', 'atcks', 'atksc', 'atkcs', 'acstk', 
'acskt', 'actsk', 'actks', 'ackst', 'ackts', 'akstc', 'aksct', 'aktsc', 'aktcs', 
'akcst', 'akcts', 'cstak', 'cstka', 'csatk', 'csakt', 'cskta', 'cskat', 'ctsak', 
'ctska', 'ctask', 'ctaks', 'ctksa', 'ctkas', 'castk', 'caskt', 'catsk', 'catks', 
'cakst', 'cakts', 'cksta', 'cksat', 'cktsa', 'cktas', 'ckast', 'ckats', 'kstac', 
'kstca', 'ksatc', 'ksact', 'kscta', 'kscat', 'ktsac', 'ktsca', 'ktasc', 'ktacs', 
'ktcsa', 'ktcas', 'kastc', 'kasct', 'katsc', 'katcs', 'kacst', 'kacts', 'kcsta', 
'kcsat', 'kctsa', 'kctas', 'kcast', 'kcats']
    120
    120
    [Finished in 0.3s]

Answer 3

def permute(seq):
    if not seq:
        yield seq
    else:
        for i in range(len(seq)):
            rest = seq[:i]+seq[i+1:]
            for x in permute(rest):
                yield seq[i:i+1]+x

print(list(permute('stack')))

Answer 4

from itertools import permutations
perms = [''.join(p) for p in permutations('ABC')]

perms = [''.join(p) for p in permutations('stack')]

Answer 5

def perm(string):
   res=[]
   for j in range(0,len(string)):
       if(len(string)>1):
           for i in perm(string[1:]):
               res.append(string[0]+i)
       else:
           return [string];
       string=string[1:]+string[0];
   return res;
l=set(perm("abcde"))

This is one way to generate permutations with recursion, you can understand the code easily by taking strings 'a','ab' & 'abc' as input.

You get all N! permutations with this, without duplicates.

Answer 6

Everyone loves the smell of their own code. Just sharing the one I find the simplest:

def get_permutations(word):
    if len(word) == 1:
        yield word

    for i, letter in enumerate(word):
        for perm in get_permutations(word[:i] + word[i+1:]):
            yield letter + perm