Finding All Possible Permutations of a Given String in Python

Finding all possible permutations of a given string in python

The itertools module has a useful method called permutations(). The documentation says:

itertools.permutations(iterable[, r])

Return successive r length permutations of elements in the iterable.

If r is not specified or is None, then r defaults to the length of the
iterable and all possible full-length permutations are generated.

Permutations are emitted in lexicographic sort order. So, if the input
iterable is sorted, the permutation tuples will be produced in sorted
order.

You'll have to join your permuted letters as strings though.

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

['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']

If you find yourself troubled by duplicates, try fitting your data into a structure with no duplicates like a set:

>>> perms = [''.join(p) for p in permutations('stacks')]
>>> len(perms)
720
>>> len(set(perms))
360

Thanks to @pst for pointing out that this is not what we'd traditionally think of as a type cast, but more of a call to the set() constructor.

Generate all permutations of a string in Python without using itertools

One easy way to go about this problem is to think of the characters in your string as digits in an unusual number system. The length of the string is the base. So the permutations (with repetition) of 'abc' correspond to the numbers from 0 to 3**3-1 in base 3, where 'a' is the digit 0, 'b' is 1 and 'c' is 2.

def permutations_with_repetition(s):
    base = len(s)
    for n in range(base**base):
        yield "".join(s[n // base**(base-d-1) % base] for d in range(base))

Sample run:

>>> for p in permutations_with_repetition("abc"):
    print(p)

aaa
aab
aac
aba
abb
abc
aca
acb
acc
baa
bab
bac
bba
bbb
bbc
bca
bcb
bcc
caa
cab
cac
cba
cbb
cbc
cca
ccb
ccc

If you were allowed to use itertools, you'd want itertools.product with a repeat keyword argument: itertools.product("abc", repeat=3)

How do I generate all permutations of a list?

Use itertools.permutations from the standard library:

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

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Adapted from here is a demonstration of how itertools.permutations might be implemented:

def permutations(elements):
    if len(elements) <= 1:
        yield elements
        return
    for perm in permutations(elements[1:]):
        for i in range(len(elements)):
            # nb elements[0:1] works in both string and list contexts
            yield perm[:i] + elements[0:1] + perm[i:]

A couple of alternative approaches are listed in the documentation of itertools.permutations. Here's one:

def permutations(iterable, r=None):
    # permutations('ABCD', 2) --> AB AC AD BA BC BD CA CB CD DA DB DC
    # permutations(range(3)) --> 012 021 102 120 201 210
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    if r > n:
        return
    indices = range(n)
    cycles = range(n, n-r, -1)
    yield tuple(pool[i] for i in indices[:r])
    while n:
        for i in reversed(range(r)):
            cycles[i] -= 1
            if cycles[i] == 0:
                indices[i:] = indices[i+1:] + indices[i:i+1]
                cycles[i] = n - i
            else:
                j = cycles[i]
                indices[i], indices[-j] = indices[-j], indices[i]
                yield tuple(pool[i] for i in indices[:r])
                break
        else:
            return

And another, based on itertools.product:

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

Finding all possible permutations of a hash when given list of grouped elements

If you make a list of the original hash value's elements and replace each element with a list of all its possible variations, you can use itertools.product to get the Cartesian product across these sublists. Transforming each element of the result back to a string with '_'.join() will get you the list of possible hashes:

from itertools import product

def possible_hashes(original_hash, grouped_elements):
    
    hash_list = original_hash.split('_')
    variations = list(set().union(*grouped_elements))
    
    var_list = hash_list.copy()
    for i, h in enumerate(hash_list):
        if h in variations:
            for g in grouped_elements:
                if h in g:
                    var_list[i] = g
                    break
        else:
            var_list[i] = [h]
                    
    return ['_'.join(h) for h in product(*var_list)]

possible_hashes("1_2_3_4_5", [["1", "1a", "1b"], ["3", "3a"]]) 

['1_2_3_4_5',
 '1_2_3a_4_5',
 '1a_2_3_4_5',
 '1a_2_3a_4_5',
 '1b_2_3_4_5',
 '1b_2_3a_4_5']

To use this function on various original hash values stored in a dataframe column, you can do something like this:

df['hash'].apply(lambda x: possible_hashes(x, grouped_elements))

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