all permutations of string without using itertools

问题

All possible strings of any length that can be formed from a given string

Input:

abc

Output:

a b c abc ab ac bc bac bca 
         cb ca ba cab cba acb

I have tried using this but it's limited to string abc, I want to generalize it like if input 'abcd' i will provide me output for the same.

def perm_main(elems):
    perm=[]
    for c in elems:
        perm.append(c)
    for i in range(len(elems)):
        for j in range(len(elems)):
            if perm[i]!= elems[j]:
                perm.append(perm[i]+elems[j])
    level=[elems[0]]
    for i in range(1,len(elems)):
        nList=[]
        for item in level:
            #print(item)
            nList.append(item+elems[i])
                #print(nList)
            for j in range(len(item)):
                #print(j)
                nList.append(item[0:j]+ elems[i] + item[j:])
                #print(nList)
        level=nList
    perm = perm + nList
    return perm

回答1:

You may not need itertools, but you have the solution in the documentation, where itertools.permutations is said to be roughly equivalent to:

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 = list(range(n))
    cycles = list(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

Or using 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)

They are both generators so you will need to call list(permutations(x)) to retrieve an actual list or substitute the yields for l.append(v) where l is a list defined to accumulate results and v is the yielded value.

For all the possible sizes ones, iterate over them:

from itertools import chain
check_string = "abcd"
all = list(chain.from_iterable(permutations(check_string , r=x)) for x in range(len(check_string )))

回答2:

Partial recursive solution. You just need to make it work for different lengths:

def permute(pre, str):
    n = len(str)
    if n == 0:
        print(pre)
    else:
        for i in range(0,n):
            permute(pre + str[i], str[0:i] + str[i+1:n])

You can call it using permute('', 'abcd'), or have another method to simplify things

def permute(str):
    permute('', str)

Answer borrowed from here.

In general, you will have better luck translating code from C/Cpp/Java solutions to Python because they generally implement things from scratch and do things without much need of libraries.

UPDATE

Full solution:

def all_permutations(given_string):
    for i in range(len(given_string)):
        permute('', given_string, i+1)

def permute(prefix, given_string, max_len):
    if len(given_string) <= 0 or len(prefix) >= max_len:
        print(prefix)
    else:
        for i in range(len(given_string)):
            permute(prefix + given_string[i], given_string[:i] + given_string[i+1:], max_len)

>>> all_permutations('abc')
a
b
c
ab
ac
ba
bc
ca
cb
abc
acb
bac
bca
cab
cba

来源：https://stackoverflow.com/questions/56425594/all-permutations-of-string-without-using-itertools

标签

python-3.x

permutation