Get all possible variations with a list without modules-CodePudding

I need to know how to get all possible variations with a list without modules. I will be adding a string and I know how to convert it to a list. I did like this:

def dpro(string):
    convertedlist = list(string.split(" "))
    return convertedlist
  
print(dpro("this will convert to List"))

This will output a list. I need a loop that will make all different possible outcomes without any modules.

The same process as this question has: Create all possible variations of string Python : but without itertools or any other modules.

CodePudding user response：

The function below will return all iterations of a string without considering spaces.

We iterate of all possible iterations, given by

total

where N is the number of characters is the string, and then each iteration is given by the number i.

The first index (index 1 in maths, index 0 in code) is given by:

first

The last index (index N in maths, index N-1 in code) is given by:

last

And the intermediate indices (in maths from 2 to N-1, in code from 1 to N-2) are given by:

inter

These equations creates a list of indices from which the function can select characters from the original string:

def dpro(string: str):
    """Similar to itertools.product for all characters in a string"""

    # get length of string
    N = len(string)

    # iterate over all iterations given by N ** N
    total_iterations = N ** N
    for i in range(total_iterations):

        # get the first character
        first_idx = i // N ** (N - 1)

        # get the last character
        last_idx = i % N

        # get the intermediate characters
        inter_idx = [i // N ** (N - 2 - j) % N for j in range(N - 2)]

        # create a list of
        indices = [first_idx]   inter_idx   [last_idx]

        yield ''.join(string[i] for i in indices)

string = 'abc'

for entry in dpro(string):
    print(entry)

This prints the following to the console:

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

I have tested the code for strings up to length 5 but the output is quite long so I stuck with 3 characters for this example.

CodePudding user response：

I found the answer from this docs:

https://docs.python.org/3/library/itertools.html#itertools.permutations

Itertools use this function to loop the list with permutations.

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

All together with a string:

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


def themainstring(words):
    return [''.join(i) for i in permutations(words)]

thevarone = themainstring("abc");

print(thevarone)

This will print the possible variations as list. Thanks to @Henry for sharing the docs.