Answers for "python string all possible combinations"

4

python print combinations of string

test_str = "abc"
res = [test_str[i: j] for i in range(len(test_str)) 
          for j in range(i + 1, len(test_str) + 1)]
print(res)#['a', 'ab', 'abc', 'b', 'bc', 'c']
Posted by: Guest on May-26-2021
3

python print combinations of string

import itertools
 
if __name__ == '__main__':
 
    nums = list("ABC")
    permutations = list(itertools.permutations(nums))
 
    # Output: ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']
    print([''.join(permutation) for permutation in permutations])
Posted by: Guest on May-26-2021
4

how to get all possible combinations in python

all_combinations = [list(zip(each_permutation, list2)) for each_permutation in itertools.permutations(list1, len(list2))]
Posted by: Guest on April-06-2020
-1

python combinations

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 = 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
Posted by: Guest on July-04-2020

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