使用生成器生成排列

Question

I was writing a program to generate all the permutations of a string:

def print_permutations_wrapper(str):
    strList = str.split()
    print_permutations(strList, 0, len(strList))


def print_permutations(strList: list, start: int, end: int):
    if start >= end - 1:
        print(strList)
        return

    print_permutations(strList, start+1, end)
    for i in range(start+1, end):
        strList[start], strList[i] = strList[i], strList[start]
        print_permutations(strList, start+1, end)
        strList[i], strList[start] = strList[start], strList[i]


def main():
    str = 'a b c'
    print_permutations_wrapper(str)


if __name__ == "__main__":
    main()

Its working correctly, but rather than printing it, I wanted to return it lazily using yield :

def print_permutations_wrapper(str):
    strList = str.split()
    yield from print_permutations(strList, 0, len(strList))


def print_permutations(strList: list, start: int, end: int):
    if start >= end - 1:
        yield strList
        return

    yield from print_permutations(strList, start+1, end)
    for i in range(start+1, end):
        strList[start], strList[i] = strList[i], strList[start]
        yield from print_permutations(strList, start+1, end)
        strList[i], strList[start] = strList[start], strList[i]


def main():
    str = 'a b c'
    x = print_permutations_wrapper(str)
    print(list(x))

if __name__ == "__main__":
    main()

The output that I get is:

[['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c'], ['a', 'b', 'c']]

rather than all the permutations.
How to correct this?

I am using Python 3.7.

Answer 1

Taking your second program, but add a print(strList) will show that the generator function yielded what you expected, but the final output is clearly not what you expected. This is due to the fact that the program as structured takes a list, but does all the operation to this same copy (assumed to limit memory usage). You can also verify this by

>>> strList = ['a', 'b', 'c'] 
>>> items = list(print_permutations(strList, 0, 3))
>>> items[0] is items[1]
True
>>> items[0] is strList
True

Clearly every single item inside the items are of the same original strList that got passed in. Given the simplicity of the problem, this may be avoided by yielding a shallow copy of the list instead. Thus the relevant yield part of the function will become

def print_permutations(strList: list, start: int, end: int):
    if start >= end - 1:
        yield list(strList)
        return

Now running it should produce:

>>> strList = ['a', 'b', 'c'] 
>>> items = list(print_permutations(strList, 0, 3))
>>> items[0] is items[1]
False

Also, as an aside, permutation is in fact a part of the standard library, available via itertools.permutation .

Related: "Least Astonishment" and the Mutable Default Argument