Missing a sequence when using itertools combonatio

问题:

from itertools import combinations_with_replacement

x = 'opo'
v = combinations_with_replacement(x, len(x))
ans = [''.join(map(str, x)) for x in v]
print(" ".join(set(ans)))

I'm not sure why im missing the sequence pop here. Why does pop not show but ppo and opp do .

expected output opp ppp poo ppo ooo opo oop pop

actual output opp ppp poo ppo ooo opo oop

回答1:

Consider this:

>>> x = 'abc'
>>> v = itertools.combinations_with_replacement(x, len(x))
>>> ans = [''.join(map(str, x)) for x in v]
>>> ans
['aaa', 'aab', 'aac', 'abb', 'abc', 'acc', 'bbb', 'bbc', 'bcc', 'ccc']

The values in the sequence are irrelevant to what combinations_with_replacement does; only the positions within the sequence count. Your question is the same as asking why 'bab' and 'cac' don't show up in my example. Hint: the name of the function isn't permutations_with_replacement ;-)

回答2:

It is documented correctly here -

itertools.combinations_with_replacement(iterable, r)

Return r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

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

Elements are treated as unique based on their position, not on their value. So if the input elements are unique, the generated combinations will also be unique.

Emphasis mine.