Why does itertools.permutations() return a list of characters or digits for each permutation, instead of just returning a string?
For example:
>&g
Because it expects an iterable as a parameter and doesn't know, it's a string. The parameter is described in the docs.
http://docs.python.org/library/itertools.html#itertools.permutations
Perumatation can be done for strings and list also, below is the example..
x = [1,2,3]
if you need to do permutation the above list
print(list(itertools.permutations(x, 2)))
# the above code will give the below..
# [(1,2),(1,3),(2,1)(2,3),(3,1),(3,2)]
I have not tried, but most likely should work
comb = itertools.permutations("1234",4)
for x in comb:
''.join(x)
itertools.permutations()
simply works this way. It takes an arbitrary iterable as an argument, and always returns an iterator yielding tuples. It doesn't (and shouldn't) special-case strings. To get a list of strings, you can always join the tuples yourself:
list(map("".join, itertools.permutations('1234')))