从函数返回多个输出

Question

I am not sure that this is possible, but...

I am trying to write a function that will output lists of numbers so that they can be input into another function. I'm at a dead end though, as neither return or yield (used in this context) can give me what I want.

def ITERATOR():
    for number in range(1,3):
        for item in itertools.permutations(range(15),number):
            yield item

Obviously, the function breaks whenever return is used, and yield returns everything at once. What I ideally want it to output is what is printed on each line by:

def ITERATOR():
    for number in range(2,4):
        for item in itertools.permutations(range(15),number):
            print(item)

ie:

(0, 1)
(0, 2)
(0, 3)
(0, 4)
(0, 5)
(0, 6)
(0, 7)
(0, 8)
(0, 9)
(0, 10)
(0, 11)
(0, 12)
(0, 13)
(0, 14)
etc...

Is there any way of achieving this?

Answer 1

If you want to return a list, you could just do it:

def fct():
    return [item for item in itertools.permutations(range(15),number) for number in range(2,4)]

And for a generator

def fct():
    return (item for item in itertools.permutations(range(15),number) for number in range(2,4))

Answer 2

Solution

new_tuples = []
for i in range(0, 5):
    for j in range(1, 16):
        new_tuples.append((i, j))

for i in new_tuples:
    print(i)

Output

 (xenial)vash@localhost:~/python$ python3 text_seg.py (0, 1) (0, 2) (0, 3) (0, 4) (0, 5) (0, 6) (0, 7) (0, 8) (0, 9) (0, 10) (0, 11) (0, 12) (0, 13) (0, 14) (0, 15) (1, 1) 

Comments

Would this approach satisfy your needs? You could interchange the numbers as desired but this, will allow you to have a list with the tuples store and be able to call them to print .

Answer 3

You can return the list through the argument of the function, and use the return of the function to indicate the number of permutations founded:

import itertools

permut = []

def ITERATOR(item):
    n_permuts= 0
    for number in range(2,4):
        for item in itertools.permutations(range(15),number):
            permut.append(item)
            n_permuts+= 1
    return n_permuts

print (ITERATOR(permut))
for item in permut:
    print(item)