Question

I need to write generator which yield all possible 8 symbols strings.
From array of symbols like this:

leters = ['1','2','3','4','5','6','7','8','9','0','q','w','e','r','t','y','u','i','o','p','a','s','d','f','g','h','j','k','l','z','x','c','v','b','n','m']

The skeleton looks like this:

def generator():
    """
    here algorithm
    """
    yield string

suppose to return list like this ['00000001','00000002','00000003', ......'mmmmmmmm']

Answer 1

itertools.combinations() and itertools.combinations_with_replacement() return a generator

>>> letters = ['a', 'b', 'c']
>>> from itertools import combinations

I am using print() in the examples to illustrate the output. Substitute it with yield, to get a generator.

>>> for c in combinations(letters, 2): 
        print(c)
... 
('a', 'b')
('a', 'c')
('b', 'c')

>>> for c in combinations(letters, 2): 
        print(''.join(c))
... 
ab
ac
bc
>>> 

>>> for c in itertools.combinations_with_replacement(letters, 2): 
        print(''.join(c))
... 
aa
ab
ac
bb
bc
cc

If you brute force it for all 8 letter passwords containing english letters and digits, you’re looking to iterate over ~ 2.8 trillion strings

EDIT
If you somehow know there are no repeated elements, use permutations

>>> for c in itertools.permutations(letters, 2): 
        print(''.join(c))
... 
ab
ac
ba
bc
ca
cb

this gives you both ab and ba

For the most general brute force sequence use itertools.product() as in Cosmologicon’s solution