为什么zip函数没有在python中给出预期的结果

Question

a = enumerate('abcdef')
b = enumerate('ghi')

for i, j in zip(a, b):
        print(i, j)
        if i[0] == 0:
            next(itertools.islice(zip(a, b), 1, 1), None)

For the above code, I expect the following result as I advance zip(a, b) by 1:

 ((0, 'a'), (0, 'g'))
 ((2, 'c'), (2, 'i')) 

However, it still gives the same result as the following code:

iter1 = enumerate('abcdef')
iter2 = enumerate('ghi')

for i, j in zip(a, b):
        print(i, j)

output:

((0, 'a'), (0, 'g'))
((1, 'b'), (1, 'h'))
((2, 'c'), (2, 'i'))

why the statement next(itertools.islice(zip(a, b), 1, 1), None) does not advance zip(a, b)?

Answer 1

The 3.6 zip returns iterators and it works as you expect:

a = enumerate('abcdef')
b = enumerate('ghi')

for i, j in zip(a, b):
        print(i, j)

        if i[0] == 0:
            next(itertools.islice(zip(a, b), 1, 1), None)

it will skip the (1,) tuples as zip returns iterators.

The 2.7 zip returns a list of tuples, and both statements are unrelated as the zip(a,b) are seperate list, both using unrelated enumeration sequences.

So for 2.7 they are not skipping the (1,) tuples.

Output 3.6:

(0, 'a') (0, 'g')
(2, 'c') (2, 'i')

Output 2.7:

((0, 'a'), (0, 'g'))
((1, 'b'), (1, 'h'))
((2, 'c'), (2, 'i'))

You are running 2.7 from your demo output.

https://docs.python.org/3.6/library/functions.html#zip https://docs.python.org/2.7/library/functions.html#zip