Why this examples of `zip()` code have different outputs?

Question

As far, as I am able to understand, we are unpacking 3 identical iterators in both cases, but output is different.
In first case, it, for some reason, similar to list(zip(c[::3], c[1::3], c[2::3])), in second, it behaves like normal zip().

c = [[a, b] for a, b in itertools.product(list(range(3)), list(range(3)))]

# first example
list(zip(*[iter(c)] * 3))
>[([0, 0], [0, 1], [0, 2]), ([1, 0], [1, 1], [1, 2]), ([2, 0], [2, 1], [2, 2])]

# second example
list(zip(*[iter(c), iter(c), iter(c)]))
>[([0, 0], [0, 0], [0, 0]),
 ([0, 1], [0, 1], [0, 1]),
 ([0, 2], [0, 2], [0, 2]),
 ([1, 0], [1, 0], [1, 0]),
 ([1, 1], [1, 1], [1, 1]),
 ([1, 2], [1, 2], [1, 2]),
 ([2, 0], [2, 0], [2, 0]),
 ([2, 1], [2, 1], [2, 1]),
 ([2, 2], [2, 2], [2, 2])]

Read https://stackoverflow.com/questions/240178/list-of-lists-changes-reflected-across-sublists-unexpectedly — timgeb, Oct 02 '18 at 12:16
First shares references to the same iterator. The second has separate iterators. — pylang, Oct 02 '18 at 22:20

hiro protagonist · Accepted Answer · 2018-10-02T12:23:10.197

3

your first example is the same as

it = iter(c)
print(list(zip(*[it, it, it])))  # == list(zip(it, it, it))

this is not the same as your second example list(zip(iter(c), iter(c), iter(c)) where you create 3 fresh iterators.

every time next is called on it this has an influence on what happens the next time it is used.

in the second example the 3 iterators run independently.

edited Oct 02 '18 at 12:23

answered Oct 02 '18 at 12:16

hiro protagonist

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Why this examples of `zip()` code have different outputs?

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