Multiple iterators (using enumerate) for the same iterable, what is going on?

Question

Consider the following example:

s = 'abc'
[(c1, c2) for j, c2 in enumerate(s) for i, c1 in enumerate(s)]

Output:

[('a', 'a'),
 ('b', 'a'),
 ('c', 'a'),
 ('a', 'b'),
 ('b', 'b'),
 ('c', 'b'),
 ('a', 'c'),
 ('b', 'c'),
 ('c', 'c')]

I would expected the same output if enumerate is called outside the list comprehension and the iterators are assigned to variables:

it1, it2 = enumerate(s), enumerate(s)
[(c1, c2) for j, c2 in it1 for i, c1 in it2]

But I get:

[('a', 'a'), ('b', 'a'), ('c', 'a')]

What is going on? I use Python 3.6.9.

Answer 1

What is happening is that the inner iterator gets exhausted after the first iteration of the outer iterator:

s = 'abc'
it1 = enumerate(s)
it2 = enumerate(s)

for i, x in it1:
    for j, y in it2:  # <-- gets consumed when i = 0 and stays empty
        ...

By contrast:

s = 'abc'

for i, x in enumerate(s):
    for j, y in enumerate(s):  # <-- gets recreated at each iteration
        ....

---

If you need persistence, enclose it in a list or tuple:

itr = list(enumerate(s))
print([(c1, c2) for j, c2 in itr for i, c1 in itr])
# [('a', 'a'), ('b', 'a'), ('c', 'a'), ('a', 'b'), ('b', 'b'), ('c', 'b'), ('a', 'c'), ('b', 'c'), ('c', 'c')]

although note the different memory footprint of using enumerate() multiple times versus having it enclosed in a list or tuple.

Answer 2

The difference is that in:

s = 'abc'
[(c1, c2) for j, c2 in enumerate(s) for i, c1 in enumerate(s)]

A new c1 enumerator is created for each value yielded on the first for. While on your second example, the same enumerator is used (it2) - and it gets exausted once it reaches "c" - when the first for advances to the next iteration (c2 = "b") and tries to iterate "it2", it is already exhausted - and the whole expression ends.