Nils :
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.
norok2 :
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
.
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