Python izip memory error with izip.count()

Question

I ran the following code:

import itertools
my_list = ['a', 'b', 'c', 'd', 'e']
for i in itertools.izip([x for x in my_list], [y for y in itertools.count()]):
        print i

Based on the documentation for izip, I expected it to return a series of tuples, ending when my_list ran out: (a,0) (b,1) (c,2) (d,3) (e,4)

Instead, I get the memory error copied below. It seems the itertools.count() function is returning infinite data, rather than being limited to the 5 records in my_list?

line 50, in main
    for i in itertools.izip([x for x in range(10)], [y for y in itertools.count()]): MemoryError

score 4 · Accepted Answer · 2014-06-19 01:59:21Z

Yes, itertools.count was designed to never stop counting. That is why it returns an iterator, which produces items on demand rather than all at once.

However, this part of your code:

[y for y in itertools.count()]

effectively tells Python to place the infinite amount of numbers produced by itertools.count into a single list. Thus, a MemoryError is raised because this operation immediately consumes all available memory.

You can fix your problem by simply removing the list comprehension¹:

>>> import itertools
>>> my_list = ['a', 'b', 'c', 'd', 'e']
>>> for i in itertools.izip(my_list, itertools.count()):
...     print i
...
('a', 0)
('b', 1)
('c', 2)
('d', 3)
('e', 4)
>>>

With this change, itertools.izip will iterate over the iterator returned by itertools.count(), which will now only produce as many numbers as are needed.

¹I removed your other list comprehension too because it is unnecessary: my_list is already iterable because it is a list.

Arkady · Accepted Answer · 2014-06-19 01:54:05Z

0

Why not just this?

for a,b in enumerate(my_list):
    print (b, a)

answered Jun 19, 2014 at 1:54

Arkady

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