Do iterators save memory in Python?

Question

I don\'t quite understand how iterators have memory in Python.

>>> l1 = [1, 2, 3, 4, 5, 6]
>>> l2 = [2, 3, 4, 5, 6, 7]
>>> iz = iz         


        

Answer 1

Your examples are too simplistic. Consider this:

nums = [1, 2, 3, 4, 5, 6]
nums_it = (n for n in nums)

nums_it is a generator that returns all items unmodified from nums. Clearly you do not have any advantage. But consider this:

squares_it = (n ** 2 for n in nums)

and compare it with:

squares_lst = [n ** 2 for n in nums]

With squares_it, we are generating the squares of nums on the fly only when requested. With squares_lst, we are generating all of them at once and storing them in a new list.

So, when you do:

for n in squares_it:
    print(n)

it's like if you were doing:

for n in nums:
    print(n ** 2)

But when you do:

for n in squares_lst:
    print(n)

it's like if you were doing:

squares_lst = []
for n in nums:
    squares_lst.append(n ** 2)
for n in squares_lst:
    print(n)

If you don't need (or don't have) the list nums, then you can save even more space by using:

squares_it = (n ** 2 for n in xrange(1, 7))

Generators and iterators also provide another significant advantage (which may actually be a disadvantage, depending on the situation): they are evaluated lazily.

Also, generators and iterators may yield an infinite number of elements. An example is itertools.count() that yields 0, 1, 2, 3, ... without never ending.