Fastest (most Pythonic) way to consume an iterator

Question

I am curious what the fastest way to consume an iterator would be, and the most Pythonic way.

For example, say that I want to create an iterator with the map

Answer 1

While you shouldn't be creating a map object just for side effects, there is in fact a standard recipe for consuming iterators in the itertools docs:

def consume(iterator, n=None):
    "Advance the iterator n-steps ahead. If n is None, consume entirely."
    # Use functions that consume iterators at C speed.
    if n is None:
        # feed the entire iterator into a zero-length deque
        collections.deque(iterator, maxlen=0)
    else:
        # advance to the empty slice starting at position n
        next(islice(iterator, n, n), None)

For just the "consume entirely" case, this can be simplified to

def consume(iterator):
    collections.deque(iterator, maxlen=0)

Using collections.deque this way avoids storing all the elements (because maxlen=0) and iterates at C speed, without bytecode interpretation overhead. There's even a dedicated fast path in the deque implementation for using a maxlen=0 deque to consume an iterator.

Timing:

In [1]: import collections

In [2]: x = range(1000)

In [3]: %%timeit
   ...: i = iter(x)
   ...: for _ in i:
   ...:     pass
   ...: 
16.5 µs ± 829 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [4]: %%timeit
   ...: i = iter(x)
   ...: collections.deque(i, maxlen=0)
   ...: 
12 µs ± 566 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Of course, this is all based on CPython. The entire nature of interpreter overhead is very different on other Python implementations, and the maxlen=0 fast path is specific to CPython. See abarnert's answer for other Python implementations.