All but the last N elements of iterator in Python

Question

What is the best way to get all but the last N elements of an iterator in Python? Here is an example of it in theoretical action:

>>> list(all_but_the_last_n(range(10), 0))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> list(all_but_the_last_n(range(10), 2))
[0, 1, 2, 3, 4, 5, 6, 7]

Answer 1

Just for the fun of it, here's a variation on Ignacio's solution that doesn't require a deque.

>>> def truncate(it, n):
...     cache = [next(it) for i in range(n)]
...     index = 0
...     for val in it:
...         val, cache[index] = cache[index], val
...         index = (index + 1) % n
...         yield val

I wasn't especially concerned with speed when I wrote the above... but perhaps this would be a tad faster:

def truncate(it, n):
    cache = [next(it) for i in range(n)]
    index = 0
    for val in it:
        yield cache[index]
        cache[index] = val
        index = (index + 1) % n

Answer 2

Use a collections.deque. Push N items from the source on the first invocation. On each subsequent invocation, pop an item out, push an item in from the source, and yield the popped item.

Answer 3

Based on Ignacio Vazquez-Abrams's description:

from collections import deque

def all_but_the_last_n(iterable, count):
    q = deque()
    i = iter(iterable)
    for n in range(count):
        q.append(i.next())
    for item in i:
        q.append(item)
        yield q.popleft()

I wondered whether it was better to use the deque right to left (append, popleft) or left to right (appendleft, pop). So i timed it with python 2.5.2 and found that rtl was 3.59 usec while ltr was 3.53 usec. The difference of 0.06 usec is not significant. the test was to append a single item and pop a single item.