Simplest way to get the first n elements of an iterator

Question

How can I get the first n elements of an iterator (generator) in the simplest way? Is there something simpler than, e. g.

def firstN(iterator, n):
  for i in range(n):
    yield iterator.next()

print list(firstN(it, 3))

I can't think of a nicer way, but maybe there is? Maybe a functional form?

Answer 1

Use itertools.islice():

from itertools import islice
print(list(islice(it, 3)))

This will yield the next 3 elements from it, then stop.

Answer 2

Without using itertools:

(t[0] for t in zip(L, range(3)))