Length of a finite generator

Question

I have these two implementations to compute the length of a finite generator, while keeping the data for further processing:

def count_generator1(generator):         


        

Answer 1

I ran Windows 64-bit Python 3.4.3 timeit on a few approaches I could think of:

>>> from timeit import timeit
>>> from textwrap import dedent as d
>>> timeit(
...     d("""
...     count = -1
...     for _ in s:
...         count += 1
...     count += 1
...     """),
...     "s = range(1000)",
... )
50.70772041983173
>>> timeit(
...     d("""
...     count = -1
...     for count, _ in enumerate(s):
...         pass
...     count += 1
...     """),
...     "s = range(1000)",
... )
42.636973504498656
>>> timeit(
...     d("""
...     count, _ = reduce(f, enumerate(range(1000)), (-1, -1))
...     count += 1
...     """),
...     d("""
...     from functools import reduce
...     def f(_, count):
...         return count
...     s = range(1000)
...     """),
... )
121.15513102540672
>>> timeit("count = sum(1 for _ in s)", "s = range(1000)")
58.179126025925825
>>> timeit("count = len(tuple(s))", "s = range(1000)")
19.777029680237774
>>> timeit("count = len(list(s))", "s = range(1000)")
18.145157531932
>>> timeit("count = len(list(1 for _ in s))", "s = range(1000)")
57.41422175998332

Shockingly, the fastest approach was to use a list (not even a tuple) to exhaust the iterator and get the length from there:

>>> timeit("count = len(list(s))", "s = range(1000)")
18.145157531932

Of course, this risks memory issues. The best low-memory alternative was to use enumerate on a NOOP for-loop:

>>> timeit(
...     d("""
...     count = -1
...     for count, _ in enumerate(s):
...         pass
...     count += 1
...     """),
...     "s = range(1000)",
... )
42.636973504498656

Cheers!