[New-bugs-announce] [issue2672] speed of set.update([])

[John Arbash Meinel]

New submission from John Arbash Meinel <john at arbash-meinel.com>:

I was performance profiling some of my own code, and I ran into
something unexpected. Specifically,
set.update(empty_generator_expression) was significantly slower than
set.update(empty_list_expression).

I double checked my findings with timeit:

With python 2.4.3:

$ python -m timeit -s 'x = set(range(10000))' 'x.update([])'
1000000 loops, best of 3: 0.296 usec per loop
$ python -m timeit -s 'x = set(range(10000))' 'x.update(y for y in [])'
1000000 loops, best of 3: 0.837 usec per loop
$ python -m timeit -s 'x = set(range(10000))' 'x.update([y for y in []])'
1000000 loops, best of 3: 0.462 usec per loop

With 2.5.1 (on a different machine)
$ python -m timeit -s 'x = set(range(10000))' 'x.update([])'
1000000 loops, best of 3: 0.265 usec per loop
$ python -m timeit -s 'x = set(range(10000))' 'x.update(y for y in [])'
1000000 loops, best of 3: 0.717 usec per loop
$ python -m timeit -s 'x = set(range(10000))' 'x.update([y for y in []])'
1000000 loops, best of 3: 0.39 usec per loop

So generally, it is about 2x faster to create the empty list expression
and pass it in than to use an empty generator.

----------
components: Interpreter Core
messages: 65694
nosy: jameinel
severity: normal
status: open
title: speed of set.update([])
versions: Python 2.4, Python 2.5

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<http://bugs.python.org/issue2672>
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