New submission from John Arbash Meinel <[EMAIL PROTECTED]>: 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 __________________________________ Tracker <[EMAIL PROTECTED]> <http://bugs.python.org/issue2672> __________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
