[New-bugs-announce] [issue13653] reorder set.intersection parameters for better performance
Andrew Dalke
report at bugs.python.org
Fri Dec 23 01:54:13 CET 2011
New submission from Andrew Dalke <dalke at dalkescientific.com>:
In Issue3069, Arnaud Delobelle proposed support for multiple values to set.intersection() and set.union(), writing "Intersection is optimized by sorting all sets/frozensets/dicts in increasing order of size and only iterating over elements in the smallest."
Raymond Hettinger commented therein that he had just added support for multiple parameters. However, he did not pick up the proposed change in the attached patch which attempts to improve the intersection performance.
Consider the attached benchmark, which constructs an inverted index mapping a letter to the set of words which contain that letter. (Rather, to word index.) Here's the output:
## Example output:
# a has 144900 words
# j has 3035 words
# m has 62626 words
# amj takes 5.902/1000 (verify: 289)
# ajm takes 0.292/1000 (verify: 289)
# jma takes 0.132/1000 (verify: 289)
Searching set.intersection(inverted_index["j"], inverted_index["m"], inverted_index["a"]) is fully 44 times faster than searching "a", "m", "j"!
Of course, the set.intersection() supports any iterable, so would only be an optimization for when all of the inputs are set types.
BTW, my own experiments suggest that sorting isn't critical. It's more important to find the most anti-correlated set to the smallest set, and the following does that dynamically by preferentially choosing sets which are likely to not match elements of the smallest set:
def set_intersection(*input_sets):
N = len(input_sets)
min_index = min(range(len(input_sets)), key=lambda x: len(input_sets[x]))
best_mismatch = (min_index+1)%N
new_set = set()
for element in input_sets[min_index]:
# This failed to match last time; perhaps it's a mismatch this time?
if element not in input_sets[best_mismatch]:
continue
# Scan through the other sets
for i in range(best_mismatch+1, best_mismatch+N):
j = i % N
if j == min_index:
continue
# If the element isn't in the set then perhaps this
# set is a better rejection test for the next input element
if element not in input_sets[j]:
best_mismatch = j
break
else:
# The element is in all of the other sets
new_set.add(element)
return new_set
Using this in the benchmark gives
amj takes 0.972/1000 (verify: 289)
ajm takes 0.972/1000 (verify: 289)
jma takes 0.892/1000 (verify: 289)
which clearly shows that this Python algorithm is still 6 times faster (for the worst case) than the CPython code.
However, the simple sort solution:
def set_intersection_sorted(*input_sets):
input_sets = sorted(input_sets, key=len)
new_set = set()
for element in input_sets[0]:
if element in input_sets[1]:
if element in input_sets[2]:
new_set.add(element)
return new_set
gives times of
amj takes 0.492/1000 (verify: 289)
ajm takes 0.492/1000 (verify: 289)
jma takes 0.422/1000 (verify: 289)
no doubt because there's much less Python overhead than my experimental algorithm.
----------
components: Interpreter Core
files: set_intersection_benchmark.py
messages: 150124
nosy: dalke
priority: normal
severity: normal
status: open
title: reorder set.intersection parameters for better performance
type: enhancement
versions: Python 3.4
Added file: http://bugs.python.org/file24081/set_intersection_benchmark.py
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<http://bugs.python.org/issue13653>
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