New submission from Antony Lee <anntzer....@gmail.com>:
Consider the following example
from enum import Flag
F = Flag("F", list("abcdefghijklm"))
for idx in range(2**len(F) - 1):
F(idx)
creating all possible combos of 13 flags, so 8192 instances (yes, I know the
instances are cached later, but the initial cost is still significant).
Locally, this takes over 4.5s to execute; profiling shows that ~30% of that
time is spent executing enum._is_power_of_two(!):
def _power_of_two(value):
if value < 1:
return False
return value == 2 ** _high_bit(value)
Indeed, replacing the implementation of _power_of_two by
import math
_powers_of_two = {
2 ** i for i in range(math.ceil(math.log2(sys.maxsize)) + 1)}
def _power_of_two(value):
return (value in _powers_of_two if value <= sys.maxsize
else value == 2 ** _high_bit(value))
shaves off ~30% of the runtime (obviously this won't help for huge
(>sys.maxsize) flag values, but these are rarer I would think).
An even better fix, I think, would be for Flag to cache `flags_to_check`
(updating it at the same time as `_value2member_map_`) instead of recomputing
it each time in _decompose
flags_to_check = [
(m, v)
for v, m in list(flag._value2member_map_.items())
if m.name is not None or _power_of_two(v)
]
as this actually needlessly introduces quadratic complexity when iterating over
all possible combos (because the size of _value2member_map_ is growing).
(Also, it's not actually clear to me how one can end up with unnamed
power-of-two instances in _value2member_map?)
----------
messages: 351244
nosy: Antony.Lee
priority: normal
severity: normal
status: open
title: Flag instance creation is slow
versions: Python 3.8
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Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue38045>
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