( )⚙ D4505 util: lower water mark when removing nodes after cost limit reached

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Differential D4505

util: lower water mark when removing nodes after cost limit reached
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Authored by indygreg on Sep 6 2018, 9:17 PM.

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rHGf296c0b366c8: util: lower water mark when removing nodes after cost limit reached
Summary

See the inline comment for the reasoning here. This is a pretty
common strategy for garbage collectors, other cache-like primtives.

The performance impact is substantial:

$ hg perflrucachedict --size 4 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 100
! inserts w/ cost limit
! wall 1.659181 comb 1.650000 user 1.650000 sys 0.000000 (best of 7)
! wall 1.722122 comb 1.720000 user 1.720000 sys 0.000000 (best of 6)
! mixed w/ cost limit
! wall 1.139955 comb 1.140000 user 1.140000 sys 0.000000 (best of 9)
! wall 1.182513 comb 1.180000 user 1.180000 sys 0.000000 (best of 9)

$ hg perflrucachedict --size 1000 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 10000
! inserts
! wall 0.679546 comb 0.680000 user 0.680000 sys 0.000000 (best of 15)
! sets
! wall 0.825147 comb 0.830000 user 0.830000 sys 0.000000 (best of 13)
! inserts w/ cost limit
! wall 25.105273 comb 25.080000 user 25.080000 sys 0.000000 (best of 3)
! wall 1.724397 comb 1.720000 user 1.720000 sys 0.000000 (best of 6)
! mixed
! wall 0.807096 comb 0.810000 user 0.810000 sys 0.000000 (best of 13)
! mixed w/ cost limit
! wall 12.104470 comb 12.070000 user 12.070000 sys 0.000000 (best of 3)
! wall 1.190563 comb 1.190000 user 1.190000 sys 0.000000 (best of 9)

$ hg perflrucachedict --size 1000 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 10000 --mixedgetfreq 90
! inserts
! wall 0.711177 comb 0.710000 user 0.710000 sys 0.000000 (best of 14)
! sets
! wall 0.846992 comb 0.850000 user 0.850000 sys 0.000000 (best of 12)
! inserts w/ cost limit
! wall 25.963028 comb 25.960000 user 25.960000 sys 0.000000 (best of 3)
! wall 2.184311 comb 2.180000 user 2.180000 sys 0.000000 (best of 5)
! mixed
! wall 0.728256 comb 0.730000 user 0.730000 sys 0.000000 (best of 14)
! mixed w/ cost limit
! wall 3.174256 comb 3.170000 user 3.170000 sys 0.000000 (best of 4)
! wall 0.773186 comb 0.770000 user 0.770000 sys 0.000000 (best of 13)

$ hg perflrucachedict --size 100000 --gets 1000000 --sets 1000000 --mixed 1000000 --mixedgetfreq 90 --costlimit 5000000
! gets
! wall 1.191368 comb 1.190000 user 1.190000 sys 0.000000 (best of 9)
! wall 1.195304 comb 1.190000 user 1.190000 sys 0.000000 (best of 9)
! inserts
! wall 0.950995 comb 0.950000 user 0.950000 sys 0.000000 (best of 11)
! inserts w/ cost limit
! wall 1.589732 comb 1.590000 user 1.590000 sys 0.000000 (best of 7)
! sets
! wall 1.094941 comb 1.100000 user 1.090000 sys 0.010000 (best of 9)
! mixed
! wall 0.936420 comb 0.940000 user 0.930000 sys 0.010000 (best of 10)
! mixed w/ cost limit
! wall 0.882780 comb 0.870000 user 0.870000 sys 0.000000 (best of 11)

This puts us ~2x slower than caches without cost accounting. And for
read-heavy workloads (the prime use cases for caches), performance is
nearly identical.

In the worst case (pure write workloads with cost accounting enabled),
we're looking at ~1.5us per insert on large caches. That seems "fast
enough."

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Event Timeline

indygreg created this revision.Sep 6 2018, 9:17 PM
Closed by commit rHGf296c0b366c8: util: lower water mark when removing nodes after cost limit reached (authored by indygreg). · Explain WhySep 12 2018, 10:53 AM
This revision was automatically updated to reflect the committed changes.

Revision Contents
Changeset List

PathPackages
Mmercurial/util.py (12 lines)
Mtests/test-lrucachedict.py (4 lines)

Diff 10824

mercurial/util.py

if len(self) <= 1 or self.totalcost <= self.maxcost: if len(self) <= 1 or self.totalcost <= self.maxcost:
return return
# This is logically equivalent to calling popoldest() until we # This is logically equivalent to calling popoldest() until we
# free up enough cost. We don't do that since popoldest() needs # free up enough cost. We don't do that since popoldest() needs
# to walk the linked list and doing this in a loop would be # to walk the linked list and doing this in a loop would be
# quadratic. So we find the first non-empty node and then # quadratic. So we find the first non-empty node and then
# walk nodes until we free up enough capacity. # walk nodes until we free up enough capacity.
#
# If we only removed the minimum number of nodes to free enough
# cost at insert time, chances are high that the next insert would
# also require pruning. This would effectively constitute quadratic
# behavior for insert-heavy workloads. To mitigate this, we set a
# target cost that is a percentage of the max cost. This will tend
# to free more nodes when the high water mark is reached, which
# lowers the chances of needing to prune on the subsequent insert.
targetcost = int(self.maxcost * 0.75)
n = self._head.prev n = self._head.prev
while n.key is _notset: while n.key is _notset:
n = n.prev n = n.prev
while len(self) > 1 and self.totalcost > self.maxcost: while len(self) > 1 and self.totalcost > targetcost:
del self._cache[n.key] del self._cache[n.key]
self.totalcost -= n.cost self.totalcost -= n.cost
n.markempty() n.markempty()
n = n.prev n = n.prev
def lrucachefunc(func): def lrucachefunc(func):
'''cache most recent results of function calls''' '''cache most recent results of function calls'''
cache = {} cache = {}

tests/test-lrucachedict.py

d.insert('b', 'vb', cost=21) d.insert('b', 'vb', cost=21)
d.insert('c', 'vc', cost=7) d.insert('c', 'vc', cost=7)
d.insert('d', 'vc', cost=50) d.insert('d', 'vc', cost=50)
self.assertEqual(d.totalcost, 87) self.assertEqual(d.totalcost, 87)
# Inserting new element should free multiple elements so we hit # Inserting new element should free multiple elements so we hit
# low water mark. # low water mark.
d.insert('e', 'vd', cost=25) d.insert('e', 'vd', cost=25)
self.assertEqual(len(d), 3) self.assertEqual(len(d), 2)
self.assertNotIn('a', d) self.assertNotIn('a', d)
self.assertNotIn('b', d) self.assertNotIn('b', d)
self.assertIn('c', d) self.assertNotIn('c', d)
self.assertIn('d', d) self.assertIn('d', d)
self.assertIn('e', d) self.assertIn('e', d)
if __name__ == '__main__': if __name__ == '__main__':
silenttestrunner.main(__name__) silenttestrunner.main(__name__)