util: teach lrucachedict to enforce a max total cost

Now that lrucachedict entries can have a numeric cost associated
with them and we can easily pop the oldest item in the cache, it
now becomes relatively trivial to implement support for enforcing
a high water mark on the total cost of items in the cache.

This commit teaches lrucachedict instances to have a max cost
associated with them. When items are inserted, we pop old items
until enough "cost" frees up to make room for the new item.

This feature is close to zero cost when not used (modulo the insertion
regressed introduced by the previous commit):

$ ./hg perflrucachedict --size 4 --gets 1000000 --sets 1000000 --mixed 1000000
! gets
! wall 0.607444 comb 0.610000 user 0.610000 sys 0.000000 (best of 17)
! wall 0.601653 comb 0.600000 user 0.600000 sys 0.000000 (best of 17)
! inserts
! wall 0.678261 comb 0.680000 user 0.680000 sys 0.000000 (best of 14)
! wall 0.685042 comb 0.680000 user 0.680000 sys 0.000000 (best of 15)
! sets
! wall 0.808770 comb 0.800000 user 0.800000 sys 0.000000 (best of 13)
! wall 0.834241 comb 0.830000 user 0.830000 sys 0.000000 (best of 12)
! mixed
! wall 0.782441 comb 0.780000 user 0.780000 sys 0.000000 (best of 13)
! wall 0.803804 comb 0.800000 user 0.800000 sys 0.000000 (best of 13)

$ hg perflrucachedict --size 1000 --gets 1000000 --sets 1000000 --mixed 1000000
! init
! wall 0.006952 comb 0.010000 user 0.010000 sys 0.000000 (best of 418)
! gets
! wall 0.613350 comb 0.610000 user 0.610000 sys 0.000000 (best of 17)
! wall 0.617415 comb 0.620000 user 0.620000 sys 0.000000 (best of 17)
! inserts
! wall 0.701270 comb 0.700000 user 0.700000 sys 0.000000 (best of 15)
! wall 0.700516 comb 0.700000 user 0.700000 sys 0.000000 (best of 15)
! sets
! wall 0.825720 comb 0.830000 user 0.830000 sys 0.000000 (best of 13)
! wall 0.837946 comb 0.840000 user 0.830000 sys 0.010000 (best of 12)
! mixed
! wall 0.821644 comb 0.820000 user 0.820000 sys 0.000000 (best of 13)
! wall 0.850559 comb 0.850000 user 0.850000 sys 0.000000 (best of 12)

I reckon the slight slowdown on insert is due to added if checks.

For caches with total cost limiting enabled:

$ hg perflrucachedict --size 4 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 100
! gets w/ cost limit
! wall 0.598737 comb 0.590000 user 0.590000 sys 0.000000 (best of 17)
! inserts w/ cost limit
! wall 1.694282 comb 1.700000 user 1.700000 sys 0.000000 (best of 6)
! mixed w/ cost limit
! wall 1.157655 comb 1.150000 user 1.150000 sys 0.000000 (best of 9)

$ hg perflrucachedict --size 1000 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 10000
! gets w/ cost limit
! wall 0.598526 comb 0.600000 user 0.600000 sys 0.000000 (best of 17)
! inserts w/ cost limit
! wall 37.838315 comb 37.840000 user 37.840000 sys 0.000000 (best of 3)
! mixed w/ cost limit
! wall 18.060198 comb 18.060000 user 18.060000 sys 0.000000 (best of 3)

$ hg perflrucachedict --size 1000 --gets 1000000 --sets 1000000 --mixed 1000000 --costlimit 10000 --mixedgetfreq 90
! gets w/ cost limit
! wall 0.600024 comb 0.600000 user 0.600000 sys 0.000000 (best of 17)
! inserts w/ cost limit
! wall 37.154547 comb 37.120000 user 37.120000 sys 0.000000 (best of 3)
! mixed w/ cost limit
! wall 4.381602 comb 4.380000 user 4.370000 sys 0.010000 (best of 3)

The functions we're benchmarking are slightly different, which could
move numbers by a few milliseconds. But the slowdown on insert is too
great to be explained by that. The slowness is due to insert heavy
operations needing to call popoldest() repeatedly when the cache is
at capacity. The next commit will address this.

Differential Revision: https://phab.mercurial-scm.org/D4503

contrib/perf.py

@@ -1869,18 +1869,23 b' def perfloadmarkers(ui, repo):'
     fm.end()
 
 @command(b'perflrucachedict', formatteropts +
-    [(b'', b'size', 4, b'size of cache'),
+    [(b'', b'costlimit', 0, b'maximum total cost of items in cache'),
+     (b'', b'mincost', 0, b'smallest cost of items in cache'),
+     (b'', b'maxcost', 100, b'maximum cost of items in cache'),
+     (b'', b'size', 4, b'size of cache'),
      (b'', b'gets', 10000, b'number of key lookups'),
      (b'', b'sets', 10000, b'number of key sets'),
      (b'', b'mixed', 10000, b'number of mixed mode operations'),
      (b'', b'mixedgetfreq', 50, b'frequency of get vs set ops in mixed mode')],
     norepo=True)
-def perflrucache(ui, size=4, gets=10000, sets=10000, mixed=10000,
-                 mixedgetfreq=50, **opts):
+def perflrucache(ui, mincost=0, maxcost=100, costlimit=0, size=4,
+                 gets=10000, sets=10000, mixed=10000, mixedgetfreq=50, **opts):
     def doinit():
         for i in xrange(10000):
             util.lrucachedict(size)
 
+    costrange = list(range(mincost, maxcost + 1))
+
     values = []
     for i in xrange(size):
         values.append(random.randint(0, sys.maxint))
@@ -1899,16 +1904,34 b' def perflrucache(ui, size=4, gets=10000,'
             value = d[key]
             value # silence pyflakes warning
 
+    def dogetscost():
+        d = util.lrucachedict(size, maxcost=costlimit)
+        for i, v in enumerate(values):
+            d.insert(v, v, cost=costs[i])
+        for key in getseq:
+            try:
+                value = d[key]
+                value # silence pyflakes warning
+            except KeyError:
+                pass
+
     # Set mode tests insertion speed with cache eviction.
     setseq = []
+    costs = []
     for i in xrange(sets):
         setseq.append(random.randint(0, sys.maxint))
+        costs.append(random.choice(costrange))
 
     def doinserts():
         d = util.lrucachedict(size)
         for v in setseq:
             d.insert(v, v)
 
+    def doinsertscost():
+        d = util.lrucachedict(size, maxcost=costlimit)
+        for i, v in enumerate(setseq):
+            d.insert(v, v, cost=costs[i])
+
     def dosets():
         d = util.lrucachedict(size)
         for v in setseq:
@@ -1923,12 +1946,14 b' def perflrucache(ui, size=4, gets=10000,'
         else:
             op = 1
 
-        mixedops.append((op, random.randint(0, size * 2)))
+        mixedops.append((op,
+                         random.randint(0, size * 2),
+                         random.choice(costrange)))
 
     def domixed():
         d = util.lrucachedict(size)
 
-        for op, v in mixedops:
+        for op, v, cost in mixedops:
             if op == 0:
                 try:
                     d[v]
@@ -1937,13 +1962,35 b' def perflrucache(ui, size=4, gets=10000,'
             else:
                 d[v] = v
 
+    def domixedcost():
+        d = util.lrucachedict(size, maxcost=costlimit)
+
+        for op, v, cost in mixedops:
+            if op == 0:
+                try:
+                    d[v]
+                except KeyError:
+                    pass
+            else:
+                d.insert(v, v, cost=cost)
+
     benches = [
         (doinit, b'init'),
+    ]
+
+    if costlimit:
+        benches.extend([
+            (dogetscost, b'gets w/ cost limit'),
+            (doinsertscost, b'inserts w/ cost limit'),
+            (domixedcost, b'mixed w/ cost limit'),
+        ])
+    else:
+        benches.extend([
         (dogets, b'gets'),
         (doinserts, b'inserts'),
         (dosets, b'sets'),
         (domixed, b'mixed')
-    ]
+        ])
 
     for fn, title in benches:
         timer, fm = gettimer(ui, opts)

mercurial/util.py

@@ -1240,8 +1240,14 b' class lrucachedict(object):'
     Items in the cache can be inserted with an optional "cost" value. This is
     simply an integer that is specified by the caller. The cache can be queried
     for the total cost of all items presently in the cache.
+
+    The cache can also define a maximum cost. If a cache insertion would
+    cause the total cost of the cache to go beyond the maximum cost limit,
+    nodes will be evicted to make room for the new code. This can be used
+    to e.g. set a max memory limit and associate an estimated bytes size
+    cost to each item in the cache. By default, no maximum cost is enforced.
     """
-    def __init__(self, max):
+    def __init__(self, max, maxcost=0):
         self._cache = {}
 
         self._head = head = _lrucachenode()
@@ -1250,6 +1256,7 b' class lrucachedict(object):'
         self._size = 1
         self.capacity = max
         self.totalcost = 0
+        self.maxcost = maxcost
 
     def __len__(self):
         return len(self._cache)
@@ -1279,6 +1286,10 b' class lrucachedict(object):'
             node.cost = cost
             self.totalcost += cost
             self._movetohead(node)
+
+            if self.maxcost:
+                self._enforcecostlimit()
+
             return
 
         if self._size < self.capacity:
@@ -1301,6 +1312,9 b' class lrucachedict(object):'
         # is already self._head.prev.
         self._head = node
 
+        if self.maxcost:
+            self._enforcecostlimit()
+
     def __setitem__(self, k, v):
         self.insert(k, v)
 
@@ -1331,7 +1345,7 b' class lrucachedict(object):'
 
         self._cache.clear()
 
-    def copy(self, capacity=None):
+    def copy(self, capacity=None, maxcost=0):
         """Create a new cache as a copy of the current one.
 
         By default, the new cache has the same capacity as the existing one.
@@ -1343,7 +1357,8 b' class lrucachedict(object):'
         """
 
         capacity = capacity or self.capacity
-        result = lrucachedict(capacity)
+        maxcost = maxcost or self.maxcost
+        result = lrucachedict(capacity, maxcost=maxcost)
 
         # We copy entries by iterating in oldest-to-newest order so the copy
         # has the correct ordering.
@@ -1445,6 +1460,13 b' class lrucachedict(object):'
         self._size += 1
         return node
 
+    def _enforcecostlimit(self):
+        # This should run after an insertion. It should only be called if total
+        # cost limits are being enforced.
+        # The most recently inserted node is never evicted.
+        while len(self) > 1 and self.totalcost > self.maxcost:
+            self.popoldest()
+
 def lrucachefunc(func):
     '''cache most recent results of function calls'''
     cache = {}

tests/test-lrucachedict.py

@@ -133,6 +133,22 b' class testlrucachedict(unittest.TestCase'
         for key in ('a', 'b', 'c', 'd'):
             self.assertEqual(d[key], 'v%s' % key)
 
+        d = util.lrucachedict(4, maxcost=42)
+        d.insert('a', 'va', cost=5)
+        d.insert('b', 'vb', cost=4)
+        d.insert('c', 'vc', cost=3)
+        dc = d.copy()
+        self.assertEqual(dc.maxcost, 42)
+        self.assertEqual(len(dc), 3)
+
+        # Max cost can be lowered as part of copy.
+        dc = d.copy(maxcost=10)
+        self.assertEqual(dc.maxcost, 10)
+        self.assertEqual(len(dc), 2)
+        self.assertEqual(dc.totalcost, 7)
+        self.assertIn('b', dc)
+        self.assertIn('c', dc)
+
     def testcopydecreasecapacity(self):
         d = util.lrucachedict(5)
         d.insert('a', 'va', cost=4)
@@ -217,5 +233,93 b' class testlrucachedict(unittest.TestCase'
         d['a'] = 'va'
         self.assertEqual(d.popoldest(), ('b', 'vb'))
 
+    def testmaxcost(self):
+        # Item cost is zero by default.
+        d = util.lrucachedict(6, maxcost=10)
+        d['a'] = 'va'
+        d['b'] = 'vb'
+        d['c'] = 'vc'
+        d['d'] = 'vd'
+        self.assertEqual(len(d), 4)
+        self.assertEqual(d.totalcost, 0)
+
+        d.clear()
+
+        # Insertion to exact cost threshold works without eviction.
+        d.insert('a', 'va', cost=6)
+        d.insert('b', 'vb', cost=4)
+
+        self.assertEqual(len(d), 2)
+        self.assertEqual(d['a'], 'va')
+        self.assertEqual(d['b'], 'vb')
+
+        # Inserting a new element with 0 cost works.
+        d['c'] = 'vc'
+        self.assertEqual(len(d), 3)
+
+        # Inserting a new element with cost putting us above high
+        # water mark evicts oldest single item.
+        d.insert('d', 'vd', cost=1)
+        self.assertEqual(len(d), 3)
+        self.assertEqual(d.totalcost, 5)
+        self.assertNotIn('a', d)
+        for key in ('b', 'c', 'd'):
+            self.assertEqual(d[key], 'v%s' % key)
+
+        # Inserting a new element with enough room for just itself
+        # evicts all items before.
+        d.insert('e', 've', cost=10)
+        self.assertEqual(len(d), 1)
+        self.assertEqual(d.totalcost, 10)
+        self.assertIn('e', d)
+
+        # Inserting a new element with cost greater than threshold
+        # still retains that item.
+        d.insert('f', 'vf', cost=11)
+        self.assertEqual(len(d), 1)
+        self.assertEqual(d.totalcost, 11)
+        self.assertIn('f', d)
+
+        # Inserting a new element will evict the last item since it is
+        # too large.
+        d['g'] = 'vg'
+        self.assertEqual(len(d), 1)
+        self.assertEqual(d.totalcost, 0)
+        self.assertIn('g', d)
+
+        d.clear()
+
+        d.insert('a', 'va', cost=7)
+        d.insert('b', 'vb', cost=3)
+        self.assertEqual(len(d), 2)
+
+        # Replacing a value with smaller cost won't result in eviction.
+        d.insert('b', 'vb2', cost=2)
+        self.assertEqual(len(d), 2)
+
+        # Replacing a value with a higher cost will evict when threshold
+        # exceeded.
+        d.insert('b', 'vb3', cost=4)
+        self.assertEqual(len(d), 1)
+        self.assertNotIn('a', d)
+
+    def testmaxcostcomplex(self):
+        d = util.lrucachedict(100, maxcost=100)
+        d.insert('a', 'va', cost=9)
+        d.insert('b', 'vb', cost=21)
+        d.insert('c', 'vc', cost=7)
+        d.insert('d', 'vc', cost=50)
+        self.assertEqual(d.totalcost, 87)
+
+        # Inserting new element should free multiple elements so we hit
+        # low water mark.
+        d.insert('e', 'vd', cost=25)
+        self.assertEqual(len(d), 3)
+        self.assertNotIn('a', d)
+        self.assertNotIn('b', d)
+        self.assertIn('c', d)
+        self.assertIn('d', d)
+        self.assertIn('e', d)
+
 if __name__ == '__main__':
     silenttestrunner.main(__name__)