fastest way to find the intersection of n lists of sets

[Prateek]

I have 3 variable length lists of sets. I need to find the common
elements in each list (across sets) really really quickly.

Here is some sample code:

# Doesn't make sense to union the sets - we're going to do
intersections later anyway
l1 = reduce(operator.add, list(x) for x in l1)
l2 = reduce(operator.add, list(x) for x in l2)
l3 = reduce(operator.add, list(x) for x in l3)

# Should I do this in two steps? Maybe by intersecting the two
shortest lists first?
s = frozenset(l1) & frozenset(l2) & frozenset(l3)

I'm assuming frozensets are (somehow) quicker than sets because
they're immutable.

Any code suggestions? Maybe using something in the new fancy-schmancy
itertools module?

Thanks,
Prateek

 

[James Stroud]

I have 3 variable length lists of sets. I need to find the common
elements in each list (across sets) really really quickly.

Here is some sample code:

# Doesn't make sense to union the sets - we're going to do
intersections later anyway
l1 = reduce(operator.add, list(x) for x in l1)
l2 = reduce(operator.add, list(x) for x in l2)
l3 = reduce(operator.add, list(x) for x in l3)

# Should I do this in two steps? Maybe by intersecting the two
shortest lists first?
s = frozenset(l1) & frozenset(l2) & frozenset(l3)

I'm assuming frozensets are (somehow) quicker than sets because
they're immutable.

Any code suggestions? Maybe using something in the new fancy-schmancy
itertools module?

Thanks,
Prateek

I don't understand why you cast to list. I would propose:

lists_of_sets = [l1, l2, l3]

reduce(set.intersection, (reduce(set.union, x) for x in lists_of_sets))

Since this is simplest, I'm guessing it should be fastest because I'm
also guessing that set impelmentation is as optimized as list--I think
this would hold true especially for large sets with sparse overlap
between lists. So it might be reasonable consider the content of your
sets and lists and write your code based on the content or on assuming a
particular composition.

James

 

[John Nagle]

I have 3 variable length lists of sets. I need to find the common
elements in each list (across sets) really really quickly.

Here is some sample code:

# Doesn't make sense to union the sets - we're going to do
intersections later anyway
l1 = reduce(operator.add, list(x) for x in l1)
l2 = reduce(operator.add, list(x) for x in l2)
l3 = reduce(operator.add, list(x) for x in l3)

# Should I do this in two steps? Maybe by intersecting the two
shortest lists first?
s = frozenset(l1) & frozenset(l2) & frozenset(l3)

I'm assuming frozensets are (somehow) quicker than sets because
they're immutable.

Any code suggestions? Maybe using something in the new fancy-schmancy
itertools module?

Thanks,
Prateek

I don't understand why you cast to list. I would propose:

lists_of_sets = [l1, l2, l3]

reduce(set.intersection, (reduce(set.union, x) for x in lists_of_sets))

Since this is simplest, I'm guessing it should be fastest because I'm
also guessing that set impelmentation is as optimized as list--I think
this would hold true especially for large sets with sparse overlap
between lists. So it might be reasonable consider the content of your
sets and lists and write your code based on the content or on assuming a
particular composition.

Python sets are hashes, like dictionaries, not trees. Intersection
is implemented by iterating over the smallest set and trying all its keys
on the other set. The Python implementation compares the length of two
sets being intersected. This is OK (it's about O(N log N), maybe better).

For the above example, it's worth sorting lists_of_sets by the
length of the sets, and doing the short ones first.

How big are the sets? If they're small, but you have a lot of
them, you may be better off with a bit-set representation, then
using AND operations for intersection. If they're huge (tens of millions
of entries), you might be better off doing sorts and merges on the
sets.

When you ask questions like this, it's helpful to give some
background. We don't know whether this is a homework assignment, or
some massive application that's slow and you need to fix it, even
if it requires heavy implementation effort.

John Nagle

 

[Prateek]

For the above example, it's worth sorting lists_of_sets by the

length of the sets, and doing the short ones first.

Thanks. I thought so - I'm doing just that using a simple Decorate-
Sort-Undecorate idiom.

How big are the sets? If they're small, but you have a lot of
them, you may be better off with a bit-set representation, then
using AND operations for intersection. If they're huge (tens of millions
of entries), you might be better off doing sorts and merges on the
sets.

I have either 2 or 3 sets (never more) which can be arbitrarily large.
Most of them are small (between 0 and few elements - say less that 5).
A few will be megamonstrous ( > 100,000 items)

When you ask questions like this, it's helpful to give some
background. We don't know whether this is a homework assignment, or
some massive application that's slow and you need to fix it, even
if it requires heavy implementation effort.

Its definitely not a homework assignment - its part of a commercial
database query engine. Heavy implementation effort is no problem.

Prateek

 

[Prateek]

I don't understand why you cast to list. I would propose:

The reason why I'm casting to a list first is because I found that
creating a long list which I convert to a set in a single operation is
faster (although probably less memory efficient - which I can deal
with) than doing all the unions.

Prateek

 

[John Nagle]

Thanks. I thought so - I'm doing just that using a simple Decorate-
Sort-Undecorate idiom.




I have either 2 or 3 sets (never more) which can be arbitrarily large.
Most of them are small (between 0 and few elements - say less that 5).
A few will be megamonstrous ( > 100,000 items)



Its definitely not a homework assignment - its part of a commercial
database query engine. Heavy implementation effort is no problem.

Prateek

If you're intersecting a set of 5 vs a set of 100,000, the
intersection time won't be the problem. That's just five lookups.
It's building a set of 100,000 items that may be time consuming.

Does the big set stay around for a while, or do you have to pay
that cost on each query?

Those really aren't big data sets on modern machines.

John Nagle

 

[Prateek]

If you're intersecting a set of 5 vs a set of 100,000, the
intersection time won't be the problem. That's just five lookups.
It's building a set of 100,000 items that may be time consuming.

Does the big set stay around for a while, or do you have to pay
that cost on each query?

Those really aren't big data sets on modern machines.

John Nagle

100,000 is an arbitrary number - that is potentially equivalent to the
number of unique cells in all tables of a typical database (thats the
best analogy I can come up with since this isn't a typical RDBMS).

The big set does stay around for a while - I've implemented an LRU
based caching algorithm on the code that does the I/O. Since the db is
transactioned, I keep one copy in the current transaction cache (which
is a simple dictionary) and one in the main read cache (LRU based)
(which obviously survives across transactions). Since the largest sets
also tend to be the most frequently used, this basically solves my I/O
caching issue.

My problem is that I had ugly code doing a lot of unnecessary list <->
set casting. Here is what I've got now:

from itertools import chain
ids1 = [...], ids2 = [...], ids3 = [...]
_efs = frozenset()
# dbx.get calls return sets
l1 = frozenset(chain(*[db1.get(x, _efs) for x in ids1])
l2 = frozenset(chain(*[db2.get(x, _efs) for x in ids2])
l3 = frozenset(chain(*[db3.get(x, _efs) for x in ids3])

decorated = [(len(x), x) for x in [l1, l2, l3]]
decorated.sort()
result = reduce(set.intersection, [x[1] for x in decorated])

What do you think?

Prateek

 

[Paul Rubin]

The big set does stay around for a while - I've implemented an LRU
based caching algorithm on the code that does the I/O. Since the db is
transactioned, I keep one copy in the current transaction cache (which
is a simple dictionary) and one in the main read cache (LRU based)
(which obviously survives across transactions). Since the largest sets
also tend to be the most frequently used, this basically solves my I/O
caching issue.

The best approach varies from instance to instance. Some big
databases often will do stuff like statistically sample the sets from
a given query, try a few different strategies on the samples in order
to figure out which one works best on them, then use the apparent best
strategy on the full sets.

 

[Alex Martelli]

Thanks. I thought so - I'm doing just that using a simple Decorate-
Sort-Undecorate idiom.

Use, instead, the DSU that is now build into Python:

listofsets.sort(key=len)

this will be faster (as well as more readable &c) than programming your
own DSU, and is exactly the reason the key= parameter was added.

I also suggest avoiding reduce in favor of a simple explicit loop.


Alex

 

[Raymond Hettinger]

[Prateek]

I have 3 variable length lists of sets. I need to find the common
elements in each list (across sets) really really quickly. .. . .
l1 = reduce(operator.add, list(x) for x in l1)
l2 = reduce(operator.add, list(x) for x in l2)
l3 = reduce(operator.add, list(x) for x in l3)
s = frozenset(l1) & frozenset(l2) & frozenset(l3)

I would write it like this:

def multi_union(listofsets):
result = set()
for s in listofsets:
result |= s
return result

def multi_intersection(listofsets):
return reduce(set.intersection, sorted(listofsets, key=len))

s = multi_intersection(map(multi_union, [l1, l2, l3]))

I'm assuming frozensets are (somehow) quicker than sets because
they're immutable.

Frozensets share the same implementation code as regular sets.
They are not more efficient.

Any code suggestions? Maybe using something in the new fancy-schmancy
itertools module?

The sets.py module shows how to implement set operations using
itertools.
In general, real set objects should do as well or better than anything
you can cook-up using itertools.

Real set objects have the advantage of knowing the hash values of
their
elements. Accordingly, binary set operations can run without any
calls to element.__hash__().


Raymond Hettinger

 

[Raymond Hettinger]

[Prateek]

The reason why I'm casting to a list first is because I found that
creating a long list which I convert to a set in a single operation is
faster (although probably less memory efficient - which I can deal
with) than doing all the unions.

That would be a surprising result because set-to-set operations do
not have to recompute hash values. Also, underneath-the-hood,
both approaches share the exact same implementation for inserting
new values one the hash value is known.

If you've seen an unfavorable speed comparison, then you most likely
had code that built new intermediate sets between step:

common = s1 | s2 | s3 | s4 | s5 | s6 | s7 | s8 | s9

Instead, it is faster to build-up a single result set:

common = set()
for s in s1, s2, s3, s4, s5, s6, s7, s8, s9:
common |= s


Raymond Hettinger

 

[Prateek]

[Prateek]

The reason why I'm casting to a list first is because I found that
creating a long list which I convert to a set in a single operation is
faster (although probably less memory efficient - which I can deal
with) than doing all the unions.

That would be a surprising result because set-to-set operations do
not have to recompute hash values. Also, underneath-the-hood,
both approaches share the exact same implementation for inserting
new values one the hash value is known.

If you've seen an unfavorable speed comparison, then you most likely
had code that built new intermediate sets between step:

common = s1 | s2 | s3 | s4 | s5 | s6 | s7 | s8 | s9

Instead, it is faster to build-up a single result set:

common = set()
for s in s1, s2, s3, s4, s5, s6, s7, s8, s9:
common |= s

Raymond Hettinger

Thanks Raymond,

This was my old code:
self.lv is a dictionary which retrieves data from the disk or cache
v_r = reduce(operator.or_, [self.lv[x.id] for x in v], set())

This code ran faster:
v_r = reduce(operator.add, [list(self.lv.get(x.id, [])) for x in v],
[])
v_r = set(v_r)

I was doing 3 of these and then intersecting them.

Now, I'm doing...
v_r = set()
_efs = frozenset()
for y in [self.lv.get(x.id, _efs) for x in v]:
v_r |= y

Since the number of sets is always 2 or 3, I just do the intersections
explicitly like so:
if len(list_of_unioned_sets) == 3:
result = list_of_unioned_sets[0]
result &= list_of_unioned_sets[1]
result &= list_of_unioned_sets[2]
elif len(list_of_unioned_sets) == 2:
result = list_of_unioned_sets[0]
result &= list_of_unioned_sets[1]
else:
# Do something else...

Sorry for the relatively non-descript variable names.
Prateek