How to check if a value is present in any of given sets

Question

Say I have different sets (they have to be different, I cannot join them as per the kind of data I am working with):

r = set([1,2,3])
s = set([4,5,6])
t = set([7,8,9])

What is the best way to check if a given variable is present in any of them?

I am using:

if myvar in r \
   or myvar in s \
   or myvar in t:

But I wonder if this can be reduced somehow by using set's properties such as union.

The following works, but I can't find a way to define multiple unions:

if myvar in r.union(s)
   or myvar in t:

I am also wondering if this union will somehow affect performance, since I guess a temporary set will be created on the fly.

Answer 1

Just use any:

if any(myvar in x for x in  (r,s,t))

set lookups are 0(1) so creating a union to check if the variable is in any set is totally unnecessary instead of simply checking using in with any which will short circuit as soon as a match is found and does not create a new set.

And I am also wondering if this union will affect somehow performance

Yes of course unioning the sets affects performance, it adds to the complexity, you are creating a new set every time which is O(len(r)+len(s)+len(t)) so you can say goodbye to the real point of using sets which are efficient lookups.

So the bottom line is that is you want to keep efficient lookups you will have to union the set once and keep them in memory creating a new variable then using that to do your lookup for myvar so the initial creation will be 0(n) and lookups will be 0(1) thereafter.

If you don't every time you want to do a lookup first creating the union you will have a linear solution in the length of r+s+t -> set.union(*(r, s, t)) as opposed to at worst three constant(on average) lookups. That also means always adding or removing any elements from the new unioned set that are added/removed from r,s or t.

Some realistic timings on moderately large sized sets show exactly the difference:

In [1]: r = set(range(10000))

In [2]: s = set(range(10001,20000))

In [3]: t = set(range(20001,30000))

In [4]: timeit any(29000 in st for st in (r,s,t))
1000000 loops, best of 3: 869 ns per loop  

In [5]: timeit 29000 in r | s | t
1000 loops, best of 3: 956 µs per loop

In [6]: timeit 29000 in reduce(lambda x,y :x.union(y),[r,s,t])
1000 loops, best of 3: 961 µs per loop

In [7]: timeit 29000 in r.union(s).union(t)
1000 loops, best of 3: 953 µs per loop

Timing the union shows that pretty much all the time is spent in the union calls:

In [8]: timeit r.union(s).union(t)
1000 loops, best of 3: 952 µs per loop

Using larger sets and getting the element in the last set:

In [15]: r = set(range(1000000))

In [16]: s = set(range(1000001,2000000))

In [17]: t = set(range(2000001,3000000))


In [18]: timeit any(2999999 in st for st in (r,s,t))
1000000 loops, best of 3: 878 ns per loop

In [19]: timeit 2999999 in reduce(lambda x,y :x.union(y),[r,s,t])
1 loops, best of 3: 161 ms per loop

In [20]: timeit 2999999 in r | s | t
10 loops, best of 3: 157 ms per loop

There is literally no difference no matter how large the sets get using any but as the set sizes grow so does the running time using union.

The only way to make it faster would be to stick to or but we are taking the difference of a few hundred nanoseconds which is the cost of creating the generator expression and the function call:

In [22]: timeit 2999999 in r or 2999999 in s or 2999999 in t
10000000 loops, best of 3: 152 ns per loop

To union sets set.union(*(r, s, t)) is also the fastest as you don't build intermediary sets:

In [47]: timeit 2999999 in set.union(*(r,s,t))
10 loops, best of 3: 108 ms per loop
In [49]: r | s | t  == set.union(*(r,s,t))
Out[49]: True

Answer 2

You can use builtin any:

r = set([1,2,3])
s = set([4,5,6])
t = set([7,8,9])
if any(myvar in x for x in [r,s,t]):
    print "I'm in one of them"

any will short circuit on the first condition that returns True so you can get around constructing a potentially huge union or checking potentially lots of sets for inclusion.

And I am also wondering if this union will affect somehow performance, since I guess a temporary set will be created on the fly.

According to wiki.python.com s|t is O(len(s)+len(t)) while lookups are O(1) .

For n sets with l elements each , doing union iteratively to construct the set will result in:

a.union(b).union(c).union(d) .... .union(n)

Which is equivalent to O(l+l) for a.union(b) and O(2l+2l+l) a.union(b).union(c) and so on which sums up to O(n*(n+1)/2)*l).

O(n^2*l) is quadratic and voids the performance advantage of using sets.

The lookup in n sets with any will perform at O(n)

Answer 3

| is a union operator of sets in python. You can define union over multiple sets using | as:

>>> r = set([1,2,3])
>>> s = set([4,5,6])
>>> t = set([7,8,9])
>>> r | s | t
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

Answer 4

You can use reduce function to apply function of two arguments cumulatively to the items of iterable:

>>> r = set([1,2,3])
>>> s = set([4,5,6])
>>> t = set([7,8,9])
>>> 
>>> reduce(lambda x,y :x.union(y),[r,s,t])
set([1, 2, 3, 4, 5, 6, 7, 8, 9])

And for checking the membership in either of them you can use a generator expression within any that is more efficient here because python use hash table for storing the sets and checking the member ship has O(1) in such data structures like dictionaries or frozenset.Also for check the membership in all of you sets use all.

if any(i in item for item in [r,s,t]):
    #do stuff

But in this case (Not for large sets) using or operator is faster.

value in r|s|t 

This is a benchmark on all the ways :

~$ python -m timeit "r = set([1,2,3]);s = set([4,5,6]);t = set([7,8,9]);3 in reduce(lambda x,y :x.union(y),[r,s,t])"
1000000 loops, best of 3: 1.55 usec per loop
~$ python -m timeit "r = set([1,2,3]);s = set([4,5,6]);t = set([7,8,9]);3 in r|s|t"
1000000 loops, best of 3: 1.11 usec per loop
~$ python -m timeit "r = set([1,2,3]);s = set([4,5,6]);t = set([7,8,9]);any(3 in item for item in [r,s,t])"
1000000 loops, best of 3: 1.24 usec per loop
~$ python -m timeit "r = set([1,2,3]);s = set([4,5,6]);t = set([7,8,9]);3 in r.union(s).union(t)"
1000000 loops, best of 3: 1.19 usec per loop

Note that as @Padraic Cunningham mentioned for large sets using a any is very much efficient!

Answer 5

You can simply do

if myvar in r.union(s).union(t)

And you needn't worry about performance here. Yes it creates a temporary set on the fly but as it isn't stored gets garbage collected.