Python set Union and set Intersection operate differently?

Bilal Akil picture Bilal Akil · Oct 25, 2013 · Viewed 94.2k times · Source

I'm doing some set operations in Python, and I noticed something odd..

>> set([1,2,3]) | set([2,3,4])
set([1, 2, 3, 4])
>> set().union(*[[1,2,3], [2,3,4]])
set([1, 2, 3, 4])

That's good, expected behaviour - but with intersection:

>> set([1,2,3]) & set([2,3,4])
set([2, 3])
>> set().intersection(*[[1,2,3], [2,3,4]])
set([])

Am I losing my mind here? Why isn't set.intersection() operating as I'd expect it to?

How can I do the intersection of many sets as I did with union (assuming the [[1,2,3], [2,3,4]] had a whole bunch more lists)? What would the "pythonic" way be?

Answer

BrenBarn picture BrenBarn · Oct 25, 2013

When you do set() you are creating an empty set. When you do set().intersection(...) you are intersecting this empty set with other stuff. The intersection of an empty set with any other collection of sets is empty.

If you actually have a list of sets, you can get their intersection similar to how you did it.

>>> x = [{1, 2, 3}, {2, 3, 4}, {3, 4, 5}]
>>> set.intersection(*x)
set([3])

You can't do this directly with the way you're doing it, though, because you don't actually have any sets at all in your example with intersection(*...). You just have a list of lists. You should first convert the elements in your list to sets. So if you have

x = [[1,2,3], [2,3,4]]

you should do

x = [set(a) for a in x]