Check for any values in set

Question

Suppose I have the following set:

things = {'foo', 'bar', 'baz'}

I would like to find out if either foo or bar exists in the set. I've tried:

>>> 'foo' in things or 'bar' in things
True

This works, but is there not a more Pythonic way of doing this check without having multiple or statements? I can't find anything in the standard Python set operations that can achieve this. Using {'foo', 'bar'} <= things checks for both, but I want to check for either of them.

Answer 1

As long as you're using sets, you could use:

if {'foo','bar'} & things:
    ...

& indicates set intersection, and the intersection will be truthy whenever it is nonempty.

Answer 2

Talking sets, what you actually want to know is if the intersection is nonempty:

if things & {'foo', 'bar'}:
    # At least one of them is in

And there is always any():

any(t in things for t in ['foo', 'bar'])

Which is nice in case you have a long list of things to check. But for just two things, I prefer the simple or.

Answer 3

You are looking for the intersection of the sets:

things = {'foo', 'bar', 'baz'}

things.intersection({'foo', 'other'})
# {'foo'}

things.intersection('none', 'here')
#set

So, as empty sets are falsy in boolean context, you can do:

if things.intersection({'foo', 'other'}):
    print("some common value")
else:
    print('no one here')

Answer 4

You can use set.isdisjoint:

if not things.isdisjoint({'foo', 'bar'}):
    ...

Or set.intersection:

if things.intersection({'foo', 'bar'}):
    ...

Or any:

if any(thing in things for thing in ['foo', 'bar']):
    ...

Or stick with or, because very often that's actually the most readable solution:

if 'foo' in things or 'bar' in things:
    ...

Answer 5

things = {'foo', 'bar', 'baz'}
any([i in things for i in ['foo', 'bar']])