Structural Pattern Matching Python - Matching a Set / Frozenset

Question

I've been playing around with Structural Pattern Matching in Python 3.10 and can't figure out how to get it to match a set. For example I've tried:

a = {1,2,3}

match a:
    case set(1,2,3):
        print('matched')

and I've tried:

a = {1,2,3}

match a:
    case set([1,2,3]):
        print('matched')

As well as:

a = {1,2,3}

match a:
    case [1,2,3] if isinstance(a, set):
        print('matched')

I'm guessing there is a way to do this since we can match other objects and I'm just missing the correct syntax but I can't think of what else to try. Any help would be appreciated! Thanks!

Answer 1

This isn't really how structural pattern matching is meant to be used; the pattern you're matching is more about value than structure. Because of this, I think you'll find that the equivalent if form is much more readable:

if a == {1, 2, 3}:
    print('matched')

With that said...

Python 3.10 doesn't have syntactic support for matching sets; it only has dedicated "display" syntax for sequences and mappings. I think we briefly considered this, but ultimately dropped it because wasn't very useful or intuitive (and it's easy enough to add in a later version).

Fortunately, it's possible to match any value by equality using a qualified (or "dotted") name. If you need to match a set as part of a larger pattern or match block, that's probably the best way to do it:

class Constants:
    SET_123 = {1, 2, 3}

match a:
    case Constants.SET_123:
        print('matched')

It can also be combined with a class pattern if you only want to match sets (and not, for example, frozensets):

match a:
    case set(Constants.SET_123):
        print('matched')

Answer 2

Since the match construct doesn't use set equality to compare sets, you will need to use a guard to do it explicitly:

a = {1,2,3}

match a:
    case _ if a == set([1,2,3]): 
        print('matched')

It's unintutive that sets aren't compared with set equality by default.