… set twice

for item_a in set_a:
    for item_b in set_b:
        if <some condition:True>:
            set_a.remove(item_a)

As you can see, if there are multiple items in set_b for which the condition evaluates to True, set_a.remove(item_a) will run twice, which will cause KeyError for second time onwards.

To solve this problem, I am checking whether the item_a is there in the set_a before removing. If it is, then remove, if it isn't then continue without removing as we have already removed it