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Description
If something is not subset, then it cannot be sure is superset or not, it must be tested
orderedset/lib/orderedset/_orderedset.pyx
Line 505 in 9b83075
| return not ret |
>>> {1}.issubset({2, 3})
False
>>> {1} <= {2, 3}
False
>>> OrderedSet([1]).issubset(OrderedSet([2, 3]))
False
>>> OrderedSet([1]) <= OrderedSet([2, 3])
False
>>> {1}.issuperset({2, 3})
False
>>> {1} >= {2, 3}
False
>>> {1} > {2, 3}
False
>>> OrderedSet([1]).issuperset(OrderedSet([2, 3]))
False
>>> OrderedSet([1]) >= OrderedSet([2, 3])
True
>>> OrderedSet([1]) > OrderedSet([2, 3])
True
Edit: Oh, this may be solved by pull #22
Compare between OrderedSet and set
>>> OrderedSet([1]).issubset({3})
False
>>> OrderedSet([1]).issuperset({3}) # What is return {3} <= OrderedSet([1]), but…
True
>>> {3} <= OrderedSet([1])
True
>>> {3}.issubset(OrderedSet([1]))
False
>>> {3}.issuperset(OrderedSet([1]))
False
Compare between OrderedSet and list
>>> OrderedSet([1]).issuperset([3])
True
>>> OrderedSet([1]).issubset([3])
True
orderedset/lib/orderedset/_orderedset.pyx
Line 495 in 9b83075
| if isinstance(other, Set): |
It may be:
isinstance(other, (Set, list))