Michael Lee wrote:
> If your instinct is to assume "+" means "concatenation", then it would be
> natural to assume that {"a": 1, "b": 2} + {"c": 3, "b": 4} would be identical
> to {"a": 1, "b": 2, "c": 3, "b": 4} -- literally concat the key-value pairs
> into a new dict.
> But of course, you can't have duplicate keys in Python. So, you would either
> recall or look up how duplicate keys are handled when constructing a dict and
> learn that the rule is that the right-most key wins. So the natural
> conclusion is that "+" would follow this existing rule -- and you end up with
> exactly the behavior described in the PEP.
This is a nice argument. And well presented. And it gave me surprise,
that taught me something. Here goes:
>>> {'a': 0}
{'a': 0}
>>> {'a': 0, 'a': 0}
{'a': 0}
>>> {'a': 0, 'a': 1}
{'a': 1}
>>> {'a': 1, 'a': 0}
{'a': 0}
This surprised me quite a bit. I was expecting to get an exception. However
>>> dict(a=0)
{'a': 0}
>>> dict(a=0, a=0)
SyntaxError: keyword argument repeated
does give an exception.
I wonder, is this behaviour of {'a': 0, 'a': 1} documented (or tested)
anywhere? I didn't find it in these URLs:
https://docs.python.org/3/library/stdtypes.html#mapping-types-dict
https://docs.python.org/3/tutorial/datastructures.html#dictionaries
I think this behaviour might give rise to gotchas. For example, if we
define inverse_f by
>>> inverse_f = { f(a): a, f(b): b }
then is the next statement always true (assuming a <> b)?
>>> inverse_f[ f(a) ] == a
Well, it's not true with these values
>>> a, b = 1, 2
>>> def f(n): pass # There's a bug here, f(n) should be a bijection.
A quick check that len(inverse) == 2 would provide a sanity check. Or
perhaps better, len(inverse_f) == len(set(a, b)). (I don't have an
example of this bug appearing 'in the wild'.)
Once again, I thank Michael for his nice, instructive and
well-presented example.
--
Jonathan
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