Rémi Lapeyre schrieb am 01.03.19 um 15:50:

Le 1 mars 2019 à 15:41:52, Stefan Behnel a écrit:

...

Rémi Lapeyre schrieb am 01.03.19 um 15:06:

...

I’m having issues to understand the semantics of d1 + d2.

I think mappings are more complicated than sequences it some things seems not obvious to me.

What would be OrderedDict1 + OrderedDict2, in which positions would be the resulting keys, which value would be used if the same key is present in both?

The only reasonable answer I can come up with is:

1) unique keys from OrderedDict1 are in the same order as before 2) duplicate keys and new keys from OrderedDict2 come after the keys from d1, in their original order in d2 since they replace keys in d1.

Basically, the expression says: "take a copy of d1 and add the items from d2 to it". That's exactly what you should get, whether the mappings are ordered or not (and dict are ordered by insertion in Py3.6+).

Thanks Stefan for your feedback, unless I’m mistaken this does not work like Rhodri suggested, he said:

I can tell it's pretty straightforward:

d = d1 + d2 is equivalent to:

...

...

...

d = d1.copy() d.update(d2)

But doing this:

...

...

...

d1 = OrderedDict({"a": 1, "b": 2, "c": 3}) d2 = OrderedDict({"d": 4, "b": 5}) d = d1.copy() d.update(d2) d OrderedDict([('a', 1), ('b', 5), ('c', 3), ('d', 4)])

It looks like that the semantics are either not straightforward or what you proposed is not the only reasonable answer. Am I missing something?

No, I was, apparently. In Py3.7:

...

...

d1 = {"a": 1, "b": 2, "c": 3} d1 {'a': 1, 'b': 2, 'c': 3} d2 = {"d": 4, "b": 5} d = d1.copy() d.update(d2) d {'a': 1, 'b': 5, 'c': 3, 'd': 4}

I think the behaviour makes sense when you know how it's implemented (keys are stored separately from values). I would have been less surprised if the keys had also been reordered, but well, this is how it is now in Py3.6+, so this is how it's going to work also for the operator. No *additional* surprises here. ;) Stefan