"foo" + "bar" != "bar" + "foo"
On Wed, Feb 27, 2019, 12:35 PM George Castillo <gmcas...@gmail.com> wrote:
> The key conundrum that needs to be solved is what to do for `d1 + d2` when
>> there are overlapping keys. I propose to make d2 win in this case, which is
>> what happens in `d1.update(d2)` anyways. If you want it the other way,
>> simply write `d2 + d1`.
>
>
> This would mean that addition, at least in this particular instance, is
> not a commutative operation. Are there other places in Python where this
> is the case?
>
> ~ George
>
>
> On Wed, Feb 27, 2019 at 10:06 AM Guido van Rossum <gu...@python.org>
> wrote:
>
>> On Wed, Feb 27, 2019 at 8:50 AM Rhodri James <rho...@kynesim.co.uk>
>> wrote:
>>
>>> On 27/02/2019 16:25, João Matos wrote:
>>> > I would like to propose that instead of using this (applies to Py3.5
>>> and upwards)
>>> > dict_a = {**dict_a, **dict_b}
>>> >
>>> > we could use
>>> > dict_a = dict_a + dict_b
>>> >
>>> > or even better
>>> > dict_a += dict_b
>>>
>>> While I don't object to the idea of concatenating dictionaries, I feel
>>> obliged to point out that this last is currently spelled
>>> dict_a.update(dict_b)
>>>
>>
>> This is likely to be controversial. But I like the idea. After all, we
>> have `list.extend(x)` ~~ `list += x`. The key conundrum that needs to be
>> solved is what to do for `d1 + d2` when there are overlapping keys. I
>> propose to make d2 win in this case, which is what happens in
>> `d1.update(d2)` anyways. If you want it the other way, simply write `d2 +
>> d1`.
>>
>> --
>> --Guido van Rossum (python.org/~guido)
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