"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) >> _______________________________________________ >> Python-ideas mailing list >> Python-ideas@python.org >> https://mail.python.org/mailman/listinfo/python-ideas >> Code of Conduct: http://python.org/psf/codeofconduct/ >> > _______________________________________________ > Python-ideas mailing list > Python-ideas@python.org > https://mail.python.org/mailman/listinfo/python-ideas > Code of Conduct: http://python.org/psf/codeofconduct/ >
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