[Python-ideas] Vectorization [was Re: Add list.join() please]

[Adrien Ricocotam]

I personally would the first option to be the case. But then vectors
shouldn't be list-like but more generator like.

Le sam. 2 févr. 2019 à 19:26, MRAB <python at mrabarnett.plus.com> a écrit :

> On 2019-02-02 09:22, Kirill Balunov wrote:
> >
> >
> > сб, 2 февр. 2019 г. в 07:33, Steven D'Aprano <steve at pearwood.info
> > <mailto:steve at pearwood.info>>:
> >
> >
> >     I didn't say anything about a vector type.
> >
> >
> > I agree  you did not say. But since you started a new thread from the
> > one where the vector type was a little discussed, it seemed to me  that
> > it is appropriate to mention it here. Sorry about that.
> >
> >      > Therefore, it allows you to ensure that the method is present for
> >     each
> >      > element in the vector. The first given example is what numpy is
> >     all about
> >      > and without some guarantee that L consists of homogeneous data it
> >     hardly
> >      > make sense.
> >
> >     Of course it makes sense. Even numpy supports inhomogeneous data:
> >
> >     py> a = np.array([1, 'spam'])
> >     py> a
> >     array(['1', 'spam'],
> >            dtype='|S4')
> >
> >
> > Yes, numpy, at some degree, supports heterogeneous arrays. But not in
> > the way you brought it. Your example just shows homogeneous array of
> > type `'|S4'`. In the same way as `np.array([1, 1.234])` will be
> > homogeneous. Of course you can say -  np.array([1, 'spam'],
> > dtype='object'), but in this case it will also be homogeneous array, but
> > of type `object`.
> >
> >     Inhomogeneous data may rule out some optimizations, but that hardly
> >     means that it "doesn't make sense" to use it.
> >
> >
> > I did not say that it  "doesn't make sense". I only said that you should
> > be lucky to call `..method()` on collections of heterogeneous data. And
> > therefore, usually this kind of operations imply that you are working
> > with a "homogeneous data". Unfortunately, built-in containers cannot
> > provide such a guarantee without self-checking. Therefore, in my opinion
> > that at the moment such an operator is not needed.
> >
> Here's a question: when you use a subscript on a vector, does it apply
> to the vector itself, or its members?
>
> For example, given:
>
>  >>> my_strings = Vector(['one', 'two', 'three'])
>
> what is:
>
>  >>> my_strings[1 : ]
>
> ?
>
> Is it:
>
> Vector(['ne', 'wo', 'hree'])
>
> or:
>
> Vector(['two', 'three'])
>
> ?
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