On 2019-02-02 09:22, Kirill Balunov wrote:
>
>
> сб, 2 февр. 2019 г. в 07:33, Steven D'Aprano <steve@pearwood.info
> <mailto:steve@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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