ValueError vs IndexError, unpacking arguments with string.split

[Morten W. Petersen]

On Sun, Dec 2, 2018 at 12:49 PM Chris Angelico <rosuav at gmail.com> wrote:

> On Sun, Dec 2, 2018 at 10:36 PM Morten W. Petersen <morphex at gmail.com>
> wrote:
> > While we're on the subject, I did a test in my Python interpreter:
> >
> > Python 3.6.7 (default, Oct 22 2018, 11:32:17)
> > [GCC 8.2.0] on linux
> > Type "help", "copyright", "credits" or "license" for more information.
> > >>> range(0,3,100)
> > range(0, 3, 100)
> > >>> len(range(0,3,100))
> > 1
> > >>> range(0,100)
> > range(0, 100)
> > >>> len(range(0,100))
> > 100
> > >>>
> >
> > Where a range with a step, gives the length 1, while a plain range gives
> > the right length.
> >
> > I think that's confusing and inconsistent, and it would be nice to have
> some
> > "value to be calculated" for the length integer as well.
> >
>
> Possibly you're misinterpreting the arguments here.
>
> >>> list(range(0, 3, 100))
> [0]
>
> class range(object)
>  |  range(stop) -> range object
>  |  range(start, stop[, step]) -> range object
>  |
>
> The *third* argument is the step, so if you were expecting "every
> third number up to 100", you'd be looking for this:
>
> >>> len(range(0, 100, 3))
> 34
> >>> len(list(range(0, 100, 3)))
> 34
>
> To my knowledge, len(x) == len(list(x)) for any core data type that
> has a length.
>

Ah yes, of course.  I haven't had my coffee yet. :)

>>> len(range(0,100,3))
34
>>> range(0,100,3).__len__
<method-wrapper '__len__' of range object at 0x7f144d5e3210>
>>> range(0,100,3).__len__()
34
>>>

But again, I guess an integer-to-be-calculated would be suitable
for iterators where it is very expensive to do a list or other
similar calculation.

These operations take quite a bit of time:

>>> x=list(range(0,100000000,3))
>>> y=len(list(range(0,100000000,3)))
>>>

-Morten

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