I think he meant doing something like this (with fuller handling for out of
bounds, exceptions throwing, slices, etc):
role Shaped[*@dimensions] {
has @!multipliers =
@dimensions.reverse.produce(&[*])[0..*-2].reverse.Slip, 1;
method AT-POS(*@indices where all @dimensions Z>
When will we have p6’s Numpy?
On Sat, Feb 6, 2021, at 2:04 AM, Brad Gilbert wrote:
> There is a reason that you can't just ask for the dimensions of an
> unspecified multidimensional array.
> It may be multiple dimensions.
>
> [[1,2,3],
> [4,5,6,7,8,9,10]].shape
>
> If it gave a
FOSDEM_2021 is going on this weekend [Online 6 & 7 February 2021],
running on CET (Central Europe Time, UTC/GMT +1 hour). That's 6 hours
ahead of the US East Coast and 9 hours ahead of the US West Coast. The
first talk in the "Perl_&_Raku" track is at 10am Saturday on Brussels
time, making it 4am
There is a reason that you can't just ask for the dimensions of an
unspecified multidimensional array.
It may be multiple dimensions.
[[1,2,3],
[4,5,6,7,8,9,10]].shape
If it gave a result it would be something like:
(2,3|7)
On Fri, Feb 5, 2021 at 8:50 AM Theo van den Heuvel
wrote:
Shaped Arrays and Native Shaped Arrays already use one contiguous blob
to store all their data; in Shaped Arrays that's an array of pointers to
the stored objects, in a Native Shaped Array it'll be like a big array
of 32bit integers or whatever you have.
Regards
- Timo
On 05/02/2021 16:48,
This got me interested in
https://docs.raku.org/language/list#index-entry-Shaped_arrays
and I find myself wanting to implement a role "Shaped" and applying it to
List, for an immutable shaped list, whose implementation packs its elements
for o(1) lookup... on my ever-lengthening to-do list
-y
Hi gurus,
I cannot seem to find an idiomatic way to get the dimensions of a
multidimensional array,
other than by looking at the size of the first row and column, with
@m[0;*].elems and @m[*;0].elems.
Am I missing something in the docs?
Thanks,
--
Theo van den Heuvel
