mir - Help on how to transform multidimentional arrays.

[Newbie]

On Thursday, 29 April 2021 at 15:56:48 UTC, jmh530 wrote:
> On Thursday, 29 April 2021 at 15:26:15 UTC, Newbie wrote:
>> [snip]
>>
>> Forgot to add the the first array was created using the 
>> following code.
>> auto base = iota(2, 5, 3);
>
> What you're basically asking for the first one is to convert 
> for row major to column major. There doesn't seem to be a 
> specific function for that, but you can piece it together. The 
> second one is just applying allReversed to the result of that. 
> So we have:
>
> ```d
> /+dub.sdl:
> dependency "mir-algorithm" version="~>3.10.25"
> +/
>
> import std.stdio: writeln;
> import mir.ndslice.topology: iota, flattened, reshape, 
> universal;
> import mir.ndslice.dynamic: transposed, allReversed;
>
> void main() {
>     auto x = iota(2, 5, 3);
>     int err;
>     auto y = x.flattened.reshape([3, 5, 2], err).transposed!(1, 
> 2);
>     auto z = y.allReversed;
>
>     writeln(x);
>     writeln(y);
>     writeln(z);
> }
> ```

First of all thanks for your guidance and i modified your example 
to get the results I was looking for.

auto y2 = base.flattened.reshape([3, 5, 2], err).transposed!(1, 
2);
produced the following.

---------
|0 10 20|
|1 11 21|
---------
|2 12 22|
|3 13 23|
---------
|4 14 24|
|5 15 25|
---------
|6 16 26|
|7 17 27|
---------
|8 18 28|
|9 19 29|
---------

and auto y2 = base.flattened.reshape([3, 2, 5], 
err).transposed!(1, 2);
gave me the result i was looking for.

---------
|0 10 20|
|1 11 21|
|2 12 22|
|3 13 23|
|4 14 24|
---------
|5 15 25|
|6 16 26|
|7 17 27|
|8 18 28|
|9 19 29|
---------

Thanks,
Newbie.