Re: selecting 'first' element/vector/matrix of an arbitrary n-dim array

Always, 5 minutes after I put the question together I find a good solution
I missed the first dozen times.
see:
https://www.mathworks.com/matlabcentral/answers/344423-index-slice-of-nd-array-of-unknown-dimension#comment_814098

>> A = magic(3);
>> dim = 2;
>> v = repmat({':'},ndims(A),1);
>> v{dim} = 1;
>> firsts = A(v{:})
firsts =

ans(:,:,1) =

   8
   3
   4

ans(:,:,2) =

   8
   3
   4

ans(:,:,3) =

   8
   3
   4

>> dim = 3; v = repmat({':'},ndims(A),1); v{dim} = 1
v =
{
  [1,1] = :
  [2,1] = :
  [3,1] =  1
}

>> firsts = A(v{:})
firsts =

   8   1   6
   3   5   7
   4   9   2


i'm sure there's some code golf to be played here, but that seems to do the
trick.  Thanks for the sounding board!


On Mon, Dec 7, 2020 at 1:06 PM Nicholas Jankowski <[email protected]>
wrote:

> i'm sure there's a compact, efficient way to do this, but it's eluding me
> and my google-fu is failing:
>
> give a n-dimensional matrix A (n unknown apriori), and a user specified
> dimension to operate on, return the first 'part' of the array for that
> dimension, preserving shape:
>
> e.g, for a 3D array, dim 2:
>
> B = A(:,1,:)
>
> for dim 3:
>
> B = A(:,1,:)
>
> for a 4D array, dim 2:
>
> B = A(:,1,:,:)
>
> for dim 4:
>
> B = A(:,:,:,1)
>
> etc.
>
> what's an efficient way to extract that subarray for any number of
> dimensions? i assume I could do something clever directly with subsasgn and
> expanding a cell array of some number of ':' with {:}, but i'm sure I'm
> missing something cleaner, no?
>
> (does not need to be matlab compatible)
>