Indexing a matrix with an array of ranges

I was wondering if it was possible to do the following:

indexArray = [4 2 0];
in = [1:10]'; in = repmat(in, 1, length(indexArray));
out = zeros(size(in));
out(1:end-indexArray,:) = in(indexArray+1:end,:);

so that:

out(:,1)' = [5 6 7 8 9 10 0 0 0 0]
out(:,2)' = [3 4 5 6 7 8 9 10 0 0]
out(:,3)' = [1 2 3 4 5 6 7 8 9 10]

The goal is to do this without being forced to replace the 4th line above with a for loop like:

for i = 1:length(indexArray)
    out(1:end-indexArray(i),i) = in(indexArray(i)+1:end,i);
end

Thanks for your help in advance.

Answer 1

[Matt J]

My recommendation from your previous related question on this was that you precompute a table of all the desired shifts:

   n=length(in);
   Table= fliplr(toeplitz([in(n),zeros(1,n-1)], in(n:-1:1) ));
     Table(end, end+1)=0;

Then you would just do

     out=Table(:,indexArray+1)

[William]

I misunderstood what the table was when you used it in my other question and when I saw how much memory it would take (way more than I have to efficiently make use of it), I didn't look any closer.

I understand what the following line does, but I don't see the point of it...

Table(end, end+1)=0;

Why the extra column?

Thanks again for your time.

[Matt J]

I misunderstood what the table was when you used it in my other question and when I saw how much memory it would take (way more than I have to efficiently make use of it), I didn't look any closer.

So are you saying this method won't suit you? You should probably mention, then, the typical values of M=length(indexArray) and N=length(in) so that we have a feel for the data sizes involved.

Why the extra column?

You never said that indexArray(i) couldn't equal N. Presumably, in that case, you would want out(:,i) to be all zeros.

[William]

So are you saying this method won't suit you? You should probably mention, then, the typical values of M=length(indexArray) and N=length(in) so that we have a feel for the data sizes involved.

True, although I was afraid of making my question too specific. N=length(in) is at least 10e6 but M=length(indexArray) will always be smaller than 100.

You never said that indexArray(i) couldn't equal N. Presumably, in that case, you would want out(:,i) to be all zeros.

True. I figured it was for that but I wasn't sure. In my case, no, indexArray will never equal N.

[Matt J]

N=length(in) is at least 10e6 but M=length(indexArray) will always be smaller than 100.

Sounds like a for-loop would be best, but see also my other Answer.

Is "in" of type sparse?

Answer 2

[Matt J]

Another method, which may be preferable if M=length(indexArray) is short.

M=length(indexArray);
N=length(in);
e=(1:N).';

Iout=bsxfun(@gt,e,indexArray);
 Iin=bsxfun(@times,e,Iout);

out=zeros(N,M);
 out(flipud(Iout))=in(nonzeros(Iin)),