Expand vectors of matrix, but a different number of times each vector

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Hi all,
I have the following problem. Consider a matrix of the form
Z =[1 0 0 0; 1 0 0 0; 1 0 0 0; 0 0 0 1; 0 0 0 1];
Because I'll use this later, consider the vector that sums the elements of the matrix by rows:
C=[3 0 0 2];
I would like to take every column in Z and expand it according to the elements in C, when the expansion when C=0 is just the column itself, to get
D=[1 1 1 0 0 0 0; 1 1 1 0 0 0 0;1 1 1 0 0 0 0;0 0 0 0 0 1 1;0 0 0 0 0 1 1];
Some things to consider. I have a lot of matrices "Z". In particular, they are defined in a structure way. This means that, for example, I have mk(i).Z where i goes from 1 to, say, 50. This means that I also have 50 vectors "C" (mk(i).C). Finally, the Z's have different dimensions that I don't know ex-ante (they are generated randomly within the code).
Any help would be great, thanks
Fernando

Accepted Answer

Sean de Wolski
Sean de Wolski on 9 May 2012
z2 = cell2mat(cellfun(@(z,c)repmat(z,1,max(c,1)),num2cell(Z,1),num2cell(C),'uni',false))
Not sure why you need to do this though. It's going to take more memory to not get you any more information. If you could explain a little more, we might be able to help with a better way.
  1 Comment
Fernando
Fernando on 9 May 2012
Thanks! I'll try to explain what I'm trying to do. Say you have data from two markets and you store the data as mk(1).data and mk(2).data. In each market I have a different number of stores (mk(1).n and mk(2).n) and this stores have a certain ownership structure. You could say that in the example above there are three stores of chain 1, zero of chains 2 and three and two o chain four. Then, when the chains decide prices, they have to consider the effect that that a change in prices of store 1 (owned by chain 1) will have on stores 2 and 3 of the same chain. This is captured by the new rows that where added.
Thanks for your help.

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More Answers (1)

Andrei Bobrov
Andrei Bobrov on 9 May 2012
C=[3 0 0 2];
D = arrayfun(@(x)[zeros(x==0),ones(x)],C,'un',0);
D = blkdiag(D{:});
D = D(any(D,2),:);
or
D = cell2mat(arrayfun(@(ii)Z(:,ii)*ones(1,C(ii)+(C(ii)==0)),1:numel(C),'un',0));

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