# Matrix Manipulation

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Amandeep on 8 Sep 2011
Hi,
From the following matrix:
A = [1; 1; 3; 1; 2; 1; 1; 1; 3; 1; 1; 1; 1; 2; 1; 1; 1; 1; 1; 3; 1; 1; 1; 2; 1; 1; 1; 1; 1; 2; 2; 1; 1; 1; 2; 1; 1; 2; 1; 1; 1; 1; 1; 1; 3; 1; 1; 3; 2; 1; 1; 1; 1; 1; 1; 1; 2; 2; 4; 1; 1; 2; 1; 1; 1; 1; 2; 2; 1; 1; 1; 3]
I want a matrix of (row numbers minus one) where A is greater than 1. Also, if the element in A is 2 for example then B for that element in A will record the (row numbers minus one) once, if it was a 3 then just twice, and so on; therfore B will look like:-
B = [2, 2, 4, 8, 8, 13, 19, 19,....., 71, 71]
Thanks, any help will be great!

Paulo Silva on 8 Sep 2011
A = [1; 1; 3; 1; 2; 1; 1; 1; 3; 1; 1; 1; 1; 2; 1; 1; 1; 1; 1; 3; 1; 1; 1; 2; 1; 1; 1; 1; 1; 2; 2; 1; 1; 1; 2; 1; 1; 2; 1; 1; 1; 1; 1; 1; 3; 1; 1; 3; 2; 1; 1; 1; 1; 1; 1; 1; 2; 2; 4; 1; 1; 2; 1; 1; 1; 1; 2; 2; 1; 1; 1; 3];
a1=find(A==2)-1; %find the index of the 2
a2=find(A==3)-1; %find the index of the 3
a2=repmat(a2,2,1) %duplicate the 3 indexes
B=sort([a1;a2])'; %join and sort the solution, also transpose
%I transposed B because of your B example
##### 3 CommentsShow 1 older commentHide 1 older comment
Paulo Silva on 8 Sep 2011
it was on purpose, that way you need to understand my code and add your code to it ;)
Amandeep on 8 Sep 2011
lol cheers

Amandeep on 8 Sep 2011
I think I figured it usin arrayfun:
A = [1; 1; 3; 1; 2; 1; 1; 1; 3; 1; 1; 1; 1; 2; 1; 1; 1; 1; 1; 3; 1; 1; 1; 2; 1; 1; 1; 1; 1; 2; 2; 1; 1; 1; 2; 1; 1; 2; 1; 1; 1; 1; 1; 1; 3; 1; 1; 3; 2; 1; 1; 1; 1; 1; 1; 1; 2; 2; 4; 1; 1; 2; 1; 1; 1; 1; 2; 2; 1; 1; 1; 3];
missing = (A==1);
A(missing) = 0;
[row, col, vec] = find(A);
b = arrayfun(@(i) (row(i,1)-1)*(ones(1,vec(i,1)-1)),1:size(row,1),'uniform',0)
b = cat(2,b{:});;
Andrei Bobrov on 8 Sep 2011
A1=A-1;
[i1,~,v]=find(A1)
B = cell2mat(arrayfun(@(x,y)x*ones(1,y),i1'-1,v','un',0))