Matlab indexing into 2D array using - matlab

Question

I have a 2D matrix, say M=zeros(10,10);
I have another column matrix, V=[1;2;3;4;5;6;5;4;3;2];
I would like to be able to set M(i,j) = 1 for all j >= V(i)
I know I can do this in a loop
for i=1:10
M(i,V(i):10) = 1;
end
but it seems that it would be possible to use some form of Matlab indexing to avoid using a loop. For example something like :
M(:,V:10)=1;
or
M(:,V(:):10)=1;
but neither of these produces the expected result.
Is there some syntactic sugar I can use to achieve this or should I revert to looping?

Answer 1

Since you're looking for syntactic sugar, here's a sort of esoteric way of doing it. :)
Assuming the length of V is the size of both dimensions in the desired matrix M, first create an identity matrix of the same size, then index appropriately and take cumsum:
V = [1;2;3;4;5;6;5;4;3;2]; #% 10x1 vector
E = eye(length(V), length(V)); #%10x10 identity matrix
M = cumsum(E(V,:),2)
M =
1 1 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1
0 0 0 0 1 1 1 1 1 1
0 0 0 0 0 1 1 1 1 1
0 0 0 0 1 1 1 1 1 1
0 0 0 1 1 1 1 1 1 1
0 0 1 1 1 1 1 1 1 1
0 1 1 1 1 1 1 1 1 1
Ok, now: less fun, but (on my machine) faster than any of the other options that have been tested so far:
n=10000;
V = randi(n-1, 1, n); #% as in #KevinRatelle's answer (but not transposed)
tic;
Vlinear = reshape(V + (0:n-1)*n, 1, []); #% find linear indices of first "ones"
M = zeros(n);
M(Vlinear)=1;
M=cumsum(M);
toc

Answer 2

Its hardly subtle and not really better than a loop I don't think but:
[J,I] = meshgrid(1:10,1:10);
V = [1;2;3;4;5;6;5;4;3;2];
M = J>V(I);
Enjoy.

Answer 3

I tried the loop method, and the 'meshgrid' method. I was wondering about the time to compute for large matrices (since the issue with loops in matlab is usually the time).
First, I optimised the code to look like this :
V = V*ones(1,n);
N = ones(1,n)'*(1:n);
M = N>=V;
Actually, N is a meshgrid, but it seems way faster to do it this way...
I tried this :
n = 10000;
V = randi(n-1,1,n)';
tic;
M = zeros(n);
for i=1:n
M(i,V(i):n) = 1;
end
toc
tic;
[J,I] = meshgrid(1:n,1:n);
M = J>=V(I);
toc
tic;
V = V*ones(1,n);
N = ones(1,n)'*(1:n);
M = N>=V;
toc
And results were :
Elapsed time is 1.726872 seconds.
Elapsed time is 5.206657 seconds.
Elapsed time is 1.548600 seconds.
But methods using matrices instead of loops are memory consuming for large n. I would personnaly stick to the loop.

Answer 4

try this:
v = [1;2;3;4;5;6;5;4;3;2];
n = 10;
M = repmat((1:n)', 1, numel(v)) > repmat(v', n, 1);