Creating a matrix of ones and zeros with location of ones based on array value

I need to acheive the following without a loop (for speed):

I have a vertical array where the value of each row tells me which column in a matrix of zeros should be replaced with a one.

For example, if I input:

y = [2; 3];

Then I want MATLAB to produce output:

Y = [0,1,0; 0, 0, 1];

I can acheive the desired output with the following for loop:

Y = zeros(2,3);
for j = 1:3
    Y(:,j) = (y==j);
end

In reality, my matrices are much larger and I want to avoid using for loops for speed. Is there anyway I can do this without a for loop?

[madhan ravi]

@ Melissa Moore :Are you interested in a faster solution still? If yes another solution can be proposed.

Answer 1

[Kevin Rawson]

One possible way is:

    Y = zeros(length(y), max(y));
    Y(sub2ind(size(Y), (1:length(y))', y)) = 1;

Answer 2

[A. Sawas]

The best solution I can think of is to use a for loop but will be faster than the one you showed because it will do one scalar assignment in each iteration compared to a vector assignment.

for j=1:2
    Y(j,y(j)) = 1;
end

Timewise this is faster than the method propsed above by Kevin Rawson

[Kevin Rawson]

A. Sawas, your method is faster for small arrays, but not for large arrays. I'm using your 10,000x10,000 example with y assigned to the diagonal (see code above).

Sawas method (run 10000 times):

tic;
for i=1:10000
    for j=1:length(y)
        Y(j,y(j)) = 1;
    end
end; 
toc;
%Elapsed time is 1.682664 seconds.

Rawson method (run 10000 times):

tic; 
for i=1:10000
    Y(ind2sub(1:length(y), y)) = 1; 
end
toc;
%Elapsed time is 0.596671 seconds.

Answer 3

[Melissa Moore]

I ended up doing this and it worked:

I = eye(3);
Y = I(y,:);

[A. Sawas]

I would still argure that this solution is much slower than using the for loop as I showed.

y = (1:1e4)';

tic;
for i=1:length(y)
    Y(i,y(i)) = 1;
end
toc;
%Elapsed time is 0.001051 seconds.

tic;
I = eye(length(y));
Y = I(y,:);
toc;
% Elapsed time is 0.741058 seconds.

[Kevin Phung]

Is Melissa's solution a valid solution?... IMO, I is a diagonal matrix where there are already existing 1's. And this is not what you want.

A. Sawas's is valid and more efficient.

[Kevin Rawson]

Kevin Phung, as long as I = eye(max(y)), then Y = I(y,:) will return a valid solution.

However, as A. Sawas pointed out, this is an extremely slow operation, ~700 times slower than A. Sawas' method, and ~1988 times slower than my method for a 10,000x10,000 array.