Let's say I'm given a 1000x1 column vector of with values ranging from 1-10. Now I want to turn this into a 1000x10 matrix, where for each row there is a 1 in the column given by the value in the column vector. I have a correct implementation below, but is there an elegant way to do this without a for loop?
ymat = zeros(1000, 10); y = randi([1,10],1000,1); for i=1:1000 ind = y(i); ymat(i,ind)=1; end We can access any subset of the entries in ymat all at once using the "linear index" of those entries. Since we already have y that contains the column coordinates, and the row coordinates are simply 1:100, We can easily obtain their linear index value with sub2ind, and then set all the indexed values to 1 in ymat:
ymat(sub2ind([1000 10], (1:1000)', y)) = 1 However note that ymat with the proper size must be created before hand.
You have set of column indices and corresponding row index should be 1:1000.
There are two options to generate matrix from this.
accumarray
sparse
acccumarray will generate full matrix and sparse will generate sparse matrix.
In your case, the resulting matrix has density of 10% which can be considered as sparse matrix.
% Full matrix using accumarray ymat = accumarray([(1:1000).', y], ones(1000,1), [1000, 10]); % Sparse matrix using sparse ymat = sparse(1:1000, y, ones(1000,1), 1000, 10, 1000); I run the code with y = 1000 x 1 and y = 10000 x 1.
I added the code suggested by @transversality condition.
It turns out that accumarray is the fastest and sparse is the slowest.
accumarray shows better performance for small size, but the gap between accumarray and bsxfun(@eq) gets smaller as we increase the size of array.
sparse is that you don't need the ones vector to declare all of the non-zero entries as 1. A special case with sparse is if every non-zero value in the matrix is equal to the same value, you only have to specify a single constant to reflect all of these values. Therefore, just do ymat = sparse(1:1000, y, 1, ...);Sure. Use repmat and bsxfun. E.g.,
bsxfun(@eq,repmat(1:10,numel(y),1),repmat(y,1,10)) Answer:
... 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 >> 
