I have a vector v of values and a vector r of indices. I want to store the values in a matrix m as follows:
for i = 1:length(v)
m(i, r(i)) = v(i);
end
What is the fastest way to do this in a vectorized way?
I do not know if it is faster, I suppose so but the difference might be very small, but here is one way:
m(sub2ind(size(m),1:length(v),r(1:length(v))))=v;
If r is a column vector then sub2ind will complain about vectors size, you can just take its transpose and it will solve this.
Related
I want to multiply every column of a M × N matrix by corresponding element of a vector of size N.
I know it's possible using a for loop. But I'm seeking a more simple way of doing it.
I think this is what you want:
mat1=randi(10,[4 5]);
vec1=randi(10,[1 5]);
result=mat1.*repmat(vec1,[size(mat1,1),1]);
rempat will replicate vec1 along the rows of mat1. Then we can do element-wise multiplication (.*) to "multiply every column of a M × N matrix by corresponding element of a vector of size N".
Edit: Just to add to the computational aspect. There is an alternative to repmat that I would like you to know. Matrix indexing can achieve the same behavior as repmat and be faster. I have adopted this technique from here.
Observe that you can write the following statement
repmat(vec1,[size(mat1,1),1]);
as
vec1([1:size(vec1,1)]'*ones(1,size(mat1,1)),:);
If you see closely, the expression boils down to vec1([1]'*[1 1 1 1]),:); which is again:
vec1([1 1 1 1]),:);
thereby achieving the same behavior as repmat and be faster. I ran three solutions 100000 times, namely,
Solution using repmat : 0.824518 seconds
Solution using indexing technique explained above : 0.734435 seconds
Solution using bsxfun provided by #LuisMendo : 0.683331 seconds
You can observe that bsxfun is slightly faster.
Although you can do it with repmat (as in #Parag's answer), it's often more efficient to use bsxfun. It also has the advantage that the code (last line) is the same for a row and for a column vector.
%// Example data
M = 4;
N = 5;
matrix = rand(M,N);
vector = rand(1,N); %// or size M,1
%// Computation
result = bsxfun(#times, matrix, vector); %// bsxfun does an "implicit" repmat
I have N kx1 sparse vectors and I need to multiply each of them by their transpose, creating N square matrices, which I then have to sum over. The desired output is a k by k matrix. I have tried doing this in a loop and using arrayfun, but both solutions are too slow. Perhaps one of you can come up with something faster. Below are specific details about the best solution I've come up with.
mdev_big is k by N sparse matrix, containing each of the N vectors.
fun_sigma_i = #(i) mdev_big(:,i)*mdev_big(:,i)';
sigma_i = arrayfun(fun_sigma_i,1:N,'UniformOutput',false);
sigma = sum(reshape(full([sigma_i{:}]),k,k,N),3);
The slow part of this process is making sigma_i full, but I cannot reshape it into a 3d array otherwise. I've also tried cat instead of reshape (slower), ndSparse instead of full (way slower), and making fun_sigma_i return a full matrix rather than a sparse one (slower).
Thanks for the help! ,
let's say I have a big Matrix X with a lot of zeros, so of course I make it sparse in order to save on memory and CPU. After that I do some stuff and at some point I want to have the nonzero elements. My code looks something like this:
ind = M ~= 0; % Whereby M is the sparse Matrix
This looks however rather silly to me since the structure of the sparse Matrix should allow the direct extraction of the information.
To clarify: I do not look for a solution that works, but rather would like to avoid doing the same thing twice. A sparse Matrix should perdefinition already know it's nonzero values, so there should be no need to search for it.
yours magu_
The direct way to retrieve nonzero elements from a sparse matrix, is to call nonzeros().
The direct way is obviously the fastest method, however I performed some tests against logical indexing on the sparse and its full() counterparty, and the indexing on the former is faster (results depend on the sparsity pattern and dimension of the matrix).
The sum of times over 100 iterations is:
nonzeros: 0.02657 seconds
sparse idx: 0.52946 seconds
full idx: 2.27051 seconds
The testing suite:
N = 100;
t = zeros(N,3);
for ii = 1:N
s = sprand(10000,1000,0.01);
r = full(s);
% Direct call nonzeros
tic
nonzeros(s);
t(ii,1) = toc;
% Indexing sparse
tic
full(s(s ~= 0));
t(ii,2) = toc;
% Indexing full
tic
r(r~=0);
t(ii,3) = toc;
end
sum(t)
I'm not 100% sure what you're after but maybe [r c] = find(M) suits you better?
You can get to the values of M by going M(r,c) but the best method will surely be dictated by what you intend to do with the data next.
find function is recommended by MATLAB:
[row,col] = find(X, ...) returns the row and column indices of the nonzero entries in the matrix X. This syntax is especially useful when working with sparse matrices.
While find has been proposed before, I think this is an important addition:
[r,c,v] = find(M);
Gives you not only the indices r,c, but also the non-zero values v. Using the nonzeros command seems to be a bit faster, but find is in general very useful when dealing with sparse matrices because the [r,c,v] vectors describe the complete matrix (except matrix dimension).
A question about matlab and randomisation of a 3d matrix respecting the rows and columns.
I have a n x n x s matrix M and I want to mess it up a bit, but with some control.
I can achieve my wish with a for loop
for j=1:size(M,3)
r=randperm(size(M,1));
random_M(:,:,j)=M(r,r,j);
end
Is there a way to perform this without having to loop over j? I need many randomisation iterations and could afford the benefits of indexing.
Cheers!
edit: Some more thoughts following Alexandrew's comments
I have created a function that randomises a squeezed version of M:
function randomMat=randomiseMat(Mat)
[rows,cols]=size(Mat);
r=randperm(rows);
randomMat=Mat(r,r);
then, using arrayfun I seem to get what I want:
randomM=arrayfun(#(x) randomiseMat(M(:,:,x)),1:size(M,3),'UniformOutput', false)
however, randomM is now a cell array of size (1,size(M,3)) with each cell containing randomised array.
Is there a way to make it in a 3d matrix just like the input M?
You can calculate all the values for r in one go, and then use arrayfun:
[nRows,nCols,nPages] = size(M);
[~,r]=sort(rand(nRows,nPages));
%# you should test on a realistic example whether a for-loop
%# isn't faster here
outCell = arrayfun(#(x) M(r(:,x),r(:,x),x), 1:nPages,'UniformOutput',false);
randomM = cat(3,outCell{:});
I have to create a matlab matrix that is much bigger that my phisical memory, and i want to take advantage of the sparsity.
This matrix is really really sparse [say N elements in an NxN matrix], and my ram is enought for this. I create the matrix in this way:
A=sparse(zeros(N));
but it goes out of memory.
Do you know the right way to create this matrix?
zeros(N) is creating an NxN matrix, which is not sparse, hence you are running out of memory. Your code is equivalent to
temp = zeros(N)
A = sparse(temp)
Just do sparse(N,N).
Creating an all zeros sparse matrix, and then modifying it is extremely inefficient in matlab.
Instead of doing something like:
A = sparse(N,N) % or even A = sparse([],[],[],N,N,N)
A(1:N,7) = 1:N
It is much more efficient to construct the matrix in triplet form. That is,
construct the column and row indices and the nonzero entries first, then
form the matrix. For example,
i = 1:N;
j = 7*ones(1,N);
x = 1:N;
A = sparse(i,j,x,N,N);
I'd actually recommend the full syntax of sparse([],[],[],N,N,N).
It's useful to preallocate if you know the maximum number of nonzero elements as otherwise you'll get reallocs when you insert new elements.