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).
Related
I would like to multiply each sub-block of a matrix A mxn with a matrix B pxq. For example A can be divided into k sub blocks each one of size mxp.
A = [A_1 A_2 ... A_k]
The resulting matrix will be C = [A_1*B A_2*B ... A_k*B] and I would like to do it efficiently.
What I have tried until now is:
C = A*kron(eye(k),B)
Edited: Daniel I think you are right. I tried 3 different ways. Computing a kronecker product seems to be a bad idea. Even the solution with the reshape works faster than the more compact kron solution.
tic
for i=1:k
C1(:,(i-1)*q+1:i*q) = A(:,(i-1)*p+1:i*p)*B;
end
toc
tic
C2 = A*kron(eye(k),B);
toc
tic
A = reshape(permute(reshape(A,m,p,[]),[1 3 2]),m*k,[]);
C3 = A*B;
C3 = reshape(permute(reshape(C3,m,k,[]),[1 3 2]),m,[]);
toc
When I look at your matrix multiplication code, you have perfectly optimized code within the loop. You can't beat matrix multiplication. Everything you could cut down is the overhead for the iteration, but compared to the long runtime of a matrix multiplication the overhead has absolutely no influence.
What you attempted to do would be the right strategy when the operation within the loop is trivial but the loop is iterated many times. If you take the following parameters, you will notice that your permute solution has actually it's strength, but not for your problem dimensions:
q=1;p=1;n=1;m=1;
k=10^6
Kron totally fails. Your permute solution takes 0.006s while the loop takes 1.512s
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! ,
I have huge n×n matrix A, and the indices of its non-zero elements by a = find(A). I have obtained a new list a1 by deleting some elements from a. I want to have matrix A of indices in a1 without using loops. Any suggestions? Is there any function for this purpose?
Considering that your matrix is "huge" (and your question implies that it is mostly zeroes), perhaps it would be best if you represent it as a sparse matrix:
[ii, jj] = ind2sub(size(A), a1);
spA = sparse(ii, jj, A(a1), size(A, 1), size(A, 2));
There might be a significant speedup when operating on sparse matrices. If you need to obtain the full matrice back, use full:
newA = full(spA);
Use vector indexing. Without really knowing how "huge" is your matrix A, but assuming you can still handle it in one piece in matlab's memory, just:
B(size(A,1),size(A,2))=0;
B(a1)=A(a1);
Now B is the same as A with only the indexes given by a1.
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.