I have a 3d array A, e.g. A=rand(N,N,K).
I need an array B s.t.
B(n,m) = norm(A(:,:,n)*A(:,:,m)' - A(:,:,m)*A(:,:,n)','fro')^2 for all indices n,m in 1:K.
Here's the looping code:
B = zeros(K,K);
for n=1:K
for m=1:K
B(n,m) = norm(A(:,:,n)*A(:,:,m)' - A(:,:,m)*A(:,:,n)','fro')^2;
end
end
I don't want to loop through 1:K.
I can create an array An_x_mt of size NK x NK s.t.
An_x_mt equals A(:,:,n)*A(:,:,m)' for all n,m in 1:K by
An_x_mt = Ar*Ac_t;
with
Ac_t=reshape(permute(A,[2 1 3]),size(A,1),[]);
Ar=Ac_t';
How do I create an array Am_x_nt also of size NK x NK s.t.
Am_x_nt equals A(:,:,m)*A(:,:,n)' for all n,m in 1:K
so that I could do
B = An_x_mt - Am_x_nt
B = reshape(B,N,N,[]);
B = reshape(squeeze(sum(sum(B.^2,1),2)),K,K);
Thx
For those who can't/won't use mmx and want to stick to pure Matlab code, here's how you could do it. mat2cell and cell2mat functions are your friends:
[N,~,nmat]=size(A);
Atc = reshape(permute(A,[2 1 3]),N,[]); % A', N x N*nmat
Ar = Atc'; % A, N*nmat x N
Anmt_2d = Ar*Atc; % An*Am'
Anmt_2d_cell = mat2cell(Anmt_2d,N*ones(nmat,1),N*ones(nmat,1));
Amnt_2d_cell = Anmt_2d_cell'; % ONLY products transposed, NOT their factors
Amnt_2d = cell2mat(Amnt_2d_cell); % Am*An'
Anm = Anmt_2d - Amnt_2d;
Anm = Anm.^2;
Anm_cell = mat2cell(Anm,N*ones(nmat,1),N*ones(nmat,1));
d = cellfun(#(c) sum(c(:)), Anm_cell); % squared Frobenius norm of each product; nmat x nmat
Alternatively, after computing Anmt_2d_cell and Amnt_2d_cell, you could convert them to 3d with the 3rd dimension encoding the (n,m) and (m,n) indices and then do the rest of the computations in 3d. You would need the permn() utility from here https://www.mathworks.com/matlabcentral/fileexchange/7147-permn-v-n-k
Anmt_3d = cat(3,Anmt_2d_cell);
Amnt_3d = cat(3,Amnt_2d_cell);
Anm_3d = Anmt_3d - Amnt_3d;
Anm_3d = Anm_3d.^2;
Anm = squeeze(sum(sum(Anm_3d,1),2));
d = zeros(nmat,nmat);
nm=permn(1:nmat, 2); % all permutations (n,m) with repeat, by-row order
d(sub2ind([nmat,nmat],nm(:,1),nm(:,2))) = Anm;
For some reason, the 2nd option (3D arrays) is twice faster.
Hopes this helps.
Related
In my code I have a slow part of which the idea can be summarized in the following short example:
A = randi(10,5); %Random 5×5 matrix containing integers ranging from 0 to 10
B = rand(10,1); %Random 10×1 vector containing values ranging from 0 to 1
C = B(A); %New 5×5 matrix consisting of elements from B, indexed using A
In my case, the matrix A is sized 1000×1000, B is a 500×1 vector and C is also 1000×1000. Given that this 3rd line is in a for loop, where A is constant and B is updated every iteration, how can I further improve speed performance? According to the profile viewer 75% of code execution is at this single line. As expected, using a for loop for this operation is much slower (10x for a 1000×1000 matrix):
AA = A(:); %Convert matrix to vector
for k=1:length(AA) %Loop through this vector and use it as index
D(k) = B(AA(k));
end
E = reshape(D,5,5); %Reshape vector to matrix of 5x5
Any ideas to optimize this?
Edit: Script used to measure performance:
N = 1500;
A = randi(500,N);
AA = A(:);
D = zeros(N,N);
B = rand(500,1);
f1 = #() VectorIndex(A,B);
timeit(f1,1)
f2 = #() LoopIndex(AA,B,N);
timeit(f2,1)
function C = VectorIndex(A,B)
C = B(A);
end
function D = LoopIndex(AA,B,N)
D = zeros(N,N);
for k=1:length(AA)
D(k) = B(AA(k));
end
D = reshape(D,N,N);
end
I would like to generate all the possible adjacency matrices (zero diagonale) of an undirected graph of n nodes.
For example, with no relabeling for n=3 we get 23(3-1)/2 = 8 possible network configurations (or adjacency matrices).
One solution that works for n = 3 (and which I think is quite stupid) would be the following:
n = 3;
A = [];
for k = 0:1
for j = 0:1
for i = 0:1
m = [0 , i , j ; i , 0 , k ; j , k , 0 ];
A = [A, m];
end
end
end
Also I though of the following which seems to be faster but something is wrong with my indexing since 2 matrices are missing:
n = 3
C = [];
E = [];
A = zeros(n);
for i = 1:n
for j = i+1:n
A(i,j) = 1;
A(j,i) = 1;
C = [C,A];
end
end
B = ones(n);
B = B- diag(diag(ones(n)));
for i = 1:n
for j = i+1:n
B(i,j) = 0;
B(j,i) = 0;
E = [E,B];
end
end
D = [C,E]
Is there a faster way of doing this?
I would definitely generate the off-diagonal elements of the adjacency matrices with binary encoding:
n = 4; %// number of nodes
m = n*(n-1)/2;
offdiags = dec2bin(0:2^m-1,m)-48; %//every 2^m-1 possible configurations
If you have the Statistics and Machine Learning Toolbox, then squareform will easily create the matrices for you, one by one:
%// this is basically a for loop
tmpcell = arrayfun(#(k) squareform(offdiags(k,:)),1:size(offdiags,1),...
'uniformoutput',false);
A = cat(2,tmpcell{:}); %// concatenate the matrices in tmpcell
Although I'd consider concatenating along dimension 3, then you can see each matrix individually and conveniently.
Alternatively, you can do the array synthesis yourself in a vectorized way, it's probably even quicker (at the cost of more memory):
A = zeros(n,n,2^m);
%// lazy person's indexing scheme:
[ind_i,ind_j,ind_k] = meshgrid(1:n,1:n,1:2^m);
A(ind_i>ind_j) = offdiags.'; %'// watch out for the transpose
%// copy to upper diagonal:
A = A + permute(A,[2 1 3]); %// n x n x 2^m matrix
%// reshape to n*[] matrix if you wish
A = reshape(A,n,[]); %// n x (n*2^m) matrix
I want to get exp() of a large matrix (A) with values that repeat at different indices. To speed-up the exp() operation I only perform it on the unique values of A and then reassemble the matrix. However the reassembly of the matrix is quite slow. The following code provides a working example:
% defintion of a grid
gridSp = 5:5:35*5;
X = repmat(gridSp,35,1);
Z = repmat(gridSp',1,35);
% calculation of the distances
locMat = [X(:) Z(:)];
dist=sqrt(bsxfun(#minus,locMat(:,1),locMat(:,1)').^2 +...
bsxfun(#minus,locMat(:,2),locMat(:,2)').^2);
sizeDist = size(dist);
uniqueDist = unique(dist,'stable');
[~, Locb] = ismember(dist,uniqueDist);
nn_A = exp(1i*2*pi*rand(sizeDist(1),100));
H_A = zeros(size(nn_A));
freq = linspace(10^-3,10,100);
psdA = 4096*length(freq).*10.*4.*22.6./((1 + 6.*freq*22.6).^(5/3));
for jj=1:100
b = exp(-8.8*uniqueDist*sqrt((freq(jj)/15).^2 + 10^-7));
b = b.*psdA(jj);
A = b(Locb);
droptol = max(A(:))*10^-10;
if min(A(:))<droptol
A = sparse(A);
HH_A = ichol(A,struct('type','ict','shape','lower','droptol',droptol));
else
HH_A = chol(A,'lower');
end
H_A(:,jj) = HH_A*nn_A(:,jj);
end
Especially the reassembly of the matrix
A = b(Locb);
and the conversion of the matrix to sparse
A = sparse(A);
in the last for-loop take up a lot of time. Is there a quicker way to do this? Interestingly:
B = A + A;
is much faster than
A = b(Locb);
I have to perfom these operations far more often than the 100 iterations in the example.
Here a condensed version of the code up on request (below).
% defintion of a grid
gridSp = 5:5:28*5;
X = repmat(gridSp,35,1);
Z = repmat(gridSp',1,35);
% calculation of the distances
locMat = [X(:) Z(:)];
dist=sqrt(bsxfun(#minus,locMat(:,1),locMat(:,1)').^2 +bsxfun(#minus,locMat(:,2),locMat(:,2)').^2);
uniqueDist = unique(dist,'stable');
[~, Locb] = ismember(dist,uniqueDist);
for jj=1:100
b = exp(jj.*uniqueDist);
A = b(Locb);
end
In your example, the dimension of dist is just 980 x 980 in which case you would be better off to just perform a dense matrix operation, i.e.
for jj=1:100
A=exp(jj*dist);
end
which is 2 times faster than
for jj=1:100
b = exp(jj.*uniqueDist);
A = b(Locb);
end
for your given example.
I have 2 matrices = X in R^(n*m) and W in R^(k*m) where k<<n.
Let x_i be the i-th row of X and w_j be the j-th row of W.
I need to find, for each x_i what is the j that maximizes <w_j,x_i>
I can't see a way around iterating over all the rows in X, but it there a way to find the maximum dot product without iterating every time over all of W?
A naive implementation would be:
n = 100;
m = 50;
k = 10;
X = rand(n,m);
W = rand(k,m);
Y = zeros(n, 1);
for i = 1 : n
max_ind = 1;
max_val = dot(W(1,:), X(i,:));
for j = 2 : k
cur_val = dot(W(j,:),X(i,:));
if cur_val > max_val
max_val = cur_val;
max_ind = j;
end
end
Y(i,:) = max_ind;
end
Dot product is essentially matrix multiplication:
[~, Y] = max(W*X');
bsxfun based approach to speed-up things for you -
[~,Y] = max(sum(bsxfun(#times,X,permute(W,[3 2 1])),2),[],3)
On my system, using your dataset I am getting a 100x+ speedup with this.
One can think of two more "closeby" approaches, but they don't seem to give any huge improvement over the earlier one -
[~,Y] = max(squeeze(sum(bsxfun(#times,X,permute(W,[3 2 1])),2)),[],2)
and
[~,Y] = max(squeeze(sum(bsxfun(#times,X',permute(W,[2 3 1]))))')
I am interesting to replace the diagonal of matrix D to 1,2,3,4.
This is matrix D:
A=[1,2,3,4,2,3,4,5; 3,4,5,6,4,5,6,7];
D=[A;A];
D=[D D]; % size of matrix [4x16] %
To set the main diagonal to integers starting a 1 and incrementing by 1:
D(eye(4)==1) = 1:4
Or to generalize it:
n = min(size(D));
D(eye(n)==1) = 1:n;
note here that the ==1 is to convert the output of eye(n), the identity matrix, to type logical.
EDIT:
This is just a guess at what you mean by all the diagonals but here goes:
n = size(D,1);
m = size(D,2);
I = repmat(eye(min([n,m])), ceil(n/m), ceil(m/n));
I = I(1:n, 1:m)==1
d = repmat(1:min([n,m]), 1, max([ceil(n/m), ceil(m/n)]));
d = d(1:max(m,n));
D(I) = d