lrr.zip_LRR代码_LRR源码_低秩_低秩表示_低秩表示lrr

function [Z,E] = lrraffine(X,A,lambda) % This routine solves the following nuclear-norm optimization problem. % Comparing to the algorithm in "lrr.m", here the difference is that % each column of Z is enforced to sum to one. The goal of this modification % is to handle the data drawn from affine subspaces. Moreover, this % modification can improve the performance of LRR even when the % subspaces are linear. % -------------------------------- % min |Z|_*+lambda*|E|_2,1 % s.t., X = AZ+E % e^tZ = e^t (i.e., each column of Z sums to one) % inputs: % X -- D*N data matrix, D is the data dimension, and N is the number % of data vectors. % A -- D*M matrix of a dictionary, M is the size of the dictionary if nargin<3 lambda = 1; end tol = 1e-8; maxIter = 1e6; [d n] = size(X); m = size(A,2); rho = 1.1; max_mu = 1e30; mu = 1e-6; atx = A'*X; e1 = ones(m,1); e2 = ones(n,1); eet = e1*e1'; inv_a = inv(A'*A+eye(m)+eet); %% Initializing optimization variables % intialize J = zeros(m,n); Z = zeros(m,n); E = sparse(d,n); Y1 = zeros(d,n); Y2 = zeros(m,n); Y3 = zeros(1,n); %% Start main loop iter = 0; disp(['initial,rank=' num2str(rank(Z))]); while iter<maxIter iter = iter + 1; %update J temp = Z + Y2/mu; [U,sigma,V] = svd(temp,'econ'); sigma = diag(sigma); svp = length(find(sigma>1/mu)); if svp>=1 sigma = sigma(1:svp)-1/mu; else svp = 1; sigma = 0; end J = U(:,1:svp)*diag(sigma)*V(:,1:svp)'; %udpate Z Z = inv_a*(e1*e2'+atx-A'*E+J+(A'*Y1-Y2-e1*Y3)/mu); %update E xmaz = X-A*Z; temp = xmaz+Y1/mu; E = solve_l1l2(temp,lambda/mu); leq1 = xmaz-E; leq2 = Z-J; stopC = max(max(max(abs(leq1))),max(max(abs(leq2)))); leq3 = e1'*Z-e2'; stopC = max(stopC,max(abs(leq3))); if iter==1 || mod(iter,50)==0 || stopC<tol disp(['iter ' num2str(iter) ',mu=' num2str(mu,'%2.1e') ... ',rank=' num2str(rank(Z,1e-3*norm(Z,2))) ',stopALM=' num2str(stopC,'%2.3e')]); end if stopC<tol break; else Y1 = Y1 + mu*leq1; Y2 = Y2 + mu*leq2; Y3 = Y3 + mu*leq3; mu = min(max_mu,mu*rho); end end function [E] = solve_l1l2(W,lambda) n = size(W,2); E = W; for i=1:n E(:,i) = solve_l2(W(:,i),lambda); end function [x] = solve_l2(w,lambda) % min lambda |x|_2 + |x-w|_2^2 nw = norm(w); if nw>lambda x = (nw-lambda)*w/nw; else x = zeros(length(w),1); end