虽然这段代码看起来很抱歉。它是psvm（近端SVM）的代码，我很难用它来分类mnist数据集。
我有一个x_train大小（10000,784）的矩阵，我将其作为A，d作为目标，大小（10000,10）作为k-cross验证。
当我运行此代码时，矩阵维度超过错误，在ma=A(find(d==1),:); mb=A(find(d==-1),:);处出现，此处d是目标值，并且是一个热编码的形式。我试图找出它来自A(find(d==1),:);的错误，它返回索引的值，但它超出了矩阵A的维度。我不确定代码的第一部分是什么意图do。正确的文档不可用。我认为这段代码只适用于二进制分类，因为它使用find（d == 1）和find（d == - 1）
如果此代码可用于多类分类，任何人都可以给我一点帮助。代码code link
的链接 function [w,gamma, trainCorr, testCorr, cpu_time, nu, mu]=n_psvm(A,d,rr,k,nu,mu,output,bal);
% version 1.1
% last revision: 01/24/03
%==========================================================================================
% Usage: [w,gamma,trainCorr, testCorr,cpu_time,nu, mu]=n_psvm(A,d,rr,k,nu,mu,output,bal)
%
% A and d are both required, everything else has a default
% An example: [w gamma train test time nu] = n_psvm(A,d,0.5,10);
%
% Input:
% A is a matrix containing m data in n dimensions each.
% d is a m dimensional vector of 1's or -1's containing
% the corresponding labels for each example in A.
% rr: reduce rate, default is 100% -> not reduced
% k is k-fold for correctness purpose
% nu - the weighting factor.
% -1 - easy estimation
% 0 - hard estimation
% any other value - used as nu by the algorithm
% default - 0
% mu: mu in calculating kernel, 0 means take the default estimation
% output - indicates whether you want output
%
% If the input parameter bal is 1
% the algorithm weighs the classes depending on the
% number of points in each class and balance them.
% It is useful when the number of point in each class
% is very unbalanced.
%
% Output:
% w,gamma are the values defining the separating
% Hyperplane w'x-gamma=0 such that:
%
% w'x-gamma>0 => x belongs to A+
% w'x-gamma<0 => x belongs to A-
% w'x-gamma=0 => x can belongs to both classes
% nu - the estimated or specified value of nu
%
% For details refer to the paper:
% "Proximal Support Vector Machine Classifiers"
% available at: www.cs.wisc.edu/~gfung
% For questions or suggestions, please email:
% Glenn Fung, gfung@cs.wisc.edu
% Sept 2001.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[m,n]=size(A);
r=randperm(size(d,1));d=d(r,:);A=A(r,:); % random permutation
%move one point in A a little if perfectly balanced
AA=A;dd=d;
ma=A(find(d==1),:); mb=A(find(d==-1),:);
[s1 s2]=size(ma);
c1=sum(ma)/s1;
[s1 s2]=size(mb);
c2=sum(mb)/s1;
if (c1==c2)
nu=1;
A(3,:)=A(3,:)+0.01*norm(A(3,:)-c1,inf)*ones(1,n);
end
% default values for input parameters
if nargin<8
bal=0;
end
if nargin<7
output=0;
end
if nargin<6
mu=EstMu(A,d);
end
if ((nargin<5)|(nu==0))
nu = EstNuLong(A,d,m); % default is hard estimation
elseif nu==-1 % easy estimation
nu = EstNuShort(A,d);
end
if (nargin<4)
k=0;
end
if (nargin<3)
rr=1;
end
[H,v]=HV(A,d,bal); % calculate H and v
trainCorr = 0;
testCorr = 0;
if (nu==0)
nu = EstNuLong(H,d,m);
elseif nu==-1 % easy estimation
nu = EstNuShort(H,d);
end
% if k=0 no correctness is calculated, just run the algorithm
if k==0
A = calcKer(A,rr,mu,output);
[H,v]=HV(A,d,bal);
tic;
[w, gamma] = core(H,v,nu);
cpu_time = toc;
fprintf(1,'\nElapsed time: %10.2f\n\n',cpu_time);
return
end
%if k==1 only training set correctness is calculated
if k==1
[kA,Abar] = calcKer(A,rr,mu,output);
tic;
[H,v]=HV(kA,d,bal);
[w, gamma] = core(H,v,nu);
trainCorr = correctness(A,Abar,d,w,gamma,mu);
cpu_time = toc;
if output == 1
fprintf(1,'\nTraining set correctness: %3.2f%% \n',trainCorr);
fprintf(1,'\nElapse time: %10.2f\n',toc);
end
return
end
%% if k= folds
accuIter = 0;
cpu_time = 0;
indx = [0:k];
indx = floor(m*indx/k); %last row numbers for all 'segments'
% split trainining set from test set
for i = 1:k
Ctest = []; dtest = [];Ctrain = []; dtrain = [];
Ctest = A((indx(i)+1:indx(i+1)),:);
dtest = d(indx(i)+1:indx(i+1));
Ctrain = A(1:indx(i),:);
Ctrain = [Ctrain;A(indx(i+1)+1:m,:)];
dtrain = [d(1:indx(i));d(indx(i+1)+1:m,:)];
[kCtrain,Abar] = calcKer(Ctrain,rr,mu,output);
tic;
[H, v] = HV(kCtrain,dtrain,bal);
[w, gamma] = core(H,v,nu);
thisToc = toc;
tmpTrainCorr = correctness(Ctrain,Abar,dtrain,w,gamma,mu);
tmpTestCorr = correctness(Ctest,Abar,dtest,w,gamma,mu);
if output==1
fprintf(1,'________________________________________________\n');
fprintf(1,'Fold %d\n',i);
fprintf(1,'Training set correctness: %3.2f%%\n',tmpTrainCorr);
fprintf(1,'Testing set correctness: %3.2f%%\n',tmpTestCorr);
fprintf(1,'Elapse time: %10.2f\n',thisToc);
end
trainCorr = trainCorr + tmpTrainCorr;
testCorr = testCorr + tmpTestCorr;
cpu_time = cpu_time + thisToc;
end % end of for (looping through test sets)
trainCorr = trainCorr/k;
testCorr = testCorr/k;
cpu_time = cpu_time/k;
if output == 1
fprintf(1,'___________________________________________________\n');
fprintf(1,'\nAverage training set correctness: %3.2f%% \n',trainCorr);
fprintf(1,'\nAverage testing set correctness: %3.2f%% \n',testCorr);
fprintf(1,'\nAverage CPU time is: %3.2f \n',cpu_time);
end
return
%%%%%%%%%%%%%%%% core function to calcuate w and gamma %%%%%%%%
function [w, gamma]=core(H,v,nu)
n=size(H,2);
v=(speye(n)/nu+H'*H)\v;
w=v(1:n-1);gamma=v(n);
return
%%%%%%%%%%%%%%%% correctness calculation %%%%%%%%%%%%%%%%
function corr = correctness(Atest,Abar,dd,w,gamma,mu)
k = Rec_Kernel(Atest,Abar,mu);
p = sign(k*w - gamma);
corr=length(find(p==dd))/size(Atest,1)*100;
return
%%%%%%%%%%%%% EstNuLong %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% use to estimate nu
function lamda=EstNuLong(H,d,m)
if m<201
H2=H;d2=d;
else
r=rand(m,1);
[s1,s2]=sort(r);
H2=H(s2(1:200),:);
d2=d(s2(1:200));
end
lamda=1;
[vu,u]=eig(H2*H2');u=diag(u);p=length(u);
yt=d2'*vu;
lamdaO=lamda+1;
cnt=0;
while (abs(lamdaO-lamda)>10e-4)&(cnt<100)
cnt=cnt+1;
nu1=0;pr=0;ee=0;waw=0;
lamdaO=lamda;
for i=1:p
nu1= nu1 + lamda/(u(i)+lamda);
pr= pr + u(i)/(u(i)+lamda)^2;
ee= ee + u(i)*yt(i)^2/(u(i)+lamda)^3;
waw= waw + lamda^2*yt(i)^2/(u(i)+lamda)^2;
end
lamda=nu1*ee/(pr*waw);
end
value=lamda;
if cnt==100
value=1;
end
return
%%%%%%%%%%%%%%%%%EstNuShort%%%%%%%%%%%%%%%%%%%%%%%
% easy way to estimate nu if not specified by the user
function value = EstNuShort(C,d)
value = 1/(sum(sum(C.^2))/size(C,2));
return
%%% function to calculate H and v %%%%%%%%%%%%%
function [H,v]=HV(A,d,bal);
[m,n]=size(A);e=ones(m,1);
if (bal==0)
H=[A -e];
v=(d'*H)';
else
H=[A -e];
mm=e;
m1=find(d==-1);
mm(m1)=(1/length(m1));
m2=find(d==1);
mm(m2)=(1/length(m2));
mm=sqrt(mm);
N=spdiags(mm,0,m,m);
H=N*H;
%keyboard
v=(d'*N*H)';
end
%%%%%%%%%%%%%%calcKer%%%%%%%%%%%%%%%%%%%%%%%
function [A,Abar] = calcKer(A,rr,mu,output)
[sm sn]=size(A);
% calculate kernel
if output==1
fprintf(1,'\nCalculating kernel . . .\n');
end
rrows = floor(rr*sm); % reduced number of rows
indx = rand(sm,1);
[s1 s2]=sort(indx);
Abar = A(s2(1:rrows),:)';
A = Rec_Kernel(A,Abar,mu);
return;
%%%%%%%%%%%%%%%%%%%%%%%EstMu%%%%%%%%%%%%%
function mu = EstMu(A,d)
Aplus = A(find(d==1),:); Aminus=A(find(d==-1),:);
AplusRow = size(Aplus,1);
AminusRow = size(Aminus,1);
x=(sum(Aplus,1)/AplusRow + sum(Aminus,1)/AminusRow);
mu = 1/(1 + x*x');
return;
答案 0 :(得分：1)
由于d是10,000×10矩阵，因此从find返回的单个输出参数将是linear indices的数组，其中可包含1到100,000之间的任何值。由于A的第一维是10,000个元素，因此您很容易得到find的超过此值的索引。也许d应该是列向量（即10,000乘1）？