I have a feature vector of 13 dimensions for m samples. I am trying to find k nearest neighbors of each sample.
I have selected the feature vector required.
[m,~] = size(featurevector);
X = [];
Y = [];
for i = 1: m-1
if (featurevector(i,14) == 3)
X = [X;featurevector(i,1:13)];
Y = [Y;featurevector(i,1:13)];
end
end
I have tried calculating the distances for each point
d = pdist2(X,Y,'euclidean');
It works fine till here. Now I wanted to find the k = 3 nearest neighbor indexes of each and every sample, so it tried
[idx,dist] = knnsearch(X,Y,'k',3,'distance','euclidean');
but it shows an error
Error using pdist2
Too many input arguments.
Error in ExhaustiveSearcher/knnsearch (line 207)
[dist,idx] = pdist2(obj.X,Y, distMetric, arg{:}, 'smallest',numNN);
Error in knnsearch (line 144)
[idx, dist] = knnsearch(O,Y,'k',numNN, 'includeties',includeTies);
I have tried with example that mentioned in help . it works fine and I am not able to do when number of samples = 630.
Did my coding went wrong ?
I am using Matlab 2015a
my paths on command
which -all pdist2
gives
c:\toolbox\classify\pdist2.m C:\Program Files (x86)\MATLAB\MATLAB Production Server\R2015a\toolbox\stats\stats\pdist2.m % Shadowed
any help appreciated !
Related
I'm using the following code of "Computed tomography-based volumetric tool for standardized measurement of the maxillary sinus" article to measure the volume of the maxillary sinus in DCOM images:
% Read images
clear all
close all
[filename, pathname] = uigetfile('*','Select CT exam (all slices)','MultiSelect','on');
num= length(filename);
bbbb=1;
step=input('Step of image reading: \')
for aaaa = 1:step:num
xinfo=dicominfo([pathname,char(filename(aaaa))]);
pxsp=cat(2,xinfo.PixelSpacing);
x=dicomread([pathname,char(filename(aaaa))])+cat(2,xinfo.RescaleIntercept);
k=x;
k = im2bw(k,0.49);
k = imfill(k,'holes');
cc = bwconncomp(k);
stats = regionprops(cc,'Area');
A = [stats.Area];
[~,biggest] = max(A);
k(labelmatrix(cc)~=biggest) = 0;
x(k~=1)=-2000;
masccranio(:,:,bbbb)=k;
cranio(:,:,bbbb)=x;
cranio_full(:,:,bbbb)=x;
bbbb=bbbb+1;
end
at the first, we don't have any idea about the reading step input at the beginning, please help on that if you can. Our second problem is when we run the code we get the following error:
Error using ~=
Matrix dimensions must agree.
Error in Quant (line 32)
k(labelmatrix(cc)~=biggest) = 0;
I'm using Matlab 2019b and as I know this code is for 2013. Any help is appreciated.
I expect you want biggest to represent the biggest value of all region areas in this line:
k(labelmatrix(cc)~=biggest) = 0;
In that case, your problem is likely that A is not a vector, and therefore biggest is not a scalar. This causes the operation to fail as the ~= operator only works on matrices of the same size or with scalars.
I have 94 samples with 263 features for each sample. The total feature vector is 94*263 in size. There are no NaN or Inf value in the feature vectors. There are two classes (51 in class a and 43 in class b). I am using sequentialfs to select features but I am getting the following error each time:
Error using crossval>evalFun (line 480)
The function '#(XT,yT,Xt,yt)(sum(~strcmp(yt,classify(Xt,XT,yT,'quadratic'))))' generated the following error:
The input to SVD must not contain NaN or Inf.
The code is:
X = FEATUREVECTOR;
y = LABELS;
c = cvpartition(y,'k',10);
opts = statset('display','iter');
fun = #(XT,yT,Xt,yt)...
(sum(~strcmp(yt,classify(Xt,XT,yT,'quadratic'))));
[fs,history] = sequentialfs(fun,X,y,'cv',c,'options',opts)
Can you please tell me how to solve the problem?
It looks like you are calling sequentialfs with some inputs, that MAY be vaguely related to the mess of random numbers we see in your question. Beyond that, I can't read anything from your mind. If you want help you need to show what you did.
I change input data and it works well,
load fisheriris;
X = randn(150,10);
X(:,[1 3 5 7 ])= meas;
y = species;
c = cvpartition(y,'k',10);
opts = statset('display','iter');
fun = #(XT,yT,Xt,yt)...
(sum(~strcmp(yt,classify(Xt,XT,yT,'quadratic'))));
[fs,history] = sequentialfs(fun,X,y,'cv',c,'options',opts)
Your input data has problem.
After having some basics understanding of GPML toolbox , I written my first code using these tools. I have a data matrix namely data consist of two array values of total size 1000. I want to use this matrix to estimate the GP value using GPML toolbox. I have written my code as follows :
x = data(1:200,1); %training inputs
Y = data(1:201,2); %, training targets
Ys = data(201:400,2);
Xs = data(201:400,1); %possibly test cases
covfunc = {#covSE, 3};
ell = 1/4; sf = 1;
hyp.cov = log([ell; sf]);
likfunc = #likGauss;
sn = 0.1;
hyp.lik = log(sn);
[ymu ys2 fmu fs2] = gp(hyp, #infExact, [], covfunc, likfunc,X,Y,Xs,Ys);
plot(Xs, fmu);
But when I am running this code getting error 'After having some basics understanding of GPML toolbox , I written my first code using these tools. I have a data matrix namely data consist of two array values of total size 1000. I want to use this matrix to estimate the GP value using GPML toolbox. I have written my code as follows :
x = data(1:200,1); %training inputs
Y = data(1:201,2); %, training targets
Ys = data(201:400,2);
Xs = data(201:400,1); %possibly test cases
covfunc = {#covSE, 3};
ell = 1/4; sf = 1;
hyp.cov = log([ell; sf]);
likfunc = #likGauss;
sn = 0.1;
hyp.lik = log(sn);
[ymu ys2 fmu fs2] = gp(hyp, #infExact, [], covfunc, likfunc,X,Y,Xs,Ys);
plot(Xs, fmu);
But when I am running this code getting:
Error using covMaha (line 58) Parameter mode is either 'eye', 'iso',
'ard', 'proj', 'fact', or 'vlen'
Please if possible help me to figure out where I am making mistake ?
I know this is way late, but I just ran into this myself. The way to fix it is to change
covfunc = {#covSE, 3};
to something like
covfunc = {#covSE, 'iso'};
It doesn't have to be 'iso', it can be any of the options listed in the error message. Just make sure your hyperparameters are set correctly for the specific mode you choose. This is detailed more in the covMaha.m file in GPML.
Matlab (2015a) is behaving weirdly: a number of builtin functions are not responding as expected. For instance, typing
ttest([1 2], [1 2])
results in
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanstd (line 59)
tile(dim) = size(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If I do a which for each of these functions:
which size
which nanstd
which ttest
I get, respetively:
built-in (C:\Program Files\MATLAB\R2015a\toolbox\matlab\elmat\size)
C:\Program Files\MATLAB\R2015a\toolbox\stats\eml\nanstd.m
C:\Program Files\MATLAB\R2015a\toolbox\stats\stats\ttest.m
Each of these files looks fine, except that size.m has each one of its rows commented out.
What could be the problem here?
Perhaps related to your problem:
ttest for R2013a makes the following call:
sdpop = nanstd(x,[],dim);
The helpfile for R2013a version of nanstd states:
Y = nanstd(X,FLAG,DIM) takes the standard deviation along dimension DIM of X.
On the other hand, nanstd in the 2005 nansuite package downloaded off Mathworks file exchange states:
FORMAT: Y = nanstd(X,DIM,FLAG)
Notice how DIM and FLAG are reversed!
If I call R2013a's ttest such that it makes a call to the old, 2005 nansuite function nanstd, Matlab generates an error similar to yours:
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanmean (line 46)
count = size(x,dim) - sum(nans,dim);
Error in nanstd (line 54)
avg = nanmean(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If [] is passed as DIM instead of FLAG, then nanstd's call to size(x, DIM) triggers an error because [] is not a positive integer scalar. If something like this is the cause, the broader question is, what's going on with your Matlab install or setup or downloads or whatever such that you're calling archaic code? Or why is that archaic code even around? I don't know at what point in Matlab's release history that nanstd(x, FLAG, DIM) became supported (instead of simply nanstd(x, DIM))?
Archive: below is my old answer which misdiagnosed your problem
Both of your sample vectors x and y are the same (i.e. [1,2]). The estimated variance of the difference is 0, and all your stats are going to blow up with NaN.
Do the stats step by step, and it will be clear what's going on.
x = [1; 2]; % Data you used in the example.
y = [1; 2]; % Data you used in the example.
z = x - y; % Your call to ttest tests whether this vector is different from zero at a statistically significant level.
Now we do all the stats on z
r.n = length(z);
r.mu = mean(z);
r.standard_error = sqrt(var(z,1) / (r.n-1)); % For your data, this will be zero since z is constant!
r.t = r.mu ./ r.standard_error; % For your data, this will be inf because dividing by zero!
r.df = r.n - 1;
r.pvals(r.t >= 0) = 2 * (1 - tcdf(r.t(r.t>=0), r.df)); % For your data, tcdf returns NaN and this all fails...
r.pvals(r.t < 0) = 2 * tcdf(r.t(r.t<0), r.df);
etc...
This should match a call to
[h, p, ci, stats] = ttest(x-y);
I am currently using the Toolbox Graph on the Matlab File Exchange to calculate curvature on 3D surfaces and find them very helpful (http://www.mathworks.com/matlabcentral/fileexchange/5355). However, the following error message is issued in “compute_curvature” for certain surface descriptions and the code fails to run completely:
> Error in ==> compute_curvature_mod at 75
> dp = sum( normal(:,E(:,1)) .* normal(:,E(:,2)), 1 );
> ??? Index exceeds matrix dimensions.
This happens only sporadically, but there is no obvious reason why the toolbox works perfectly fine for some surfaces and not for others (of a similar topology). I also noticed that someone had asked about this bug back in November 2009 on File Exchange, but that the question had gone unanswered. The post states
"compute_curvature will generate an error on line 75 ("dp = sum(
normal(:,E(:,1)) .* normal(:,E(:,2)), 1 );") for SOME surfaces. The
error stems from E containing indices that are out of range which is
caused by line 48 ("A = sparse(double(i),double(j),s,n,n);") where A's
values eventually entirely make up the E matrix. The problem occurs
when the i and j vectors create the same ordered pair twice in which
case the sparse function adds the two s vector elements together for
that matrix location resulting in a value that is too large to be used
as an index on line 75. For example, if i = [1 1] and j = [2 2] and s
= [3 4] then A(1,2) will equal 3 + 4 = 7.
The i and j vectors are created here:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
Just wanted to add that the error I mentioned is caused by the
flipping of the sign of the surface normal of just one face by
rearranging the order of the vertices in the face matrix"
I have tried debugging the code myself but have not had any luck. I am wondering if anyone here has solved the problem or could give me insight – I need the code to be sufficiently general-purpose in order to calculate curvature for a variety of surfaces, not just for a select few.
The November 2009 bug report on File Exchange traces the problem back to the behavior of sparse:
S = SPARSE(i,j,s,m,n,nzmax) uses the rows of [i,j,s] to generate an
m-by-n sparse matrix with space allocated for nzmax nonzeros. The
two integer index vectors, i and j, and the real or complex entries
vector, s, all have the same length, nnz, which is the number of
nonzeros in the resulting sparse matrix S . Any elements of s
which have duplicate values of i and j are added together.
The lines of code where the problem originates are here:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
s = [1:m 1:m 1:m];
A = sparse(i,j,s,n,n);
Based on this information removal of the repeat indices, presumably using unique or similar, might solve the problem:
[B,I,J] = unique([i.' j.'],'rows');
i = B(:,1).';
j = B(:,2).';
s = s(I);
The full solution may look something like this:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
s = [1:m 1:m 1:m];
[B,I,J] = unique([i.' j.'],'rows');
i = B(:,1).';
j = B(:,2).';
s = s(I);
A = sparse(i,j,s,n,n);
Since I do not have a detailed understanding of the algorithm it is hard to tell whether the removal of entries will have a negative effect.