I am trying to execute this function for image reconstruction where
ra, rh, rv, rd are reconstructed coefficients. but i am facing problem in addition and subtraction.
Please help.
Xhat = ra2 + rh2 + rv2 + rd2 + rh1 + rv1 + rd1;
sprintf('Reconstruction error (using wrcoef2) = %g', max(max(abs(X-Xhat))))
OR
XXhat = waverec2(wc,s,wname);
sprintf('Reconstruction error (using waverec2) = %g', max(max(abs(X-XXhat)))
I decomposed the image using:
>> a1 = appcoef2(wc,s,wname,1);
>> h1 = detcoef2('h',wc,s,1);
>> v1 = detcoef2('v',wc,s,1);
>> d1 = detcoef2('d',wc,s,1);
>> a2 = appcoef2(wc,s,wname,2);
>> h2 = detcoef2('h',wc,s,2);
>> v2 = detcoef2('v',wc,s,2);
>> d2 = detcoef2('d',wc,s,2);
Then reconstructed using above parameters.
Now i have to comnbine them.
I'm guessing your problem is almost certainly at the abs(X-Xhat) line.
Why? You seem to be doing some sort of wavelet decomposition/reconstruction, and if you don't pass in the right parameters your output may be larger than the original image. Therefore it makes no sense to ask for X-Xhat if these are of different sizes and you will get an error message.
The best way of fixing this is to when you reconstruct (presumably using upcoef2), is to pass size as an additional parameter to truncate. A truncated (heh) and adjusted example from the docs:
load woman;
[c,s] = wavedec2(X,2,'db4');
siz = s(size(s,1),:);
ca1 = appcoef2(c,s,'db4',1);
a = upcoef2('a',ca1,'db4',1,siz);
a2 = upcoef2('a',ca1,'db4',1);
You'll see that size(X) and size(a) are both 256 x 256, but size(a2) is larger. Therefore a-X is fine and a2-X will give you a "Matrix dimensions must agree." error.
Related
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.
I asked this question in Math Stackexchange, but it seems it didn't get enough attention there so I am asking it here. https://math.stackexchange.com/questions/1729946/why-do-we-say-svd-can-handle-singular-matrx-when-doing-least-square-comparison?noredirect=1#comment3530971_1729946
I learned from some tutorials that SVD should be more stable than QR decomposition when solving Least Square problem, and it is able to handle singular matrix. But the following example I wrote in matlab seems to support the opposite conclusion. I don't have a deep understanding of SVD, so if you could look at my questions in the old post in Math StackExchange and explain it to me, I would appreciate a lot.
I use a matrix that have a large condition number(e+13). The result shows SVD get a much larger error(0.8) than QR(e-27)
% we do a linear regression between Y and X
data= [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
X = data(:,1);
Y = data(:,2);
X_1 = [ones(length(X),1),X];
%%
%SVD method
[U,D,V] = svd(X_1,'econ');
beta_svd = V*diag(1./diag(D))*U'*Y;
%% QR method(here one can also use "\" operator, which will get the same result as I tested. I just wrote down backward substitution to educate myself)
[Q,R] = qr(X_1)
%now do backward substitution
[nr nc] = size(R)
beta_qr=[]
Y_1 = Q'*Y
for i = nc:-1:1
s = Y_1(i)
for j = m:-1:i+1
s = s - R(i,j)*beta_qr(j)
end
beta_qr(i) = s/R(i,i)
end
svd_error = 0;
qr_error = 0;
for i=1:length(X)
svd_error = svd_error + (Y(i) - beta_svd(1) - beta_svd(2) * X(i))^2;
qr_error = qr_error + (Y(i) - beta_qr(1) - beta_qr(2) * X(i))^2;
end
You SVD-based approach is basically the same as the pinv function in MATLAB (see Pseudo-inverse and SVD). What you are missing though (for numerical reasons) is using a tolerance value such that any singular values less than this tolerance are treated as zero.
If you refer to edit pinv.m, you can see something like the following (I won't post the exact code here because the file is copyrighted to MathWorks):
[U,S,V] = svd(A,'econ');
s = diag(S);
tol = max(size(A)) * eps(norm(s,inf));
% .. use above tolerance to truncate singular values
invS = diag(1./s);
out = V*invS*U';
In fact pinv has a second syntax where you can explicitly specify the tolerance value pinv(A,tol) if the default one is not suitable...
So when solving a least-squares problem of the form minimize norm(A*x-b), you should understand that the pinv and mldivide solutions have different properties:
x = pinv(A)*b is characterized by the fact that norm(x) is smaller than the norm of any other solution.
x = A\b has the fewest possible nonzero components (i.e sparse).
Using your example (note that rcond(A) is very small near machine epsilon):
data = [
47.667483331 -122.1070832;
47.667483331001 -122.1070832
];
A = [ones(size(data,1),1), data(:,1)];
b = data(:,2);
Let's compare the two solutions:
x1 = A\b;
x2 = pinv(A)*b;
First you can see how mldivide returns a solution x1 with one zero component (this is obviously a valid solution because you can solve both equations by multiplying by zero as in b + a*0 = b):
>> sol = [x1 x2]
sol =
-122.1071 -0.0537
0 -2.5605
Next you see how pinv returns a solution x2 with a smaller norm:
>> nrm = [norm(x1) norm(x2)]
nrm =
122.1071 2.5611
Here is the error of both solutions which is acceptably very small:
>> err = [norm(A*x1-b) norm(A*x2-b)]
err =
1.0e-11 *
0 0.1819
Note that use mldivide, linsolve, or qr will give pretty much same results:
>> x3 = linsolve(A,b)
Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.159326e-16.
x3 =
-122.1071
0
>> [Q,R] = qr(A); x4 = R\(Q'*b)
x4 =
-122.1071
0
SVD can handle rank-deficiency. The diagonal matrix D has a near-zero element in your code and you need use pseudoinverse for SVD, i.e. set the 2nd element of 1./diag(D) to 0 other than the huge value (10^14). You should find SVD and QR have equally good accuracy in your example. For more information, see this document http://www.cs.princeton.edu/courses/archive/fall11/cos323/notes/cos323_f11_lecture09_svd.pdf
Try this SVD version called block SVD - you just set the iterations equal to the accuracy you want - usually 1 is enough. If you want all the factors (this has a default # selected for factor reduction) then edit the line k= to the size(matrix) if I recall my MATLAB correctly
A= randn(100,5000);
A=corr(A);
% A is your correlation matrix
tic
k = 1000; % number of factors to extract
bsize = k +50;
block = randn(size(A,2),bsize);
iter = 2; % could set via tolerance
[block,R] = qr(A*block,0);
for i=1:iter
[block,R] = qr(A*(A'*block),0);
end
M = block'*A;
% Economy size dense SVD.
[U,S] = svd(M,0);
U = block*U(:,1:k);
S = S(1:k,1:k);
% Note SVD of a symmetric matrix is:
% A = U*S*U' since V=U in this case, S=eigenvalues, U=eigenvectors
V=real(U*sqrt(S)); %scaling matrix for simulation
toc
% reduced randomized matrix for simulation
sims = 2000;
randnums = randn(k,sims);
corrrandnums = V*randnums;
est_corr_matrix = corr(corrrandnums');
total_corrmatrix_difference =sum(sum(est_corr_matrix-A))
After retrieving image coeffients (both approximate and detailed) in wavelet decomposition, i want quadtree structure for appoximate and detailed coefficients. but every time i am getting same error. Please help.
I tried to make it power of 3 instead of 2 but it coudnt help. may be i am going wrong somehwhere.
code for image decomposition
X=imread('abc.tif');
close all
clf
image(X)
colormap('default')
axis ('image'); set(gca,'XTick',[],'YTick',[]); title('Original')
pause
dwtmode('sym')
wname = 'bior4.4'
t = wtree(X,2,'bior4.4');
plot(t)
pause
close(2)
[wc,s] = wavedec2(X,5,wname);
a1 = appcoef2(wc,s,wname,1);
h1 = detcoef2('h',wc,s,1);
v1 = detcoef2('v',wc,s,1);
d1 = detcoef2('d',wc,s,1);
a2 = appcoef2(wc,s,wname,2);
h2 = detcoef2('h',wc,s,2);
v2 = detcoef2('v',wc,s,2);
d2 = detcoef2('d',wc,s,2);
a3 = appcoef2(wc,s,wname,3);
h3 = detcoef2('h',wc,s,3);
v3 = detcoef2('v',wc,s,3);
d3 = detcoef2('d',wc,s,3);
a4 = appcoef2(wc,s,wname,4);
h4 = detcoef2('h',wc,s,4);
v4 = detcoef2('v',wc,s,4);
d4 = detcoef2('d',wc,s,4);
a5 = appcoef2(wc,s,wname,5);
h5 = detcoef2('h',wc,s,5);
v5 = detcoef2('v',wc,s,5);
d5 = detcoef2('d',wc,s,5);
For quadtree i am using:
S = qtdecomp(I,.27); %I is image in greyscale.
blocks = repmat(uint8(0),size(S));
for dim = [512 256 128 64 32 16 8 4 2 1];
numblocks = length(find(S==dim));
if (numblocks > 0)
values = repmat(uint8(1),[dim dim numblocks]);
values(2:dim,2:dim,:) = 0;
blocks = qtsetblk(blocks,S,dim,values);
end
end
blocks(end,1:end) = 1;
blocks(1:end,end) = 1;
imshow(I), figure, imshow(blocks,[])
the error it is showing is:
??? Error using ==> qtdecomp>ParseInputs at 229
MAXDIM / MINDIM is not a power of 2
Error in ==> qtdecomp at 88
[A, func, params, minDim, maxDim] = ParseInputs(varargin{:});
or
??? Error using ==> qtdecomp>ParseInputs at 145
A must be two-dimensional
Error in ==> qtdecomp at 88
[A, func, params, minDim, maxDim] = ParseInputs(varargin{:});
The error is pretty clear. Your image must have dimensions that belong to a power of two. As such, you need to resize your image so that it follows this. qtdecomp computes the quadtree decomposition algorithm, and a pre-condition for this function is that it NEEDS to have an image that has each dimension as a power of two as input.
You asked something similar in another question that I provided an answer to. What is curious is that this other question you asked was 6 days ago before this one, and you accepted my answer for that other one. The issue that you were experiencing is more or less the same thing.
Quadtree decomposition
i have some experimental data and a theoretical model which i would like to try and fit. i have made a function file with the model - the code is shown below
function [ Q,P ] = RodFit(k,C )
% Function file for the theoretical scattering from a Rod
% R = radius, L = length
R = 10; % radius in Å
L = 1000; % length in Å
Q = 0.001:0.0001:0.5;
fun = #(x) ( (2.*besselj(1,Q.*R.*sin(x)))./...
(Q.*R.*sin(x)).*...
(sin(Q.*L.*cos(x)./2))./...
(Q.*L.*cos(x)./2)...
).^2.*sin(x);
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
end
with Q being the x-values and P being the y-values. I can call the function fine from the matlab command line and it works fine e.g. [Q,P] = RodFit(1,0.001) gives me a result i can plot using plot(Q,P)
But i cannot figure how to best find the fit to some experimental data. Ideally, i would like to use the optimization toolbox and lsqcurvefit since i would then also be able to optimize the R and L parameters. but i do not know how to pass (x,y) data to lsqcurvefit. i have attempted it with the code below but it does not work
File = 30; % the specific observation you want to fit the model to
ydata = DataFiles{1,File}.data(:,2)';
% RAdius = linspace(10,1000,length(ydata));
% LEngth = linspace(100,10000,length(ydata));
Multiplier = linspace(1e-3,1e3,length(ydata));
Constant = linspace(0,1,length(ydata));
xdata = [Multiplier; Constant]; % RAdius; LEngth;
L = lsqcurvefit(#RodFit,[1;0],xdata,ydata);
it gives me the error message:
Error using *
Inner matrix dimensions must agree.
Error in RodFit (line 15)
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
Error in lsqcurvefit (line 199)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Caused by:
Failure in initial user-supplied objective function evaluation. LSQCURVEFIT cannot continue.
i have tried i) making all vectors/matrices the same length and ii) tried using .* instead. nothing works and i am giving the same error message
Any kind of help would be greatly appreciated, whether it is suggestion regading what method is should use, suggestions to my code or something third.
EDIT TO ANSWER Osmoses:
A really good point but i do not think that is the problem. just checked the size of the all the vectors/matrices and they should be alright
>> size(Q)
ans =
1 1780
>> size(P)
ans =
1 1780
>> size(xdata)
ans =
2 1780
>> size([1;0.001]) - the initial guess/start point for xdata (x0)
ans =
2 1
>> size(ydata)
ans =
1 1780
UPDATE
I think i have identified the problem. the function RodFit works fine when i specify the input directly e.g. [Q,P] = RodFit(1,0.001);.
however, if i define x0 as x0 = [1,0.001] i cannot pass x0 to the function
>> x0 = [1;0.001]
x0 =
1.0000
0.0010
>> RodFit(x0);
Error using *
Inner matrix dimensions must agree.
Error in RodFit (line 15)
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
The same happens if i use x0 = [1,0.001]
clearly, matlab is interpreting x0 as input for k only and attempts to multiplay a vector of length(ydata) and a vector of length(x0) which obviously fails.
So my problem is that i need to code so that lsqcurvefit understands that the first column of xdata and x0 is the k variable and the second column of xdata and x0 is the C variable. According to the documentation - Passing Matrix Arguments - i should be able to pass x0 as a matrix to the solver. The solver should then also pass the xdata in the same format as x0.
Have you tried (that's sometimes the mistake) looking at the orientation of your input data (e.g. if xdata & ydata are both row/column vectors?). Other than that your code looks like it should work.
I have been able to solve some of the problems. One mistake in my code was that the objective function did not use of vector a variables but instead took in two variables - k and C. changing the code to accept a vector solved this problem
function [ Q,P ] = RodFit(X)
% Function file for the theoretical scattering from a Rod
% R = radius, L = length
% Q = 0.001:0.0001:0.5;
Q = linspace(0.11198,4.46904,1780);
fun = #(x) ( (2.*besselj(1,Q.*R.*sin(x)))./...
(Q.*R.*sin(x)).*...
(sin(Q.*L.*cos(x)./2))./...
(Q.*L.*cos(x)./2)...
).^2.*sin(x);
P = (integral(fun,0,pi/2,'ArrayValued',true))*X(1)+X(2);
with the code above, i can define x0 as x0 = [1 0.001];, and pass that into RodFit and get a result. i can also pass xdata into the function and get a result e.g. [Q,P] = RodFit(xdata(2,:));
Notice i have changed the orientation of all vectors so that they are now row-vectors and xdata has size size(xdata) = 1780 2
so i thought i had solved the problem completely but i still run into problems when i run lsqcurvefit. i get the error message
Error using RodFit
Too many input arguments.
Error in lsqcurvefit (line 199)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Caused by:
Failure in initial user-supplied objective function evaluation. LSQCURVEFIT cannot continue.
i have no idea why - does anyone have any idea about why Rodfit recieves to many input arguments when i call lsqcurvefit but not when i run the function manual using xdata?
I am using Gonzalez frdescp function to get Fourier descriptors of a boundary. I use this code, and I get two totally different sets of numbers describing two identical but different in scale shapes.
So what is wrong?
im = imread('c:\classes\a1.png');
im = im2bw(im);
b = bwboundaries(im);
f = frdescp(b{1}); // fourier descriptors for the boundary of the first object ( my pic only contains one object anyway )
// Normalization
f = f(2:20); // getting the first 20 & deleting the dc component
f = abs(f) ;
f = f/f(1);
Why do I get different descriptors for identical - but different in scale - two circles?
The problem is that the frdescp code (I used this code, that should be the same as referred by you) is written also in order to center the Fourier descriptors.
If you want to describe your shape in a correct way, it is mandatory to mantain some descriptors that are symmetric with respect to the one representing the DC component.
The following image summarize the concept:
In order to solve your problem (and others like yours), I wrote the following two functions:
function descriptors = fourierdescriptor( boundary )
%I assume that the boundary is a N x 2 matrix
%Also, N must be an even number
np = size(boundary, 1);
s = boundary(:, 1) + i*boundary(:, 2);
descriptors = fft(s);
descriptors = [descriptors((1+(np/2)):end); descriptors(1:np/2)];
end
function significativedescriptors = getsignificativedescriptors( alldescriptors, num )
%num is the number of significative descriptors (in your example, is was 20)
%In the following, I assume that num and size(alldescriptors,1) are even numbers
dim = size(alldescriptors, 1);
if num >= dim
significativedescriptors = alldescriptors;
else
a = (dim/2 - num/2) + 1;
b = dim/2 + num/2;
significativedescriptors = alldescriptors(a : b);
end
end
Know, you can use the above functions as follows:
im = imread('test.jpg');
im = im2bw(im);
b = bwboundaries(im);
b = b{1};
%force the number of boundary points to be even
if mod(size(b,1), 2) ~= 0
b = [b; b(end, :)];
end
%define the number of significative descriptors I want to extract (it must be even)
numdescr = 20;
%Now, you can extract all fourier descriptors...
f = fourierdescriptor(b);
%...and get only the most significative:
f_sign = getsignificativedescriptors(f, numdescr);
I just went through the same problem with you.
According to this link, if you want invariant to scaling, make the comparison ratio-like, for example by dividing every Fourier coefficient by the DC-coefficient. f*1 = f1/f[0], f*[2]/f[0], and so on. Thus, you need to use the DC-coefficient where the f(1) in your code is not the actual DC-coefficient after your step "f = f(2:20); % getting the first 20 & deleting the dc component". I think the problem can be solved by keeping the value of the DC-coefficient first, the code after adjusted should be like follows:
% Normalization
DC = f(1);
f = f(2:20); % getting the first 20 & deleting the dc component
f = abs(f) ; % use magnitudes to be invariant to translation & rotation
f = f/DC; % divide the Fourier coefficients by the DC-coefficient to be invariant to scale