Matt J
Professional Interests: medical image processing, optimization algorithms
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Content Feed
how convert the vector to matrix ? unknown number of columns
No, the solution is ill-defined. Another result with the same number of rows is: A=[1 2 3 4 5 6 7 8 9]
alrededor de 9 horas ago | 0
Applying vectorization techniques to speedup the performance of dividing a 3D matrix by a 2D matrix
Another idea. clc; clear all; % Test Data % I'm trying to remove the for loop for j in the code below N = 10; M = 10; Nx...
alrededor de 12 horas ago | 1
Applying vectorization techniques to speedup the performance of dividing a 3D matrix by a 2D matrix
On the GPU (i.e. if A and b are gpuArrays), the for-loop can be removed: Uhat = permute( pagefun(@mldivide,A,reshape(b,[],1,Nx)...
1 día ago | 0
Finding the vertical offset of a gaussian fit
This File Exchange routine does gaussian+constant fitting: https://www.mathworks.com/matlabcentral/fileexchange/69116-gaussfitn...
1 día ago | 0
| accepted
How to set a custom equation to fit 5 points in space by fitsurface?
I can't find "fitsurface" in the Mathworks documentation, but the fit is easy enough to do algebraically. x=x(:); y=y(:); z=z(:...
1 día ago | 0
Why is x(:) so much slower than reshape(x,N,1) with complex arrays?
The following simple test seems to support @Bruno Luong's conjecture that (:) results in data copying. The data of B1 resulting ...
2 días ago | 1
Is it possible to correctly perform a multi-dimensional FFT on a 1D linearised version of a 3D array?
Reshaping to and from 3D format should not add significant cost: Nx = 256; Ny = 256; Nz = 128; N = Nx*Ny*Nz; A0 = rand(N,1)...
2 días ago | 0
| accepted
Question
Why is x(:) so much slower than reshape(x,N,1) with complex arrays?
The two for loops below differ only in the flattening operation used to obtain A_1D . Why is the run time so much worse with A_3...
2 días ago | 3 answers | 2
3
answerssee the 3d image and see the coordinates while moving the mouse over the image
You can use imshow() to display any particular slice of the image and then use the axis datatip button to get the in-slice 2D co...
3 días ago | 0
How do i find the coordinates of a boundary
How about simply, [m,n]=size(Image); I=boundary(:,1); J=boundary(:,2); corner1 = [1, max(J(I==1))]; %row/column coordinat...
3 días ago | 0
How to Draw an ellipsoid between two 3d coordinates
Another way would be to use the ellipsoidalFit.groundTruth method in this File Exchange submission. https://www.mathworks.com/m...
4 días ago | 0
Trying to count the number of pixels in an image over a specific RBG value
results.area=nnz(imstd>90);
4 días ago | 0
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Normally distributed random numbers with fixed sum
Why not simply, x(1:N-1)=randn(1,N-1); x(N)=fixedValue-sum(x(1:N-1))
4 días ago | 0
Only analyze what is outlined in black in an image
If you have a binary mask of the circles, you can just do numBlack = nnz( circles & ~BW ); numWhite = nnz( circles & BW);
4 días ago | 0
If the quadprog problem is infeasible, how i can solve it with a certain tolerance?
It never really makes sense to have more equality constraints (here 36) than you do unknowns ( here 6). In order for the problem...
6 días ago | 0
| accepted
How to construct a "ones" matrix with zero blocks in the diagonal
repelem(1-eye(4),4,4)
6 días ago | 1
Iterative solution of non-linear equation with a provided guess value?
You could use fzero's OutputFcn option https://www.mathworks.com/help/optim/ug/fzero.html#btn3z5s-options to access the sequen...
6 días ago | 0
Iterative solution of non-linear equation with a provided guess value?
Using fsolve() instead of fzero(), you could solve the problem as a system of two equations in two unknowns. The two unknowns ar...
6 días ago | 0
Failure in initial user-supplied nonlinear constraint function evaluation
One problem: nonlcon = @simple_constraint; There is no function called 'simple_constraint' in your attached script. There is a...
7 días ago | 1
| accepted
for loop for permutations of for length of array
List=nchoosek([a,b,c],2); for i=1:size(List,1) p1=List(i,1); p2=List(i,2); ... end
7 días ago | 1
| accepted
How do I use an if-else structure inside a for loop to investigate if each component of a row vector is inside or outside the boundaries?
Hint: for i=1:numel(vec) v=vec(i); end
8 días ago | 0
2*2 Matrix factorization/decomposition
I also found out what I wanted may have something to do with PLU(permuted lower upper decomposition), but am clueless to how to ...
8 días ago | 0
How can I call a script within a for loop, looping over multiple files?
result = scriptname(ID{nSubject},Visit{nVisit});
8 días ago | 0
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I am trying to find the exact distance of a curvy line.
Why not as follows? I2 = imread('image.jpeg'); blue=bwareafilt( I2(:,:,1)<=50 & I2(:,:,2)<=50 & I2(:,:,3)>=200, 1); Len...
8 días ago | 0
| accepted
Number of branches per branch point
Perhaps as follows: k=ones(3); k(5)=0; neighborCount=conv2(double(BW),k,'same'); neighborCount=neighborCount.*(neighborCo...
9 días ago | 0
| accepted
Rank deficient error.
Are you attempting to do matrix division? if not use ./ instead.
9 días ago | 1
| accepted
Merge struct with matrix
You cannot avoid a for-loop altogether, but you can avoid writing your own for-loop as follows: tmp=num2cell(modelOutputSignal,...
9 días ago | 0
| accepted
Unable to detect Iris boundary using imfind circles
Increasing the lower radius bound as suggested by @Jonas as well as increasing the sensitivity setting seems to help: A=any( im...
9 días ago | 0
| accepted
adjusting the code from optimization output from minimum to maximum
fun=str2func(FUNCTION_NAME); fun=@(x) -fun(x); [Scaled_x_Single,fval,exitflag,output,population,scores] = ga(,3,[],[],[],[],...
9 días ago | 0
Question
Re-import a binary map into Image Segmenter or Color Thresholder
I have generated a binary mask of an image using the Image Segmenter app. I am wondering if it is possible to re-import that mas...
10 días ago | 1 answer | 0
