Professional Interests: medical image processing, optimization algorithms

PLEASE NOTE: I do not read email sent through my author page. Please post questions about FEX submissions in their respective Comments section.

Answered

vectorize a for loop

This should be pretty well optimized already, r=pdist([xc,yc]);

vectorize a for loop

This should be pretty well optimized already, r=pdist([xc,yc]);

ongeveer 5 uur ago | 1

| accepted

Answered

Function with one input and no outputs

printmsg("Hello World") function printmsg(msg) disp(msg) end

Function with one input and no outputs

printmsg("Hello World") function printmsg(msg) disp(msg) end

ongeveer 5 uur ago | 0

| accepted

Answered

What is the sampling frequency after convolution?

I want to calculate the convolution of the two signals with conv If you want to convolve with conv() then the signals definitel...

What is the sampling frequency after convolution?

I want to calculate the convolution of the two signals with conv If you want to convolve with conv() then the signals definitel...

ongeveer 5 uur ago | 0

Answered

detecting rectangle in real images

This works for the one given image. I would need more examples to see if it's a reliable solution. It uses the FEX contribution ...

detecting rectangle in real images

This works for the one given image. I would need more examples to see if it's a reliable solution. It uses the FEX contribution ...

ongeveer 9 uur ago | 0

Answered

Plotting a 3d cone

the general cone generators like 'cylinder([0 1])' are not usable for me. I think they are. You just have to reposition the co...

Plotting a 3d cone

the general cone generators like 'cylinder([0 1])' are not usable for me. I think they are. You just have to reposition the co...

ongeveer 10 uur ago | 0

Answered

Randomly remove percentage of values above mean in array

A = [set of values]; A_rem=A; index1 = find(A>mean(A)); N=numel(index1); index2 = index1( randi(N,1,round(0.2*N)) );...

Randomly remove percentage of values above mean in array

A = [set of values]; A_rem=A; index1 = find(A>mean(A)); N=numel(index1); index2 = index1( randi(N,1,round(0.2*N)) );...

ongeveer 11 uur ago | 0

| accepted

Answered

Boolean False when it should be True

dom(8,2) is only 8.3000 to 4 decimal places. You cannot see the differences beyond the 4th decimal unless you use >> format lon...

Boolean False when it should be True

dom(8,2) is only 8.3000 to 4 decimal places. You cannot see the differences beyond the 4th decimal unless you use >> format lon...

ongeveer 11 uur ago | 0

Answered

how do i get MATLAB to extract a specific value in multidimension array or matrix

idx=CHIRPS_SPI3_SM<-1 & CHIRPS_SPI3_SM>-4; extractedValues=CHIRPS_SPI3_SM(idx);

how do i get MATLAB to extract a specific value in multidimension array or matrix

idx=CHIRPS_SPI3_SM<-1 & CHIRPS_SPI3_SM>-4; extractedValues=CHIRPS_SPI3_SM(idx);

ongeveer 14 uur ago | 0

Submitted

Further tools for analyzing objects in N-dimensional images

A collection of tools to supplement bwpropfilt, bwareaopen, and the like.

ongeveer 22 uur ago | 1 download |

Question

Axis resizes inconsistently in a TiledChartLayout when you change its TileSpan

In the first example below, I obtain a TiledChartLayout in which all the plots have the same size (which is what I want). Why do...

ongeveer 23 uur ago | 0 answers | 0

Answered

set intervals with starts and ends in two arrays

Because it's homework, I've left some blanks for you to fill in. starts = [ 1 20 30 40]; ends = [3 22 34 41]; D=ends-starts...

set intervals with starts and ends in two arrays

Because it's homework, I've left some blanks for you to fill in. starts = [ 1 20 30 40]; ends = [3 22 34 41]; D=ends-starts...

1 dag ago | 0

Answered

How do I check/verify that an array is a certain dimension?

while ~isequal(size(point),[1,2])

How do I check/verify that an array is a certain dimension?

while ~isequal(size(point),[1,2])

1 dag ago | 0

Answered

Counting Number of Voxels between 2 points without drawing lines

Assuming you have a binarized image and the curve is the only object in that image, [I,J]=find( bwmorph(yourImage,'endpoints'))...

Counting Number of Voxels between 2 points without drawing lines

Assuming you have a binarized image and the curve is the only object in that image, [I,J]=find( bwmorph(yourImage,'endpoints'))...

1 dag ago | 0

Answered

How does the command: real(ifft(fftshift(Y))*N) operate?

All of those oeprations are O(N) except for the IFFT which is O(Nlog(N)). So the chain of operations is O(Nlog(N)) overall.It wo...

How does the command: real(ifft(fftshift(Y))*N) operate?

All of those oeprations are O(N) except for the IFFT which is O(Nlog(N)). So the chain of operations is O(Nlog(N)) overall.It wo...

1 dag ago | 0

| accepted

Answered

how can i remove objects that are on the edge of another object?

You can use imopen().

how can i remove objects that are on the edge of another object?

You can use imopen().

1 dag ago | 0

Answered

How can concatenate matrices with dimension expansion.

permute( cat(3,img1,im2), [3,1,2]);

How can concatenate matrices with dimension expansion.

permute( cat(3,img1,im2), [3,1,2]);

2 dagen ago | 0

Answered

Plotting a 3D mode shape by revolving 2D datapoints (in the r-z-plane) of an axisymmetric geometry around the z-axis

I can't follow most of what you posted, but if you just want to generate a surface/solid of revolution, you can use cylinder() ...

Plotting a 3D mode shape by revolving 2D datapoints (in the r-z-plane) of an axisymmetric geometry around the z-axis

I can't follow most of what you posted, but if you just want to generate a surface/solid of revolution, you can use cylinder() ...

2 dagen ago | 0

Answered

making a 2d image from 3d shape

It seems like it would be better if you would just generate your spheres direclty as a 3D image volume. Then you could just use...

making a 2d image from 3d shape

It seems like it would be better if you would just generate your spheres direclty as a 3D image volume. Then you could just use...

2 dagen ago | 0

Answered

Zero points on graph

xIntercept1=roots(TL1); xintercept2=roots(TL2); yIntercept1=polyval(TL1,0); yIntercept2=polyval(TL2,0);

Zero points on graph

xIntercept1=roots(TL1); xintercept2=roots(TL2); yIntercept1=polyval(TL1,0); yIntercept2=polyval(TL2,0);

2 dagen ago | 0

| accepted

Answered

Extract objects from binary image

Something like this perhaps, BW=bwareafilt(~yourBinarizedImage,3); regions=regionprops(BW,'Image');

Extract objects from binary image

Something like this perhaps, BW=bwareafilt(~yourBinarizedImage,3); regions=regionprops(BW,'Image');

2 dagen ago | 0

Answered

Changing the same property of some axis children using subsasgn

I assume you know that subsasgn is called implicitly whenever you make an assignment statement, as in the following: [toHide.Vi...

Changing the same property of some axis children using subsasgn

I assume you know that subsasgn is called implicitly whenever you make an assignment statement, as in the following: [toHide.Vi...

3 dagen ago | 1

| accepted

Answered

Area of a implicit curve

Here is a very similar strategy based on alphaShape instead of polyshape. It is slower, but IMO handles the bounding rectangle a...

Area of a implicit curve

Here is a very similar strategy based on alphaShape instead of polyshape. It is slower, but IMO handles the bounding rectangle a...

3 dagen ago | 2

Answered

Perimeter of a implicit curve

Perhaps as follows (see also this thread) plotRange=[-1,+1, -1,+1]*8; plotCorners=[-1 1;1 1;1 -1; -1 -1]*8; f=@(x,y) (sin(x...

Perimeter of a implicit curve

Perhaps as follows (see also this thread) plotRange=[-1,+1, -1,+1]*8; plotCorners=[-1 1;1 1;1 -1; -1 -1]*8; f=@(x,y) (sin(x...

3 dagen ago | 0

Answered

The determinant of a unitary matrix is 0

Matlab's det involves taking the product of long matrix diagonals. This can overflow or underflow very easily. The following mig...

The determinant of a unitary matrix is 0

Matlab's det involves taking the product of long matrix diagonals. This can overflow or underflow very easily. The following mig...

3 dagen ago | 1

Answered

The determinant of a unitary matrix is 0

Determinant calculations for large matrices are numerically delicate (which is why they're often avoided). There are other/bett...

The determinant of a unitary matrix is 0

Determinant calculations for large matrices are numerically delicate (which is why they're often avoided). There are other/bett...

3 dagen ago | 0

Answered

curve shifting on horizontal axis

For example, x=linspace(-pi,pi,1000); y=exp(x); plot(x(:)+(0:5)*10,y); axis([0,100,0,10]); legend

curve shifting on horizontal axis

For example, x=linspace(-pi,pi,1000); y=exp(x); plot(x(:)+(0:5)*10,y); axis([0,100,0,10]); legend

3 dagen ago | 0

Answered

Area of a implicit curve

Perhaps as follows? plotRange=[-1,+1, -1,+1]*8; plotCorners=[-1 1;1 1;1 -1; -1 -1]*8; f=@(x,y) (sin(x).*sin(y)-0.5); fp=fi...

Area of a implicit curve

Perhaps as follows? plotRange=[-1,+1, -1,+1]*8; plotCorners=[-1 1;1 1;1 -1; -1 -1]*8; f=@(x,y) (sin(x).*sin(y)-0.5); fp=fi...

3 dagen ago | 2

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Answered

Matrix Dimension Must Agree

theta and l do indeed have different lengths, so it is not clear what you are trying to do in your calculation of Eteta. n = 37...

Matrix Dimension Must Agree

theta and l do indeed have different lengths, so it is not clear what you are trying to do in your calculation of Eteta. n = 37...

4 dagen ago | 0

Answered

How to find all the possible feasible solutions to a integer linear programme?

For R^20, yes. Just do an exhaustive search, [x{1:20}]=ndgrid([0,1]); x=reshape( cat(21,x{:}) ,[],20).'; feasible = x(:, al...

How to find all the possible feasible solutions to a integer linear programme?

For R^20, yes. Just do an exhaustive search, [x{1:20}]=ndgrid([0,1]); x=reshape( cat(21,x{:}) ,[],20).'; feasible = x(:, al...

4 dagen ago | 0

| accepted