This can be done using cellfun and strsplit as follows: NewValues = cellfun(@(x) strsplit(x, ‘, ‘), MyValues, ‘UniformOutput’,0); NewValues = [NewValues{:}].’ 7 solved Split a value of different data types cell array based upon a delimiter
Answers to your questions: 1 – This is a binary perceptron algorithm, working on an offline batch. 2 – as you wrote – Y is the labels vector. each label can be either be 1 or -1. 3 – The rational of testing if y*a<=0 is to check if the perceptron classified a certain sample … Read more
%8.1f is a format specifier. This should point you in the right direction for figuring out what all the parts mean. It specifies the format in which a variable will be printed. In this specific case, it is a place-holder for a floating point variable. It reserves a width of at least 8 characters and … Read more
Can I put tags on the nodes? Are you referring to the following question?Drawing a network of nodes in circular formation with links between nodes A general method to add textual labels to a graph is to use the text command. Note that it requires the coordinates for each label. It’s also recommended to make … Read more
What you said, is quite a project, not a question. I’d rather to study more about the similar projects and ask about some details in here. Although I could give you some hints: 1-Do a suitable thresholding on your image, like Otsu method (with a little change in the threshold value if it’s needed). 2-On … Read more
I’ve tried running this code for plotting the figure in matlab, but it gives the following error, what will be the solution? [closed] solved I’ve tried running this code for plotting the figure in matlab, but it gives the following error, what will be the solution? [closed]
You have put the k=k+1 at a wrong place. It is the correct code: stab1=zeros(1,4991); k=0; for ii=-0.6:0.01:-.3 m=0; for jj=0:0.01:1.6 m=m+1; if …. (some condition) stab1(k*161+m)=1; end end k=k+1; end Now it has 4991 elements. solved How many elements the output matrix should have in this code? [closed]
n1=rgb2gray(imread(‘fin.jpg’)); imshow(n1); F=fft2(double(n1)); %Calculate Size of Image [M,N]=size(F); %Distance Radius Size D0=50; n=2; u=0:(M-1); v=0:(N-1); idx=find(u>M/2); u(idx)=u(idx)-M; idy=find(v>N/2); v(idy)=v(idy)-N; [V,U]=meshgrid(v,u); %Distance Calculation for High Pass Filter using Distance Formula D=sqrt(U.^2+V.^2); H=double(D>D0); subplot(2,1,1); imshow(fftshift(H)); G=H.*F; g=real(ifft2(double(G))); [a,b]=size(g); for x=1:a for y=1:b sharpen_image(x,y)=(g(x,y))*(-1)^((x)+(y)); end end figure imshow(sharpen_image); OUTPUT solved Transformation with High Pass Filter [closed]
You can use mat2cell to split the image into as many splits as you want: img = imread(‘https://i.stack.imgur.com/s4wAt.png’); [h w c] = size(img); numSplits = 3; % how many splits you want sw = floor(w/numSplits); % width of split widths = repmat(sw, 1, numSplits-1); widths(numSplits) = w – sum(widths); % last one is bit wider … Read more
If you look at the feedforwardnet MATLAB help page there is this example: [x,t] = simplefit_dataset; net = feedforwardnet(10); net = train(net,x,t); view(net) y = net(x); perf = perform(net,y,t) This is almost what you want. The feedforwardnet can take an array of different hidden layer sizes, so we can do: net = feedforwardnet([2 10 2]); … Read more
You need to transpose X and ‘B’ to get the correct result. Try the following: C = A * (X.’) + B.’; 0 solved multiply two arrays with different size in matlab
You don’t plot an image, you plot data, either in the plain form, or resulting from equations, for images, you just display them, using functions like image Of course, images are in there raw form are data too, but they have special formats. solved What is means by Plotting an Image ?What is the purpose … Read more
You only need to carefully read the section B. Proposed methodology. It says something like this: Now our aim is to find the best set of values for these four parameters which can produce the optimal result and to perform this work PSO is used. P number of particles are initialized, each with four parameters … Read more
you can use regionprops for that , for example: s=regionprops(bw,’BoundingBox’,’PixelIdxList’); where bw is your binary image. The output of s.BoundingBox is a [x,y,width,height] vector you can loop over s for i=1:numel(s) ar(i) = s(i).BoundingBox(3)/s(i).BoundingBox(4) end and see if the width/height ratio ar (or whatever you define aspect ratio) is approx above 1 or not (because … Read more
Define an nxn matrix A by a(i,j)=-1+2max(i,j) and b(j)=sum(j to n) a(i,j). Then test Naive-Gauss on this system (Matlab) solved Define an nxn matrix A by a(i,j)=-1+2max(i,j) and b(j)=sum(j to n) a(i,j). Then test Naive-Gauss on this system (Matlab)