[ad_1] My first thought is to do this with arrayfun using an anonymous function that adds each scalar element of a to the full array in b. Then since you get a cell array result you can expand that cell array into the array you are looking for: >> a=[1,2,3,4], b=[5,6,7] >> result = arrayfun(@(x) … Read more
[ad_1] If you don’t have the statistics toolbox, it is not the correct function to use. If you do – then it is one of many functions you could use. You can also do something like: sz = size(A); A1 = reshape(A, [1 sz]); A2 = permute(A1, [2 1 3]); D = sqrt(sum(bsxfun(@minus, A1, A2).^2,3)); … Read more
[ad_1] In short: probably Matlab and Python achieve similar speeds in loops. The sources that you found are likely comparing old versions of Matlab, as @Eitan mentioned the new versions have proper Just In Time mechanisms which speed simple loops up a lot. There will be loops that python does faster, and those that Matlab … Read more
[ad_1] As I understand it the function (of two symmetric positive definite matrices) is JL(X,Y) = log( det( (X+Y)/2)) – log( det( X*Y))/2 = log( det( (X+Y)/2)) – (log(det(X)) + log(det(Y)))/2 I reckon the way to do this is to compute the cholesky factorisations of X, Y and (X+Y)/2, that is find lower triangular matrices … Read more
[ad_1] Looks like you are missing a closing parenthesis, at least, in the second to last line where you have .*sin((y).*sin(z), and missing a * (or another operator) between your )( in several lines. In MATLAB (A)(B) is not A*B. 8 [ad_2] solved What is the syntax error “Error: Unbalanced or unexpected parenthesis or bracket.”? … Read more
[ad_1] You are not using the function correctly. your code is this (verbatim): f = figure; hax = axes(f); af = figure(‘Name’, ‘Do Not Touch’); x = rand(1,100); y = rand(1,100); linewidth=1; plot(hax, x,y, ‘LineWidth’, linewidth); I = getframe(hax); I = I.cdata; The matrix I is not a matrix that contain peaks like the function … Read more
[ad_1] In Octave, you can decrease the size of the marker like this: x = 0:0.1:100; fx = rand(length(x), 1)’; plot(x, fx, “.”, “markersize”, 1) 2 [ad_2] solved Plotting discontinuous function in MATLAB/Octave
[ad_1] You can simply write: figure(‘Units’,’inches’); b1 = subplot(2,2,1); b2 = subplot(2,2,2); b3 = subplot(2,2,3); b4 = subplot(2,2,4); or preferably, if you want to have an array of the axes: figure(‘Units’,’inches’); b(1) = subplot(2,2,1); b(2) = subplot(2,2,2); b(3) = subplot(2,2,3); b(4) = subplot(2,2,4); or use a simple for loop: figure(‘Units’,’inches’); b(1:4) = axes; for k … Read more
[ad_1] Option Explicit Option Base 1 Sub shiftCircArray() Dim iInputArray(3) As Integer iInputArray(1) = 1 iInputArray(2) = 2 iInputArray(3) = 3 Dim iArray2() As Integer iArray2 = RotateArrayRight(iInputArray) End Sub Function RotateArrayRight(ArrayToRotate) Dim objNewArray() As Integer, iOldArrayPos As Integer, iNewArrayPos As Integer, iArrayLength As Integer Dim iPlacesToRotate As Integer ‘ Check that the array to … Read more
[ad_1] as the error says, you brackets is not in pairs. change while(norm(g_k)/(max(1,norm(v_k)) > eps && iter < iterMax) to while(norm(g_k)/max(1,norm(v_k)) > eps && iter < iterMax) [ad_2] solved Matlab While statement [closed]
[ad_1] You can do it using the reshape function. B = reshape(A,9,3); [ad_2] solved Change a multidimensional vector to two dimensional vector matlab [closed]
[ad_1] @RodyOldenhuis basically answered your question, and I’m flattered he hasn’t made this into an actual answer! Simply take your code and change the inner for loop index so that it matches the outer loop index. In other words, do this: syms l %// Why are you doing this? a = [0 1 0 0 … Read more
[ad_1] Just use the following code: num = []; % ToInitializeTheArray for i=1:5 num = input(‘Enter value or enter 100 to stop: ‘); num(end+1)=num; end I’m new to MathLab myself I hope this helped 1 [ad_2] solved Store numbers in an array in MATLAB
[ad_1] Looking at your comments, an appropriate solution would be: A = imread(‘Car.jpg’); % loads the image with a crop tool, select the region to be cropped B = imcrop(A); close % Displays the cropped region figure, imshow(B); 10 [ad_2] solved how to get a single feature vector from SIFT descriptor [closed]
[ad_1] Is there any standard rule for considering the best result or at which stage I have to stop train the data with minimum error [ad_2] solved Is there any standard rule for considering the best result or at which stage I have to stop train the data with minimum error