MATLAB Compiler encrypts and archives your MATLAB code (which remains as MATLAB .m code), and packages it in a thin executable (either .exe or .dll) wrapper. This is delivered to the end user along with the MATLAB Compiler Runtime (MCR). If you wish, the MCR can be packaged within the executable as well. The MCR … Read more
Using repmat is by far the most efficient way to preallocate structs : N = 10000; b = repmat(struct(‘x’,1), N, 1 ); This is ~10x faster using Matlab 2011a than preallocating via indexing, as in N = 10000; b(N).x = 1 The indexing method is only marginally faster than not preallocating. No preallocation: 0.075524 Preallocate … Read more
Image coordinates vs plot coordinates Used to get me every time. %# create an image with one white pixel img = zeros(100); img(25,65) = 1; %# show the image figure imshow(img); %# now circle the pixel. To be sure of the coordinate, let’s run find [x,y] = find(img); hold on %# plot a red circle… … Read more
Note that in matlab’s hist(x, vec), vec difines the bin-centers, while in matlab histc(x, vec) vec defines the bin-edges of the histogram. Numpy’s histogram seems to work with bin-edges. Is this difference important to you? It should be easy to convert from one to the other, and you might have to add an extra Inf … Read more
This is just a minor improvement over Divakar’s answer. It is a little faster because it replaces a 3D-array permute with a 2D-array permute: B = bsxfun(@times, permute(A, [1 3 2]), permute(A, [3 1 2]));
In newer versions of MATLAB (R2015a and later) the easiest way to do this is using the repelem function: B = repelem(A, 2, 2); For older versions, a short alternative to the other (largely) indexing-based solutions is to use the functions kron and ones: >> A = [2 6; 8 4]; >> B = kron(A, … Read more
The technique is called texture mapping. This is a code example from surface function (R2011b): load clown surface(peaks,flipud(X),… ‘FaceColor’,’texturemap’,… ‘EdgeColor’,’none’,… ‘CDataMapping’,’direct’) colormap(map) view(-35,45) This example loads RGB image from “peppers.png” and maps it onto cylinder: imgRGB = imread(‘peppers.png’); [imgInd,map] = rgb2ind(imgRGB,256); [imgIndRows,imgIndCols] = size(imgInd); [X,Y,Z] = cylinder(imgIndRows,imgIndCols); surface(X,Y,Z,flipud(imgInd),… ‘FaceColor’,’texturemap’,… ‘EdgeColor’,’none’,… ‘CDataMapping’,’direct’) colormap(map) view(-35,45) Things are … Read more
Ok, here is a way involving several randomization steps needed to get a “natural” non symmetrical appearance. I am posting the actual code in Mathematica, just in case someone cares translating it to Matlab. (* A preparatory step: get your image and clean it*) i = Import@”http://i.stack.imgur.com/YENhB.png”; i1 = Image@Replace[ImageData[i], {0., 0., 0.} -> {1, … Read more
You could also do this completely vectorized by using mat2cell, cellfun, then cell2mat. Assuming our matrix is stored in A, try: numBlocks = size(A,1) / 3; B = mat2cell(A, 3*ones(1,numBlocks), 3); C = cellfun(@(x) x – x([1 1 1], :), B, ‘UniformOutput’, false); D = cell2mat(C); %//Output The first line figures out how many 3 … Read more
You need to use interp1 with ‘linear’ interpolation method: >> v = [1 2 3 4 10]; >> newNum = 13; % new number of elements in the “buffed” vector >> iv = interp1( linspace(0,1,numel(v)), v, linspace(0,1,newNum) ) iv = 1.0000 1.3333 1.6667 2.0000 2.3333 2.6667 3.0000 3.3333 3.6667 4.0000 6.0000 8.0000 10.0000