Try this: A=imread(‘your image’); imshow(A); impixelinfo; [c, r, vals] = impixel; c and r are arrays of all the columns and rows of the part of the image that user clicked on and vals are the RGB values of each point. 2 solved how to extract data from image by user [closed]
[A,B] = size(v) for ii = 1:B v(:,ii) = v(:,ii)./abs(min(v(:,ii))); end In case this is the exact example you have. And just in case doubles are not good enough: v = int32(v); % or whichever integer you want bsxfun way, thanks to @rayryeng: v = bsxfun(@rdivide, v, abs(min(v,1))); 4 solved How to convert decimals in … Read more
If mat1 and mat2 are the first and second matrices you described, this should do the join you need with indexing functions. [~, I] = ismember(mat2(:, 2), mat1(:, 1)); Output = [mat2(:, 1) mat1(I, 2:end)] 1 solved Find and replace using Matlab [closed]
I tend to agree with nikie that you should be working from a book if you are at this basic level, but here is an answer anyway. I = imread(‘your_image’); # convert I to grayscale double as appropriate using rgb2gray(), double(), etc. # calculate the gradients and the formula you provided [dIx, dIy] = gradient(I); … Read more
Another approach is through bsxfun and unique. Assuming that A is a row vector, you would do the following: un = unique(A.’, ‘stable’); out = bsxfun(@times, bsxfun(@eq, A, un), un); out contains your desired result. This code deserves some explanation. The first line of code determines all unique entries in A stored in un, but … Read more
You can draw homogeneity criteria from the histogram of gray-levels or the histogram of gradient intensities. A low dispersion (small variance) reveals a quasi-constant lightness, but is sensitive to smooth variations. A low gradient average reveals a lack of detail, but is sensitive to texture/noise. 4 solved how homogeneity is calculated? [closed]
The upper bounds of your for-loops are the cause for your troubles. You specify: imh=size(ab,2); imw=size(ab,1); However, your for-loops have these conditions: j=1:imh+wh-1 and i=1:imw+ww-1 So you move past both the ‘height’ and the ‘width’ dimension. Try this instead: for j=1:imh-wh for i=1:imw-ww w1=ab(j:j+wh,i:i+wh,:); w2=salt(j:j+wh,i:i+wh,:); w3=w1-w2; end k=k+1; end 1 solved How to apply sliding … Read more
This question is about how to use these matlab functions: http://www.mathworks.com.au/help/stats/svmtrain.html http://www.mathworks.com.au/help/stats/svmclassify.html http://www.mathworks.com.au/help/bioinfo/ref/classperf.html If you are trying to classify entire videos you will have one label per video, i.e. assign 1 or 0 to “single block of 40 by 5 like 20 rows”, In which case your Training data matrix should be 20×200 (20 videos … Read more
You have to project your image coordinates to a 2d plane. For a cone you could proceed like this: For a simple approach you could ignore the fact that it’s a truncated cone and treat it like a cylinder. For each point (alpha,r,z) on the cone calculate the angle alpha, distance r and height z. … Read more
By the information in the question : In vector cell arrays it is usually the first vector used as a sparse matrix dimension The second element is a scalar specifying the default value of the sparse matrix The other vectors are used to specify the location and the value of the element in the sparse … Read more
its still a little unclear as to what you’re trying to accomplish. Is this what you’re trying to do?: x1=0.5:.01:1.6; x2=280:0.453:330; x3=-2.06:.0373:2.06; V = x1.^2+x2.^2+x3.^2+3.*x1.*x3-x2.*x3-4.*x2.*x1; subplot(3,1,1) plot(V,x1) subplot(3,1,2) plot(V,x2) subplot(3,1,3) plot(V,x3) 3 solved Plot a three variables function over the specified domain in MATLAB [closed]
Remove this line legend(‘-DynamicLegend’); and write it just before the drawnow line and write hold off after that line. So the following will be the lines at the end of your loop. legend(‘-DynamicLegend’); drawnow; hold off index=index+1; 0 solved How to Get Dynamic Legend on Linear Graph in Loop?
I would be careful about the indexing of u and y when using nbefore and nafter so that I don’t get index out of bounds error. Here is a Pythonic way – ynew = [dt * u[i] + y[i] for i in xrange(1,n+1)] unew = [dt * (k / m) * (y[i+1] + y[i-1]) for … Read more
how to construct sub-expressions with only two variables and one arithmatic operation from complex Expression Maple/Matlab? solved how to construct sub-expressions with only two variables and one arithmatic operation from complex Expression Maple/Matlab?
semilogx will scale the x-axis logarthmically, the same applies for semilogy on the y-axis. The data is exactly the same but sometimes it can be easier to view data where your axes are scaled like this. For example compare the two plots below. x = [0.01 0.1 1 10 100 1e3 10e3]; subplot(121); plot(x, ‘-o’); … Read more