bsxfun

I have a matrix a and I want to calculate the distance from one point to all other points . So really the outcome matrix should have a zero (at the point I have chosen) and should appear as some sort of circle of numbers around that specific point. This is what I have already but I cant seem to get the correct outcome. a = [1 2 3 4 5 6 7 8 9 10] for i = 2:20 a(i,:) = a(i-1,:) + 1; end N = 10 for I = 1:N for J = 1:N dx = a(I,1)-a(J,1); dy = a(I,2)-a(J,2); distance(I,J) = sqrt(dx^2 + dy^2) end end chappjc Your a matrix is a 1D vector and is incompatible with the nested loop, which computes

I need to calculate the euclidean distance between 2 matrices in matlab. Currently I am using bsxfun and calculating the distance as below( i am attaching a snippet of the code ): for i=1:4754 test_data=fea_test(i,:); d=sqrt(sum(bsxfun(@minus, test_data, fea_train).^2, 2)); end Size of fea_test is 4754x1024 and fea_train is 6800x1024 , using his for loop is causing the execution of the for to take approximately 12 minutes which I think is too high. Is there a way to calculate the euclidean distance between both the matrices faster? I was told that by removing unnecessary for loops I can reduce

As always trying to learn more from you, I was hoping I could receive some help with the following code. I need to accomplish the following: 1) I have a vector: x = [1 2 3 4 5 6 7 8 9 10 11 12] 2) and a matrix: A =[11 14 1 5 8 18 10 8 19 13 20 16] I need to be able to multiply each value from x with every value of A , this means: new_matrix = [1* A 2* A 3* A ... 12* A] This will give me this new_matrix of size (12*m x n) assuming A (mxn) . And in this case (12*4x3) How can I do this using bsxfun from matlab? and, would this method be faster than a for-loop ? Regarding my for-loop , I need some

My Question: I've noticed that a lot of good answers to Matlab questions on SO frequently use the function bsxfun . Why? Motivation: In the Matlab documentation for bsxfun , the following example is provided: A = magic(5); A = bsxfun(@minus, A, mean(A)) Of course we could do the same operation using: A = A - (ones(size(A, 1), 1) * mean(A)); And in fact a simple speed test demonstrates the second method is about 20% faster. So why use the first method? I'm guessing there are some circumstances where using bsxfun will be much faster than the "manual" approach. I'd be really interested in seeing

Suppose I have a matrix A and I sort the rows of this matrix. How do I replicate the same ordering on a matrix B (same size of course)? E.g. A = rand(3,4); [val ind] = sort(A,2); B = rand(3,4); %// Reorder the elements of B according to the reordering of A This is the best I've come up with m = size(A,1); B = B(bsxfun(@plus,(ind-1)*m,(1:m)')); Out of curiosity, any alternatives? Update: Jonas' excellent solution profiled on 2008a (XP): n = n 0.048524 1.4632 1.4791 1.195 1.0662 1.108 1.0082 0.96335 0.93155 0.90532 0.88976 n = 2m 0.63202 1.3029 1.1112 1.0501 0.94703 0.92847 0.90411 0.8849 0.8667