A=[1 2; 3 4] M = cat(3, A, A+1, A+2, A+3);
A = 1 2 3 4So the function should work like this: IDX = someFunction(M >=5);
IDX = [NaN 4; 3 2]
IDX = NaN 4 3 2And, the customer wanted this to work for arrays of higher dimensions as well. find on the entire array, as if it were a vector, get the locations and sort them into their proper locations and find the first one in the correct last dimension. And then Sean said "Max Logical". Here's what a great solution looks like.
catdim = 3; szA = size(A); [mv,midx] = max(M >= 5,,catdim); IDX = NaN(szA); IDX(mv) = midx(mv)
IDX = NaN 4 3 2Why/how did this work? Well, I use the condition I want met into function max, and it is logical input, meaning only 0s and 1s. And max returns the index where the first condition is met, and I do this along the final dimension. Then create an array of NaN values, and use logical indexing to update this index matrix. Voila!
A = randi(17, [3 2 1 3]); catdim = ndims(A)+1 M = cat(catdim, A, A+1, A+2, A+3, A+4);
catdim = 5See a slice of M
sliceOfM = M(:,:,:,1)
sliceOfM = 12 2 13 4 8 16Apply the max, logical indexing to M next.
szA = size(A) [mv,midx] = max(M >= 13,,catdim); IDX = NaN(szA); IDX(mv) = midx(mv)
szA = 3 2 1 3 IDX(:,:,1,1) = 2 NaN 1 NaN NaN 1 IDX(:,:,1,2) = NaN 1 1 NaN 4 NaN IDX(:,:,1,3) = NaN 1 1 1 NaN 1indexing that readers have found helpful. Do you have any interesting logical stories to share? Post them here.
댓글을 남기려면 링크 를 클릭하여 MathWorks 계정에 로그인하거나 계정을 새로 만드십시오.