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I have a large number of 2D image patches and a set of 2D filters. The sizes of them are same. I want to get the filter response (1x1) of an image patch (dxd) when I apply a 2D filter of size dxd. Descargar Telenovela Dona Barbara on this page. Instalacion Electrica En Casa De Adobe Ramon.

What is the fast way to do it? The image patches are stored as a 3D matrix. Size of the patches as well as the filters are 15x15. I tried conv in matlab but seems it is slow. (about 20-30 seconds for 5000 patches).

Can't I do it via some matrix multiplications? Cant I just convert each patch and each filter as a one D vector and do a multiplication? A lot of information is lacking for anyone to help you solve this problem.

How are the image patches stored? Are the image patches discontinuous (i.e. Are they samples from your image from various locations)?

How are the 2D filters stored? What have you tried? You do realize that convolution requires a weighted sum of coefficients and image pixels over local neighbourhoods, so if you have image patches that do not have locality, then the convolution of these patches combined don't make any sense. – Jan 20 '15 at 17:46 •. Don't use conv2. That is designed for general 2D signals.

However, given your description, you have 15 x 15 image patches stored in a 3D matrix, as well as a 15 x 15 filter. Imfilter is specifically designed to filter images, but this won't help you either because only full or same outputs are supported.

My guess is that with your use of conv2, you are using the valid flag so this will reduce to a 1 x 1 output. What I recommend you do is unroll your image patches so that you create a 2D matrix. Each patch fits in one column and all of the image patches are concatenated column-wise to create your 2D matrix. You would also unroll your filter so that it fits into a single column, then use to achieve the filtering over each patch. Therefore, do something like this, given that your image patches are stored in A and your filter is stored in K: B = reshape(A, 15*15, []); C = sum(bsxfun(@times, B, K(:)), 1); B reshapes your 3D matrix into a 2D matrix where each column represents one image patch.

Next, C is the output response for each image patch with the filter K. Bsxfun allows you to take your filter that is unrolled into a column vector, and it replicates the filter for as many columns as there are in B. This results in another 2D matrix which is the same size as B, but each column is the same (i.e. With this matrix, we do an element-wise multiplication, then sum along the rows for the output response of the filter.

Therefore, each element of C will give you the response of the image patch for the particular filter of interest. Note that this code is assuming that correlation is performed.

If the 2D filter is symmetric, then convolution is the same as correlation. However, if you want to do convolution and the filter is not symmetric, you will need to flip in each dimension before applying convolution. As such, you would have to do this for each of your filters: K = K(end:-1:1, end:-1:1); You would then apply the above two lines of code to achieve your filtering. Because you have multiple filters, what you can do is use a for loop that will do the above operation for you for each filter that you have. You can certainly vectorize it and use a bsxfun approach way, but my gut feeling is that this will be slower. I think it's better if you work on each filter individually, rather than trying to find the response of all filters for each image patch. Therefore, if your filters were stored in a 3D matrix.

We'll call it K again, you could do something like this: B = reshape(A, 15*15, []); num_patches = size(B, 2); num_filters = size(K, 3); C = zeros(num_filters, num_patches); for idx = 1: num_filters filt = K(:,:,idx);%filt = filt(end:-1:1, end:-1:1);%// Do this if the filter is not symmetric C(idx,:) = sum(bsxfun(@times, B, filt(:)), 1); end Here, C would be a 2D matrix where each row denotes the response of each image patch with the particular filter. @knedlsepp - Yes it could. It could actually be done by: C(idx,:) = (B.'