MATLAB: Convolution from bottom right

convolutionimage processingImage Processing Toolbox

Hi all

I want to do a convolution from the bottom right and not as usual from the top left. I think conv2 of Matlab only does from the top left.

How can I do a convolution in Matlab from the bottom right?

Thank you very much for the answers.

Best Answer

After it's done it makes absolutely no difference. There is no need whatsoever to do what you're asking. Unless you're wanting to not flip the kernel. In that case use imfilter(), but it almost certainly starts at row #1 like conv2(), the only difference is it doesn't flip the kernel.

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MATLAB: Find ones blocks in upper triangular matrix

Let me plead for you to change how you are doing things. What you are asking to do is perhaps the most silly form of information storage I have seen in a while. To understand what I mean by that, what block is decribed by the set:

[7 8 9 10 11]

That is, does this describe a 1x4 rectangle, thus with 1 row (row 7) and 4 columns (8:11)? Or perhaps a 2x3 rectangle? Or perhaps a 3x2 rectangle? Or perhaps a 4x1 rectangle?

Do you recognize this is just an information storage idea that is just begging for a future head pounding bug, where you then need to disambiguate which of those many possible boxes was intended?

A far better solution is to just indicate the box by 4 numbers, ALWAYS, thus the first and last row and the first and last column in said box.

As far as how to find those bounding boxes, this should not be that difficult. For any 1 in the array, if another element is within 1 pixel, you would necessarily merge the two corresponding boxes, though I'm not sure what rules you may have for those mergers, so I must stop here for the moment. In fact, I am pretty confidant that if Image Analyst views this question, his response will be a simple, direct solution based on the IPT. If he does, then use it! :)

MATLAB: Signal in and signal out need to be shifted to match one another…

Don't know that it'll help your run time unless you can minimize the lags over which you run it, but the basic way to overlay the two is...

c=xcorr(data(:,1),data(:,2),'coeff');        % xcorr for the two vectors
x3=[ceil(length(c)/2)+1:length(c)]';         % zero-th lag is in middle
[~,ic]=max(c(x3));                           % max/location in positive lags
y=circshift(data(:,2),ic);                   % rotate second by that much
plot([1:length(data)]',[data y])             % show the results on the data
legend('1','2','shift')                      % identify who's who...

ADDENDUM 1: Certainly if the data as given are representative, one way to speed it up would be to eliminate the data where both are zero or even at/below the threshold level that you've set. Whether you need to keep those pre- and post-trigger lengths around to adjust the overall sample length in the end depends on the application, of course.

ALTERNATIVE: Again on the presumption the data are representative; w/o the cross-correlation bottleneck. This does depend upon the rise/fall times being as clean as are here and that one can find a suitable threshold for each pair expeditiously...

It needs must be above the early noise into the fast-rise/fall area and yet must not intersect the lower middle level of the lower=magnitude signal. Also it must be a value that isn't in the dataset to make the subsequent test robust. Since it appears that the data are integer-valued, one can assure that by using a fractional value as the threshold.

th=30.5;                         % see above notes
d=diff(sign(data-th));
% look for crossings and find them; look at result of sample case
id=[find(d(:,1)==2) find(d(:,2)==2);find(d(:,1)==-2) find(d(:,2)==-2)]
id =
 610   437
 715   542
shft=diff(id,[],2)
shft =
-173
-173
shft=floor(mean(diff(id,[],2)));   % the actual shift averages
yd=circshift(data(:,2),abs(shft)); % will need the direction logic here

Quick, but more sensitive to noise by far...salt to suit! :)

Best Answer

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MATLAB: Find ones blocks in upper triangular matrix

MATLAB: Signal in and signal out need to be shifted to match one another…

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