Using hist3 and sorting

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Michael 2016-1-20

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编辑： Stephen23 2016-1-22

I'm simulating event positions and times, and histogramming then in the XY plane. I want to find for each event, the time difference to the next event in the same bin. I feel liek this should be a ~4 liner or so...

posXY = rand(nEv,2)
times = rand(nEv,1)
posBin = [2 2]; % put the events on a 2x2 mesh in X-Y plane
[hitMatrix, ~] = hist3(posXY, posBin);

then I want to find (for each event, in a given bin--I made four bins!) the time the next event occurs in that position bin. So i thought about sorting the events by their times, and using the indices to pick out the correct events that also occur in the same bin... making a logical and looping over events.. but I can't quite nail the whole sequence down. Do you have any tips?

I appreciate your input, Michael B.

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Answer 1

Stephen23 2016-1-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/264423-using-hist3-and-sorting#answer_206771

编辑：Stephen23 2016-1-20

在 MATLAB Online 中打开

The trick is to get the indices from the histogram, but unfortunately hist3 does not return the indices of the input values. Introduced in 2015b is the function histcounts2, which does return these indices. However I do not have 2015b, so I found histcn on FEX, which seems to do what you require.

I tried it like this, is three lines okay?:

nEv = 10;
posXY = rand(nEv,2);
times = rand(nEv,1);
%
[idx,~,~,loc] = histcn(posXY,2,2);
A = accumarray(loc,times,[],@(n){n});
B = cellfun(@(v)diff(sort(v)),A,'UniformOutput',false)

Where B is a cell array where the cells correspond to the bins of the histogram and contain the differences in the sorted times values corresponding to the X and Y values that were sorted into the corresponding bin. The variable A contains the time values themselves, so you can check this by hand if you want.

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Guillaume 2016-1-21

+1 about length. That should be purged from every single matlab example and the function deprecated.

Michael 2016-1-21

Thanks for all your help! I'm still parsing the original solution but your input is very much appreciated (I never looked at length help--thinking I had mastered its use... jeez I feel like a newb now) Again, your time and help are much appreciated!

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