Find at least 5 consecutive values above a certain threshold in a vector with the certain gap?

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Ahmad Bayhaqi
Ahmad Bayhaqi 2021-6-3
编辑： daniel mitchell 2022-7-14
Hi All,
I have a 3D matrix (longitude x latitude x time). I want to find the consecutive values above a certain threshold with two conditions.
first : have to 5 coonsecutive above threshold
second: neglecting the gap with the maximum =2.
For example:
A>B
A= 5 6 7 8 9 4 5 6 3 4 5 4 6 9 3 4 4 6
B= 4 4 6 7 8 3 6 4 2 3 4 7 7 8 2 3 3 4
I want the result to be :
A= 5 6 7 8 9 4 5 6 3 4 5 ( neglecting 1 gap in the 6th number which A<B. because the maximum gap is 2, so the 6th number is included in group).
The rest numbers are not included in the group because they are only 4 consecutive value after two gaps.
I have this code (thanks to Akira Agata:
% Sample data (3:lat, 3:lon, 50:days)
rng('default'); % for reproducability
a = rand(3,3,50);
b = rand(3,3,50)*0.4;
% Extract values which satisfies a > b for more than 5 consecutive days
idx = a > b;
[latGrid, lonGrid] = meshgrid(1:3,1:3);
tResult = table(latGrid(:),lonGrid(:),'VariableNames',{'lat','lon'});
tResult.extData = cell(height(tResult),1);
for kk1 = 1:height(tResult)
% break down into 1-D problem and apply the original answer
lat = tResult.lat(kk1);
lon = tResult.lon(kk1);
idx0 = bwareafilt(squeeze(idx(lat,lon,:)),[5,Inf]);
a0 = squeeze(a(lat,lon,:));
label = bwlabel(idx0);
N = cell(max(label),1);
for kk2 = 1:max(label)
N{kk2} = a0(label == kk2);
end
tResult.extData{kk1} = N;
end
However, this code does not include the second condition (neglecting gap with the maximum 2).
So, how do I put the second condition?
Thank you very much
Your help is greatly appreciated.
  5 个评论
显示 3更早的评论
Ahmad Bayhaqi
Ahmad Bayhaqi 2021-6-6
Hi @Image Analyst
Thank you before,
Yes, the example that I gave is 1D matrix. That just shows my logical thinking. I tried to implement it into the 3D matrix with the attached script.
However, the script only covers the first condition. I do not know how to put my second condition into the script.
I revised my example in @darova's comment.
MANON QUIPOURT
MANON QUIPOURT 2022-7-14
Hi @Ahmad Bayhaqi
Did you manage to resolve your code ? Cause I have the same problem, and I don't know how to deal with the 2-days gap.
Thank you

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回答（1 个）

daniel mitchell
daniel mitchell 2022-7-14
编辑：daniel mitchell 2022-7-14
Ran in:
There are nicer ways to implement but this way, it is very easy to adjust to multidimensional
a = [5 6 7 8 9 2 5 6 3 4 5 8 6 7 3 4 4 6];
b = [4 4 6 7 8 3 6 4 2 3 4 7 7 8 2 3 3 4];
temp = double(a>b);
temp = imfilter(temp,[0 0 0 0 1 1 1 1 1]);
temp = temp >= 5;
temp = double(imfilter(temp,[1 1 1 1 1 0 0 0 0]));
tempGap = imfilter(temp,[1 0 0 0 1]);
tempGap = (tempGap == 2);
r = (logical(temp) | tempGap);
a(r)
ans = 1×12
5 6 7 8 9 2 5 6 3 4 5 8

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