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

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.

[darova]

Can you explain why 2d and 3d group are included?

A= [5 6 7 8 9] [4 5 6] [3 4 5] 4 6 9 3 4 4 6

[Ahmad Bayhaqi]

Hi @darova.

I am sorry, I made mistake for giving an example

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

Here the result I want to be:

A= 5 6 7 8 9 [2 5] 6 3 4 5 8

[2 5] are included in groud because have two gaps and 5 consecutive values after that.

While the rest numbers are not included , eventhough they only have one gap [6 7], because they are followed by only four consecutive values [3 4 4 6]

[Image Analyst]

I don't understand your capital A and B. They are 1-D vectors, not 3-D volumetric arrays like little a and b. So, let's say with little a and b where you have a stack of 50 three-by-three arrays (9 columns each 50 tall), what constitutes "consecutive"? Would you take one vertical column through all 50 slices along the z direction and look at that 50 element vector and look for consecutive runs in that? And then do that for all 9 columns in the array? And what does "find" mean to you? Do you want the (row, column, slice) coordinates of any vertical 5-or-more run above the threshold? Like (for a start)

idx = a > b;
[rows, columns, slices] = size(idx);
for col = 1 : columns
    for row = 1 : rows
        thisVec = idx(row, col, :);
        % Get rid of runs less than 5
        thisVec = bwareaopen(thisVec, 5); % Keep only 5 or longer runs.
        % Get the indexes of all such runs.
        props = regionprops(thisVec, 'PixelList');
        % Now log/save them somehow.
    end

[Ahmad Bayhaqi]

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]

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

Answer 1

[daniel mitchell]

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