How to find consecutive values above a certain threshold?
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Hi.
I'm picking out values from a hourly data set, and I want to pick out values that are above 12 at least three consecutive times. Any tips on how I can do this?
example: A=[0 1 2 5 7 8 13 17 28 11 6 0 2 1 4]
I want to put the three values 13 17 28 into a vector.
Help is greatly appreciated!
- Kristine
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Accepted Answer
Azzi Abdelmalek
on 22 Jul 2015
Edited: Azzi Abdelmalek
on 22 Jul 2015
A=[0 1 2 5 7 8 13 17 28 11 6 0 2 1 40 55 88 47 4 44 ]
idx=A>12;
ii1=strfind([0 idx 0],[0 1]);
ii2=strfind([0 idx 0],[1 0])-1;
ii=(ii2-ii1+1)>=3;
out=arrayfun(@(x,y) A(x:y),ii1(ii),ii2(ii),'un',0);
celldisp(out)
9 Comments
Muhammad Shahid Iqbal
on 15 Feb 2019
Hello guys, this was very helpful but how about if we also need the index of those consective values which are greater than threshold???
Image Analyst
on 16 Feb 2019
Muhammad: See my solution below. Just ask regionprops for PixelIdxList or PixelList - that will be the indexes.
More Answers (4)
Image Analyst
on 22 Jul 2015
Kristine:
This pretty easy and straightforward if you have the Image Processing Toolbox to identify stretches where the numbers are above the threshold and measure their lengths. Then just save those stretches of numbers into cells of a cell array.
% Define sample data and a threshold value.
A=[0 1 2 5 7 8 13 17 28 11 6 0 2 1 4 91 49 37 79 9 100 101 3]
threshold = 12;
% Find logical vector where A > threshold
binaryVector = A > 12
% Label each region with a label - an "ID" number.
[labeledVector, numRegions] = bwlabel(binaryVector)
% Measure lengths of each region and the indexes
measurements = regionprops(labeledVector, A, 'Area', 'PixelValues');
% Find regions where the area (length) are 3 or greater and
% put the values into a cell of a cell array
for k = 1 : numRegions
if measurements(k).Area >= 3
% Area (length) is 3 or greater, so store the values.
ca{k} = measurements(k).PixelValues;
end
end
% Display the regions that meet the criteria:
celldisp(ca)
In the command window, this is what you'll see:
A =
0 1 2 5 7 8 13 17 28 11 6 0 2 1 4 91 49 37 79 9 100 101 3
binaryVector =
0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 1 1 1 0 1 1 0
labeledVector =
0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 2 2 2 2 0 3 3 0
numRegions =
3
ca{1} =
13 17 28
ca{2} =
91 49 37 79
4 Comments
Image Analyst
on 6 Apr 2021
I believe it would go something like this:
mask = a > b;
for i=1:length(lon)
for j = 1:length(lat)
thisSlice = mask(i, j, :);
[labeledMatrix, numRegions] = bwlabel(thisSlice);
props = regionprops(labeledMatrix, 'Area', 'PixelValues')
% Now do something with props....
end
end
Ahmad Bayhaqi
on 7 Apr 2021
Edited: Ahmad Bayhaqi
on 7 Apr 2021
Thank you @Image Analyst but, in my data, the threshold in every grid is also different, so the output would be different props in every location.
The code that you showed just like producing the one grid.
so, I tried this
mask = a > b;
for i=1:length(lon)
for j = 1:length(lat)
thisSlice(i,j,:) = mask(i, j, :);
[labeledMatrix(i,j,:), numRegions(i,j,:)] = bwlabeln(thisSlice(i,j,:));
props(i,j,:) = regionprops(labeledMatrix(i,j,:),a(i,j,:),'Area','PixelValues')
end
end
but, it ends up with the error in the props. It always said about dimension error.
Do you have any idea?
Thank you
Jan
on 22 Jul 2015
A = [0 1 2 5 7 8 13 17 28 11 6 0 2 1 4];
[B, N] = RunLength(A > 12);
B(N < 3) = false;
mask = RunLength(B, N);
Result = A(mask);
3 Comments
Image Analyst
on 22 Jul 2015
You'd need to download that "RunLength" function from the File Exchange using the link he gave you.
Stephen23
on 22 Jul 2015
Edited: Stephen23
on 26 Jul 2015
Here is a conceptually very simple method (I changed the third value to 20 as well, to provide a sequence of values > 12 but shorter then three):
>> N = 12;
>> A = [0,1,20,5,7,8,13,17,28,11,6,0,2,1,4];
>> idx = [true,A(1:end-1)>N] & A>N & [A(2:end)>N,true];
>> idx = [false,idx(1:end-1)] | idx | [idx(2:end),false];
>> A(idx)
ans =
13 17 28
4 Comments
Lane Foulks
on 15 Nov 2019
Edited: DGM
on 13 Feb 2023
A=[0 1 2 5 7 8 13 17 28 11 6 0 2 1 4 14 18 0 2 16 15 18 13 0 ] % added a passing and failing block above threshold
Adiff = diff(A>12);
ind_start = find(Adiff==1);
ind_stop = find(Adiff==-1);
block_length = ind_stop-ind_start; % list of consecutive section lengths
blocks_ind = find(block_length>2)% list of blocks above min length
for ii = 1:numel(blocks_ind) % loops through each block
A(ind_start(blocks_ind(ii))+1:ind_stop(blocks_ind(ii)))
end
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