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

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I have a vector
M = [3,3,5,7,8,8,9,7,4,3,2,1,7,6,5,2,2,2];
I want to find at least 5 consecutive values which are above the threshold, which is 4. So the vector will be
N = [5,7,8,8,9,7]
Could anyone help me with this? Thanks.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 17 Apr. 2019
mask = M > 4;
starts = strfind([0 mask], [0 1 1 1 1 1]);
stops = strfind([mask 0], [1 1 1 1 1 0]);
N = M(starts(1) : stops(1))
  2 Kommentare
HYZ
HYZ am 17 Apr. 2019
How can I modify this code to get N = [5 7 8 8 9 7]? I guess the code you provided gave only the first and last value.
>> mask = M > 4;
starts = strfind([0 mask], [0 1 1 1 1 1]);
stops = strfind([mask 0], [1 1 1 1 1 0]);
N = M(starts(1) : stops(1))
N =
5 7

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Weitere Antworten (2)

Akira Agata
Akira Agata am 17 Apr. 2019
If there are 2 or more consective values above the threshold, the following is one possible solution.
(* The following needs Image Processing Toolbox)
M = [3,3,5,7,8,8,9,7,4,3,6,5,8,5,8,2,1,7,6,2,6,2];
idx = M > 4;
idx = bwareafilt(idx,[5,Inf]);
label = bwlabel(idx);
N = cell(max(label),1);
for kk = 1:max(label)
N{kk} = M(label == kk);
end
Result:
>> N
N =
2×1 cell array
{1×6 double}
{1×5 double}
>> N{1}
ans =
5 7 8 8 9 7
>> N{2}
ans =
6 5 8 5 8
>>
  9 Kommentare
Akira Agata
Akira Agata am 9 Apr. 2021
OK. I believe the error was due to the difference between the 3D matrix in my sample code and the real data.
Is it possible to upload part of your real data and the modified code which generates the error?
Ahmad Bayhaqi
Ahmad Bayhaqi am 1 Mai 2021
Thank you for your code. It works for me.
However, since my temperature data contains date. How the result of looping includes the datetime to define when the temperature is higher than threshold?
Thank you very much

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KSSV
KSSV am 17 Apr. 2019
  1 Kommentar
HYZ
HYZ am 17 Apr. 2019
N is the vector I want after thresholding and the values must be consecutive for at least 5 times.

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