using function with loop

1 view (last 30 days)
sermet OGUTCU
sermet OGUTCU on 26 Apr 2021
Commented: sermet OGUTCU on 27 Apr 2021
function [b, n, index] = RunLength(x)
d = [true; diff(x(:)) ~= 0]; % TRUE if values change
b = x(d); % Elements without repetitions
k = find([d.', true]); % Indices of changes
n = diff(k); % Number of repetitions
index = k(1:numel(k) - 1);
if iscolumn(x)
n = n.';
index = index.';
end
end
data = [1.2;1.0;0.05;0.4;0.3;0.2;0.1;0.05;0.04;0.03;0.03;0.02;0.01;0.001;0.01;0.01;0.001 ] ;
[b, n, index] = RunLength(data < 0.1);
match = find(b & n >= 10, 1);
result = index(match)
% This function find the number of row which 10 consecutive rows from the
% this computed row are smaller than 0.10. The result equals 8th row.
I need to use this function with loop such as;
data= 605x1 array includes numeric data
I need to compute the "result" with every 121 rows within 605 rows. So I need to run RunLength function five times (to produce five different result ) using the every 121 rows of the "data". I tried the below codes but it doesn't work;
i=1:121:605
for j=5
[b(:,j), n(:,j), index(:,j)] = RunLength(abs(data(i(j))) < 0.1);
match(j) = find(b(:,j) & n(:,j) >= 10, 1);
result(j) = index(match)
end

Accepted Answer

Jan
Jan on 26 Apr 2021
Edited: Jan on 26 Apr 2021
Result = zeros(1, 5);
iResult = 0;
for k = 1:121:605
datak = data(k:k+120);
[b, n, index] = RunLength(datak < 0.1);
match = find(b & n >= 10, 1);
iResult = iResult + 1;
Result(iResult) = index(match);
end
Maybe you have to catch the case, that match is empty.
Alternatively:
Result = zeros(1, 5);
iResult = 0;
for k = 1:5
index = (k - 1)*121 + 1;
datak = data(k:k+120);
[b, n, index] = RunLength(datak < 0.1);
match = find(b & n >= 10, 1);
Result(k) = index(match);
end
  5 Comments
sermet OGUTCU
sermet OGUTCU on 27 Apr 2021
Dear @Jan
Thank you very much for your solution.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!