How to generate a loop using logical indexes intervals?
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Gustavo Oliveira
el 22 de Jun. de 2015
Comentada: Star Strider
el 23 de Jun. de 2015
Hi all,
I generated an index based on the depth position in which one sensor changes the direction at the water column, from upward to downward (Code below). My idea is to apply this index to locate the position of the MAXIMUM density between each index. For example, I would like to know the position of MAXIMUM density from index 1:ind(1), ind(1):ind(2), ind(2):ind(3)... The index interval are not regular and I am processing a 400810x1 size data. The figure below illustrate what I am planning to obtain, the values in red are the maximum differences in each group and I am interested to obtain the "Position" of the same. The position will work as index to continue the data processing.
Thank you for your time,

%Locating the position of change in direction based on depth differences
ind_change_direc=nan(size(depth_diff));
for i=1:length(depth_diff)-1
if depth_diff(i)<0 & depth_diff(i+1)>0;
ind_change_direc(i)=-9999;
end
end
index=ind_change_direc==-9999;
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Star Strider
el 23 de Jun. de 2015
Another possibility:
index = [-9999; NaN(7,1); -9999; NaN; NaN; -9999];
density = rand(12,1)*0.01; % Create Density
position = [1:12]';
change_idx = find(~isnan(index));
for k1 = 1:length(change_idx)-1
[mx,idx] = max(density(change_idx(k1):change_idx(k1+1)-1));
max_density(k1,:) = [mx, position(idx+change_idx(k1)-1)];
end
The ‘max_density’ matrix gives the maximum in the interval in the first column, and the value of ‘position’ for it in the second.
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Más respuestas (1)
Azzi Abdelmalek
el 22 de Jun. de 2015
s=[1 0 2 3 4 3 2 1 0 5 7 8 9 4 3 0 1] % Example
idx1=sign([diff(s) ])
idx1=idx1([1 1:end])
position=find(diff(idx1)~=0)+1
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