# Problem with threshold in if statement

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Siegmund Nuyts 2022 年 10 月 26 日
コメント済み: Siegmund Nuyts 2022 年 11 月 2 日
My goal is to determine the difference between troughs and peaks, and select the troughs/peaks with a difference bigger than 35.
Sometimes there is no difference bigger than 35. So now I'm writing an if statement that if there is no difference bigger than 35, that it can just be 0.
The issue that I have is that there are multiple values varying below and above the threshold so it's not correctly identifying the threshold.
It should select the first trough/peak with a difference bigger than 35 and if there is no difference bigger than 35, then it should just be 0.
Here is the code that I have:
Ps = P.Ps;
threshold = 35;
[mins, min_locs] = findpeaks(-Ps);
[maxs, max_locs] = findpeaks(Ps);
peaks = [maxs max_locs];
troughs = [mins min_locs];
sp = size (peaks);
st = size (troughs);
s = max(sp(1), st(1));
dif = [[peaks;zeros(abs([s 0] - sp))],[troughs;zeros(abs([s, 0]-st))]];
difs = [dif (dif(:,1)+dif(:,3))];
if difs(difs(:,5)>threshold,:);
[idx, ~] = find(difs(:,5)>threshold);
difs = difs(unique(idx),:);
thresh = difs(1,4);
else thresh = 0
end
thresh = 0
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Siegmund Nuyts 2022 年 10 月 27 日
The if statement should look for any data in column 5 that is above 35.
If none of the data is above 35, then the result can be 0.

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### 採用された回答

Jeffrey Clark 2022 年 10 月 29 日

@Siegmund Nuyts, since you are negating the values input to Find local maxima - MATLAB findpeaks (mathworks.com) when looking for your valleys the mins returned will be negated values that you must then negate when used. Also since you are only interested in the first difference of the find indexed difs you should just have find return the first and eliminate some code. Along with my first comment the code would change to:
Ps = P.Ps;
threshold = 35;
[mins, min_locs] = findpeaks(-Ps);
[maxs, max_locs] = findpeaks(Ps);
peaks = [maxs max_locs];
troughs = [-mins min_locs]; %negate values to match -Ps use
sp = size (peaks);
st = size (troughs);
s = max(sp(1), st(1));
dif = [[peaks;zeros(abs([s 0] - sp))],[troughs;zeros(abs([s, 0]-st))]];
difs = [dif (dif(:,1)+dif(:,3))];
idx = find(difs(:,5)>threshold,1); %only need the first
if ~isempty(idx) %saves doing the find twice
thresh = difs(idx,4);
else thresh = 0
end
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Siegmund Nuyts 2022 年 11 月 2 日
Thanks! That worked perfectly

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