# Problem with threshold in if statement

1 visualizzazione (ultimi 30 giorni)
Siegmund Nuyts il 26 Ott 2022
Commentato: Siegmund Nuyts il 2 Nov 2022
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
##### 3 CommentiMostra 1 commento meno recenteNascondi 1 commento meno recente
Jeffrey Clark il 27 Ott 2022
@Siegmund Nuyts, hard to say from what you gave but please note that the conditional if difs(difs(:,5)>threshold,:) will be true as long as none of the difs(difs(:,5)>threshold,:) are zero. With floating point data it is unlikely that any column would be zero when column 5>threshold so the if may always be executing. Did tou intend if difs(:,5)>threshold which would execute the if part when any row with column 5>threshold was found?
Siegmund Nuyts il 27 Ott 2022
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.

Accedi per commentare.

### Risposta accettata

Jeffrey Clark il 29 Ott 2022
Modificato: Jeffrey Clark il 29 Ott 2022
@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
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Siegmund Nuyts il 2 Nov 2022
Thanks! That worked perfectly

Accedi per commentare.

### Categorie

Scopri di più su Descriptive Statistics in Help Center e File Exchange

R2020b

### Community Treasure Hunt

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

Start Hunting!

Translated by