Problem with threshold in if statement

3 次查看（过去 30 天）
Siegmund Nuyts
Siegmund Nuyts 2022-10-26
评论： Siegmund Nuyts 2022-11-2
Ran in:
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:
P = load('Ps.mat');
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 个评论
显示 1更早的评论
Jeffrey Clark
Jeffrey Clark 2022-10-27
@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
Siegmund Nuyts 2022-10-27
Thanks for the reply. I added the data.
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Jeffrey Clark
Jeffrey Clark 2022-10-29
编辑：Jeffrey Clark 2022-10-29
@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:
P = load('Ps.mat');
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

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Descriptive Statistics 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by