Detect all local minimum and remove them
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Salma fathi 2022년 11월 1일
댓글: John D'Errico 2022년 11월 1일
I am trying to do some 'correction' on the curve in the attached figuer (blue in the original curve and green is the one we desire), mainly I am trying to get rid of the local minimum points that we have and instead conect between the points before and after the minimum. One issue I am having is that we would like to detect all the points that would give the valley shape in our curve not only the one considered to be local minimum. also when trying to get rid of the ones that we have we are only able to do this for the last minimum point only, any idea how we can fix this?
these are the lines I am using
[TF1,P] = islocalmin(y);
plot(x,y,x(TF1),y(TF1),'rv','LineWidth', 2, 'MarkerSize', 12)
plot(x,y,'g','LineWidth', 1, 'MarkerSize', 12)
thanks in advance.
편집: John D'Errico 2022년 11월 1일
@Matt J showed you how to use a convex hull. Very possibly that is what you want, even though you did not flag the other points. But a convex hull will also replace some points that are not a local minimum, but just a point of non-convexity. If your goal is PURELY to replace the local minima, then just test for that.
x = [280 295 310 325 340 355 370 370 385 400 415];
y = 1e12*[0.9760 1.2100 1.2530 1.4270 1.4310 1.5020 1.4550 1.4550 1.5010 1.4220 1.4230];
But one problem I see in your data, is you have TWO points at the location x==370. And those points have the same values indicated for y. So I'll arbitrarily drop one of them. (Simplest to use unique for this, unless perhaps you need to average the y values that would then be replaced.)
[x,loc] = unique(x);
y = y(loc)
n = numel(x);
ind = 2:n-1;
localmin = ind(find((y(ind) <= y(ind - 1)) & (y(ind) <= y(ind + 1))))
Those are the two points you indicated. But I carefully did not find the two other points where your data is merely non-convex. Now we can use linear interpolation to replace those values.
yhat = y;
usedata = setdiff(1:n,localmin);
yhat(localmin) = interp1(x(usedata),y(usedata),x(localmin),'linear');
Again, your choice on what you want to see done. If it were my guess, logically you want to use the convex hull. Why? Consider a scenario where you vary one data point slightly. If this is truly noise, then only if that data point becomes a local minimum will you adjust it to the linear interpolation between its neighbors. But there are several points where they are actually quite close to being local minimia, yet you choose not to adjust them. This seems quite arbitrary to me, and my mind rebels at inconsistencies. Had those other points been only slightly different, they would have been also singled out to be tweaked. In my eyes, that seems to beg for the convex hull approach.
Of course, that also begs the question of why are you not just smoothing ALL of the points, not just rhe low ones. And down this path lies madness. :) But you might have chosen to do this instead:
ysmooth = S(x);
As you can see, now it smooths the entire curve. But if it is noise that has caused the problems, then surely you wanted to smooth the noise, not just replce the local minima. Again, this is probably overstepping what you want.
Steven Lord 2022년 11월 1일
Another approach similar to John D'Errico's, but automating the detection and filling of the local minima:
x = [280 295 310 325 340 355 370 385 400 415];
y = 1e12*[0.9760 1.2100 1.2530 1.4270 1.4310 1.5020 1.4550 1.5010 1.4220 1.4230];
localMinimaLocations = islocalmin(y);
plot(x, y, '-o', 'MarkerIndices', find(localMinimaLocations))
title("Plot of original data")
Now we'll use fillmissing to fill in the 'missing' locations (the local minima) using linear interpolation.
y2 = fillmissing(y, 'linear', 'SamplePoints', x, 'MissingLocations', localMinimaLocations);
Let's plot the original data with circle markers for the local minima and the new data with x markers for the locations that were filled.
plot(x, y, 'r-o', 'MarkerIndices', find(localMinimaLocations), 'DisplayName', 'original');
plot(x, y2, 'k--x', 'MarkerIndices', find(localMinimaLocations), 'DisplayName', 'filled');
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