Deleting outliers by code
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Hi again,
I have a measurements matrix as follows:
105.993000000000 1.64178960306505e+17
106.007000000000 3.10346010252124e+16
106.046000000000 2.22784317607289e+17
106.051000000000 1.48978160280980e+17
106.061000000000 2.79186942297259e+17
106.076000000000 2.02039468852741e+17
106.080000000000 5.02562504223962e+17
the first column is the x value, and the second its the y measurement.
I want to delete the rows in which the average of the neighboring y values are much bigger (or smaller) then the local y (in this example, i want to delete the second row).
How can i do it ?
Thank you !!!
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Antworten (3)
Image Analyst
am 18 Aug. 2014
Try this, by Brett from the Mathworks:
If you want something less sophisticated, try a modified median filter where you identify outliers, for example by thresholding the signal you get from subtracting the median signal from the original signal and taking the absolute value, and then replace only those elements above the threshold with the median value.
7 Kommentare
Image Analyst
am 18 Aug. 2014
Bearbeitet: Image Analyst
am 18 Aug. 2014
Try this:
outliers = y > (10 * averaged_y) | y < 0.1 * averaged_y
% Remove outliers
y(outliers) = []
averaged_y comes from conv(). By the way, I don't think this (your algorithm) is a very robust algorithm (just think about it and you'll realize why), but might be okay for your specific set of data.
Star Strider
am 18 Aug. 2014
It depends on how you define ‘much bigger (or smaller)’, and the number of neighboring elements you want to average over.
To delete the second row is easy enough (calling your matrix ‘X’ here):
X(2,:) = [];
2 Kommentare
Star Strider
am 18 Aug. 2014
Bearbeitet: Star Strider
am 18 Aug. 2014
I implemented a linear interpolation between (y-1) and (y+1), excluding (y), instead of an average of (y-1) and (y+1). Did you mean to include (y)?
The problem is that all of your points violate the ‘power-of-ten’ exclusion criterion.
My contribution:
X = [105.993000000000 1.64178960306505e+17
106.007000000000 3.10346010252124e+16
106.046000000000 2.22784317607289e+17
106.051000000000 1.48978160280980e+17
106.061000000000 2.79186942297259e+17
106.076000000000 2.02039468852741e+17
106.080000000000 5.02562504223962e+17];
for k1 = 2:size(X,1)-1
B(:,k1) = [[1 1]' [X(k1-1,1) X(k1+1,1)]']\[[X(k1-1,2) X(k1+1,2)]'];
E(k1) = [1 X(k1,1)] * B(:,k1); % Expected From Interpolation
D(k1) = E(k1) - X(k1,2); % Difference
end
Ep = E(2:end);
Xp = X(2:end-1,1);
figure(1)
plot(X(:,1), X(:,2), '-xb') % Plot Data
hold on
plot(Xp, Ep, '-+r') % Plot Interpolated Values
hold off
Guillaume
am 18 Aug. 2014
Bearbeitet: Guillaume
am 18 Aug. 2014
averages = (m(1:end-2, 2) + m(3:end, 2)) / 2; %averages of row, row+2
m([1; abs(averages - m(2:end-1, 2)) > tolerance; end], :) = [];
Should do it
5 Kommentare
Guillaume
am 18 Aug. 2014
Sorry, should have been
m([1; abs(averages - m(2:end-1, 2)) > tolerance; end], :) = [];
I've edited my answer to correct all the typos.
Joseph Cheng
am 18 Aug. 2014
Another method would be to use the conv.
x=randi(100,1,20);
Nav= [1 0 1]/2;
test = conv(x,Nav,'valid');
then perform the subtraction from the input (here x) from index t2o to the end-1.
difference =x(2:end-1)-test;
then threshold appropriately.
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