manipulating matrix for outlier detection

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sermet
sermet el 10 de En. de 2015
Comentada: Greg Heath el 13 de En. de 2015
%ids are the point numbers, L is the measurement which need to be checked for outlier.
ids=[1;2;3;4;5;6;7;8;9;10];
L=[0.16;0.2;0.18;0.4;0.14;0.15;0.11;0.17;0.15;0.18]; %measurement
% To perform outlier detection for 0.16;
L_1=[0.2;0.18;0.4;0.14;0.15;0.11;0.17;0.15;0.18]; %without first row of L (0.16)
L_ort_1=mean(L_1);
for i=1:9 % n=9
V(i)=L_1(i)-L_ort_1;
end
V=V(:);
S=sqrt((V'*V)/7);
d=abs(L_ort_1-0.16) %0.16 is subjected to outlier detection
Sd=((S^2)/8)+S^2;
t_1=d/Sd;
t_statistics_1=tinv(0.995,7);
if t_statistics_1 > t_1
% there is no outlier for 0.16 (id=1) go on to check second number (0.20).
%L_2=[0.16;0.18;0.4;0.14;0.15;0.11;0.17;0.15;0.18] %without 0.20 (second number)
% do the same things to L2 for detection of 0.20....................
if t_statistics_1 < t_1
% there is outlier for 0.16, save its id (1) and go on to check second number (0.20)
end
end
I need to write this kind of loop for this outlier detection for each row of L.
Thanks in advance.

Respuesta aceptada

Image Analyst
Image Analyst el 10 de En. de 2015
So, what's stopping you?
If you want an outlier detection method developed by the Mathworks, see this by the amazing Brett Shoelson: http://www.mathworks.com/matlabcentral/fileexchange/3961-deleteoutliers
You might also look at Median Absolute Deviation which is a more robust method than looking at the z score, which can be influenced by the outlier itself which you want to get rid of ( not a good thing obviously).
  1 comentario
Greg Heath
Greg Heath el 13 de En. de 2015
Correct. However, the sensitivity of the zscore method can be mitigated via iterated removal or modification, plots and human intervention.
Greg

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Más respuestas (1)

Greg Heath
Greg Heath el 10 de En. de 2015
The best way to detect outliers
1. Standardize the data to zero-mean/unit variance
2. Plot
3. Find outliers where abs(x) > mean + alpha*sigma
where alpha is a user determined parameter
Hope this helps.
Thank you for formally accepting my answer
Greg

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