Effacer les filtres
Effacer les filtres

how can I Replace outliers with median of previous observations?

3 vues (au cours des 30 derniers jours)
Nicolas
Nicolas le 19 Juil 2012
Hello i have some outliers in a 206*174 dataset matrix.. I want to replace them with the median of the 5 previous observaitons using a loop..
how can i do that?
[EDITED, copied from Answer section, Jan]
i will be more clear.. outliers are observations of stationary series with absolute deviations from the median which exceed six times the interquartile range. I want to replace them with the median of the preceding five observations. thanks
  7 commentaires
Nicolas
Nicolas le 19 Juil 2012
So i will repeat.. I have a data of 206*174 observations... rows is time observations and columns is variables.. i want to find the outliers that are defined the the median absolute deviations to be greater 6 times the interquartile range in each variable series.
after that i want to replace each outlier with the median of previous 5 rows. thanks
Nicolas
Nicolas le 19 Juil 2012
% %Now we remove outliers like the paper of Stock and Watson 2005(num=data)
[t n]=size(NUM)% row size of data
X=median(NUM) %find the median of each column of NUM
X1=repmat(X,t,1)% creates a large matrix that
% each column has n times the median value of the column
NUM1=NUM-X1 %substract each row to find the Mean absolute deviation
NUM1=abs(NUM(:,:))%take the absolute value
for j=1:n
Y(:,j)=iqr(NUM(:,j))% find the value of the difference between
%3 and 1 quartile.
end
Y1=repmat(Y,t,1)
NUM2=6*Y1% multiply each value x6
outliers=NUM1-NUM2 %an outlier is when the MAD>6*Diff inquartiles
[x w]=find(outliers>0)%x is the row and w the column of each outlier
v=ones(t,n)
v(outliers>0)=0
%Note here that some problems arise for very smooth series so we remove %them for further analysis v(:,[39;84;86;92;95])=1 [x w]=find(v==0)
NUM1=zeros(size(data_st)) j=1
for i=1:t
if v(i,j)==0
NUM1(i,:)=NUM(median(NUM(i-6:i-1,:),1))
elseif v(i,:)==1
NUM1(i,:)=data_st(i,:)
end
j=j+1
if j==175
break
end
end
disp('Done')

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Réponses (1)

Miro
Miro le 19 Juil 2012
Modifié(e) : Miro le 19 Juil 2012
something like this should work
yourthreshold = 10;
Data(Data>yourthreshold) = median(median(Data));
this replaces all values being greater than 10.

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