Using is outlier with threshold
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Mande Hudson
le 29 Mai 2021
Commenté : Star Strider
le 29 Mai 2021
I need to only plot the outliers that are out of my upper and lower limits. But I get a percentile error using TF= (A,'percentiles',threshold)
Below is my code and plot with out apply threshold to outlier.
XT{1} is a cell array of data.
%% Plot the control charts:
%For all the states (states 1, 3, 10, 12, 14), plot x(t), shown in blue in the hint below.
timev=zeros(1020,1);% initalize time vector
for i= 1:1020;
timev(i)= timev(i)+int*(8*(i)+30);
end
threshold=[LCL,UCL];
[TF]= isoutlier(XT{1},) %'percentiles',threshold)
%Create a new figure
figure('Renderer', 'painters', 'Position', [10 10 1200 900])
plot( timev, XT{1}, timev(TF),XT{1}(TF),'x')
hold on
%Plot the control limits CL, UCL, and LCL as horizontal lines in green. Remember that
%these limits are the same for all plots (calculated from state 1).
yline(CL,'g')% plot the control limits
yline(UCL,'g')% plot the upper control limits
yline(LCL,'g')% plot the lower control limits

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Star Strider
le 29 Mai 2021
You have a typographical error. It should be 'percentiles', not 'percentile'.
From the documentation:
TF = isoutlier(A,'percentiles',threshold) defines outliers as points outside of the percentiles specified in threshold. The threshold argument is a two-element row vector containing the lower and upper percentile thresholds, such as [10 90].
.
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Star Strider
le 29 Mai 2021
Please check the documentation for your version. The 'percentiles' option is not available in isoutlier in R2018b, however the prctile function is.
Instead, do something like this —
X{1} = rand(1, 100);
t = 1:numel(X{1});
p = prctile(X{:},[20 80])
TF = (X{1}<=p(1)) | (X{1}>=p(2)) % Logical Index Of Values Outside The Respective Percentiles
figure
plot(t, X{1})
hold on
plot(t(TF), X{1}(TF),'x')
yline(0.2)
yline(0.8)
hold off
grid
That will identify them.
Then do whatever you want to with them after that.
.
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