Hi Mohammad,

Take a look at Andreas Lauschke's livecoding session on anomaly detection on YouTube.

Hi Jim,

Thank you so much for your useful explains. I have more than 5000 data for finding outliers. I just wanted to learn with the lists including small numbers of elements.

Hi Henrik,

Thank you so much for the interesting method.

I get different results every time I run the program.

I do believe in outliers. But I also believe that when doing science (or even making a decision for a business) one must not just look for and toss "inconvenient" observations.

Context matters. Are the dimensions in the same units? Are the dimensions of equal importance? How is the data collected? Are the "potential outliers" just in a region of space not explored as intensively as other regions?

There is no algorithm void of subject matter knowledge that will appropriately find outliers. Such algorithms ignorant of how the data was collected might help "round up the usual suspects" but each suspected outlier needs to be vetted (with vetting also for some observations that "seem OK").

And if one has less than 50 observations, one probably has no business looking for an automated outlier detection algorithm.

Not sure what you mean. In the image you annotated with outliers, {2, -3, 6} is not an outlier, so it is present in the result.

Dear Rohit, you always are great. Thank you so much.

I got a small error in the output.

Thank you so much, Jim.

If we study a three-dimensional list data, (u,v,w) separately, then we get the below results:

u = {-10, -2, 0, -3, 1, 2, 6, 14, 5, 4, 8, 11, 9, 3, 7};

{min, max} = {Mean[u] - 2*StandardDeviation[u], 
    Mean[u] + 2*StandardDeviation[u]} // N;

Select[u, Or[# > max, # < min] &];

ListPlot[u]

v = {-19, 5, 1, 1.5, 3.5, -3, 0, 7, 6, -11, 25, 17, 2, 7.5, 4};

{min, max} = {Mean[v] - 2*StandardDeviation[v], 
    Mean[v] + 2*StandardDeviation[v]} // N;

Select[v, Or[# > max, # < min] &];

ListPlot[v]

w = {-5, 8, -16, 9, 13, 6, -26, 15, 14, 6, 15, -2, 6, 3, 10};

{min, max} = {Mean[w] - 2*StandardDeviation[w], 
    Mean[w] + 2*StandardDeviation[w]} // N;

Select[w, Or[# > max, # < min] &];

Now for all of the lists in a 3D list we have :

How do I remove these three elements sublists from the data?

I appreciate your kindness and help.

Check out this post.