I'm working on dataset which isn't normally distributed.which contains three dimensions like cost, discount and profit

I'm trying to find outliers in all these dimensions. I used z score to find outliers in single dimension to find high cost causing outliers.

As a next step I tried to find outliers with high cost and high profit and low discount.

I came up with a formula of Zscore(cost)+Zscore(profit)-Zscore(discount)

negative sign because I want to find outliers with low discount.

Is this approach meaningful to do? or is there any further proven way to achieve this?

  • $\begingroup$ I guess this rather belongs on math.stackexchange.com or stats.stackexchange.com unless you have a question about the software Mathematica. $\endgroup$
    – Yves Klett
    Apr 30, 2016 at 8:37
  • 2
    $\begingroup$ I answered this question, but is it a good idea to remove my answer and respond with it to a different question titled, say, "Finding outliers in 2D and 3D numerical data"? $\endgroup$ Apr 30, 2016 at 15:38

1 Answer 1


The question title poses a good question, although the question formulation is somewhat specialized and misplaced (as mentioned in a comment).

This answer provides data and a method description answering:

How to find outliers in 3D numerical data?


It order to provide a good answer it would be better to use "real life" data. Not spending much time looking for a relevant set I found this one: "UCI Online Retail Data Set". That dataset has columns for online purchase transactions (quantity and price).

I had problems loading the data with this command:


so I downloaded the XLSX file and saved it into a CSV file, then imported it:

data = Import["~/Datasets/UCI Online Retail Data Set/Online Retail.csv", 
   "IgnoreEmptyLines" -> True, "HeaderLines" -> 0];
columnNames = data[[1]];
data = Rest[data];

Here are the dimensions of the dataset (seems "realistically" large):

(* {65499, 8} *)

Here is a summary of the quantative columns:

Grid[{RecordsSummary[N@data[[All, {4, 6}]], columnNames[[{4, 6}]]]}, Dividers -> All]

enter image description here

Adding a "discount" column

Since we have only two quantative columns in the data let us add a third one, and make it have 3 outliers.

dvec = RandomReal[SkewNormalDistribution[1, 2, 4], Dimensions[data][[1]]];
Block[{inds = RandomSample[Range[Length[dvec]], 3]}, 
  dvec[[inds]] = 10*dvec[[inds]]];
testData = MapThread[Append, {N@data[[All, {4, 6}]], dvec}];

Here is the summary:

Grid[{RecordsSummary[testData]}, Dividers -> All]

enter image description here

The data adheres to the data description in the quesion, it has non-Normal distributions of its variables:

enter image description here


It is a good idea to standardize the data. This is not necessary for the outlier finding procedure used below, but it makes the data more convenient for visualization or other exploration.

sTestData = Transpose[Standardize /@ Transpose[N@testData]];

Because of the outliers plotting the data might produce uninformative plots. We can use logarithms of the point coordinates and for that we have to shift the standardized data to be positive. This is done with this command:

Block[{offset = -2 (Min /@ Transpose[sTestData])}, 
  sTestData = Map[# + offset &, sTestData]];

Let us get the standardized data summary and visualize:

Grid[{RecordsSummary[sTestData]}, Dividers -> All]

enter image description here

opts = {PlotRange -> All, ImageSize -> Medium, 
  PlotTheme -> "Detailed"}; Grid[{{ListPointPlot3D[sTestData, opts], 
   ListPointPlot3D[Log10@sTestData, opts]}}]

enter image description here

Using Quantile Regression envelopes to find outliers

We are going to find the outliers by computing envelopes around the dataset points that contain almost all points (e.g. 99.7% of them).

The finding of directional quantile envelopes in 2D and 3D is explained in these blog posts:

  1. "Directional quantile envelopes",

  2. "Directional quantile envelopes in 3D".

I am a big fan of Quantile Regression (QR) and I have implemented a collection of functions and applications of QR. See these blog posts.

This command imports the package QuantileRegression.m :


Here is an example of an envelope found using directional quantiles over a small sample of the points:

Block[{testData = RandomSample[sTestData, 2000], qreg},
 qreg = QuantileEnvelopeRegion[testData, 0.997, 10];
 Show[{Graphics3D[{Red, Point[testData]}, Axes -> True], 

enter image description here

The outlier points (in red) are outside of the envelope.

The example (and plot) above are just for illustration purposes. We calculate a quantile evelope region using all points.

Block[{testData = sTestData},
 qreg = QuantileEnvelopeRegion[testData, 0.9997, 10];

This command makes a function to test does a point belong to the found envelope region or not:

rmFunc = RegionMember[qreg];

This calculates the membership predicates for all points:

 pred = rmFunc /@ sTestData;

(* {15.6485, Null} *)

And we can see the membership breakdown:

(* {{True, 65402}, {False, 97}} *)

and visualize it (using Pick and taking logarithms):

Graphics3D[{Gray, Point[Log10@Pick[sTestData, pred]], Red, 
  Point[Log10@Pick[sTestData, Not /@ pred]]}, Axes -> True]

enter image description here

The plot above contains both top and bottom outliers (in red). If we are intereseted only in the top outliers we can find these thresholds:

topThresholds = Quantile[#, 0.95] & /@ Transpose[testData]
(* {25, 10.95, 4.89433} *)

and use them to select the top outliers:

Select[Pick[testData, Not /@ pred], 
 Total[Thread[# > topThresholds] /. {True -> 1, False -> 0}] > 1 &]

(* {{1, 836.14, 6.48564}, {1, 16.13, 8.71455}, {5, 25.49, 8.47351}, {-1, 1126, 5.25211}, {1000, 0, 5.4212}, {1, 15.79, 9.14674}, {-1, 544.4, 6.86266}} *)

Note that the quantile regression envelope and the membership predicates were computed over the standardized data, and the predicates were used to retrieve the outliers of the original data.

  • 1
    $\begingroup$ Excellent post! I retracted my close vote even though the OP does not mention any Mathematica context... but it would be a pity to let this answer go to waste. $\endgroup$
    – Yves Klett
    Apr 30, 2016 at 16:30
  • $\begingroup$ @YvesKlett This is nice to hear -- thank you! $\endgroup$ Apr 30, 2016 at 16:37
  • 1
    $\begingroup$ @ciao Thanks! And note that no sermon was necessary, just matter of fact statements! $\endgroup$ May 1, 2016 at 1:18
  • 1
    $\begingroup$ @ciao. The answer is excellent in terms of identifying potential outliers. However, I'd argue that a sermon is especially appropriate here for labeling of outliers which would then be used to exclude data automatically. Labeling things as out of the ordinary for closer examination is fine but it is the automatic exclusion that's the problem. I do have some idea of my limits for my power of persuasion for certain folks and chose not to comment earlier. Some in this forum seem to have learned a different flavor of statistics than I did. $\endgroup$
    – JimB
    Feb 10, 2017 at 14:10
  • 2
    $\begingroup$ @JimBaldwin Completely agree. Removing outliers without explaining them is not a good idea. This procedure is too ad-hoc, which is both advantage and dis-advantage depending on the problem domain, data size, etc. $\endgroup$ Feb 10, 2017 at 14:20

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