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I am trying to draw a custom BoxWhiskerChart with alternative values determining fences. I have data that has a lot of outliers and I have to compare it to another chart that has 10-90th percentile instead of standard min/max.

How would I go about it?

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Hi ! Do you have any code you can share with us ? – Sektor Jul 16 '14 at 21:59
Hi,data = RandomVariate[NormalDistribution[0, 1], 200]; BoxWhiskerChart[data] is the function I am using. Nothing more fancy about it. I am wondering if there is a way to modify the specification of what is a fence - ie instead of min-max - I would like it to be 10-th and 90th percentile. – jps Jul 16 '14 at 22:07
I would imagine it is possible to draw boxes, I am not sure how to use mathematica function draw[] to plot fences. – jps Jul 16 '14 at 22:10

With the data beeing

data = RandomVariate[NormalDistribution[0, 1], 200];

the range of the box specified to be one sigma (approx. 68.3 %tile range) by


and a limit for the fences defined to be 10 %

fencesLimit = 0.1

we can plot a BoxWhiskerChart using:

BoxWhiskerChart[data, "Median", Method -> "BoxRange" -> (Quantile[#, {fencesLimit, (1 - sigma)/2, 1/2, (1 + sigma)/2, 1 - fencesLimit}, {{1/3, 1/3}, {0, 1}}] &)]

This BoxWhisker is median centered, has a box range of one sigma, a lower fence at the 10 percentile, and an upper fence at the 90 percentile.
You can find in the documentation of Quantile how to choose another centering.

Here a plot of this BoxWhiskerChart:

enter image description here

With some additional styling and as a function with optional arguments:

bwChart[data_, br_: Erf[1/Sqrt[2]], flimit_: 0.1] := 
   {{"MedianMarker", 1, Directive[Thickness[0.01], Blue]}, {"MedianNotch", 0.5, Gray}},
   Method -> "BoxRange" -> (Quantile[#, {flimit, (1 - br)/2, 1/2, (1 + br)/2, 1 - flimit}, {{1/3, 1/3}, {0, 1}}] &), 
   BarSpacing -> None, AspectRatio -> 0.7, BarOrigin -> Left, 
   ChartElementFunction -> "GlassBoxWhisker", ChartStyle -> 6]

enter image description here

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(+1) Thank you! Spent hours to make BoxRange work as an option :) – kglr Jul 17 '14 at 2:08
This is fantastic. Thank you for the clear explanation! – user18593 Jul 17 '14 at 22:41

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