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Are there graphing techniques (certain kinds of plots) that are similar to the Edward Tufte's box plots - shown below - but that consider the distribution of the data points?

Edward Tufte style box plots

But instead of a black line traveling vertically, I want a colored line (say red) where the intensity of the redness changes in relation to the frequency of data points at the location. What I am picturing exactly is a vertical red line that does from 1% to 99%, with horizontal grey lines at 25% and and 75%, a black line at the median, with the color intensity of the vertical red line changing with respect to the number of times each data point along the vertical location was measured.

Any ideas?

Thanks, Tom

gung - Reinstate Monica
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traggatmot
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  • This is a question for [Stack Overflow](http://stackoverflow.com/). – Nick Stauner Jan 07 '14 at 03:01
  • This question appears to be off-topic because it is about looking for code. – gung - Reinstate Monica Jan 07 '14 at 03:12
  • I really cannot understand the description of the desired graphic. "Data points at the location?" "Number of times each data point ... was measured?" An example would be be helpful--perhaps essential. In the meantime, the related thread at http://stats.stackexchange.com/questions/13875 may be of some use. – whuber Jan 07 '14 at 03:28
  • @Nick Stauner, if such a style of graph exits, then I will go looking for the code - but I haven't found the use of such a graph anywhere, ao I am wondering first if it exists, then second about code. – traggatmot Jan 07 '14 at 03:43
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    @whuber, you nailed it. That page references beanplots - which I had never heard of before!!! That's the suggestion I was looking for - a beanplot. – traggatmot Jan 07 '14 at 03:48
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    I am very pleased that did the trick. I'll close this thread as a duplicate in order to link the two questions. – whuber Jan 07 '14 at 03:50
  • Fair enough, but the graph can certainly be coded, so if the answer to your "first" *implied* question (which you might want to edit into the OP more directly)...oh never mind. Looks like you got your answer already? :) Nice one @whuber! – Nick Stauner Jan 07 '14 at 03:51
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    @Nick, you are correct. Edited for meaning/clarity. Thanks! – traggatmot Jan 08 '14 at 04:02

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