3
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If I have a list and plot it:

sample={{0.0758227, 2.76751, "C"}, {0.230704, 2.773, "B"}, {0.25679, 2.66845,
   "S"}};
ListPlot[sample[[All,{1,2}]]]

Now I would like to print each point with a colour based on the third column, say if I have an association:

<|"B" -> 1, "C" -> 2, "S" -> 3|>

then the numbers would be indices of some ColorData, say: ColorData[97] /@ Range[3], then everything that has "B" in the last column is plotted blue, "C" is plotted orange and "S" is plotted green.

Also, I should mention that the sample above is really just a sample, the real list has on the order $10^5$ rows, however, the number of distinct letters in the last column is about 20.

I have found some answers but they deal with plotting each point with different colour.

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SeedRandom[1];

sample = Table[
   {RandomReal[], RandomReal[5], 
    RandomSample[CharacterRange["A", "J"], 1][[1]]}, 10^5];

color = Association@Thread[CharacterRange["A", "J"] -> Range[10]];

SplitBy and SortBy are used rather than GatherBy to control the sequencing of the categories.

data = SplitBy[SortBy[sample, Last], Last];

Module[{categories = (#[[1, -1]] & /@ data)},
 ListPlot[
  #[[All, 1 ;; 2]] & /@ data,
  Frame -> True,
  Axes -> False,
  PlotStyle -> (ColorData[97][color[#]] & /@ categories),
  PlotLegends -> SwatchLegend[categories]]]

enter image description here

With 10^5 data points it only takes a couple of seconds; however, the data is so dense as to be meaningless unless there is some inherent structure that becomes visible.

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sample={{0.0758227, 2.76751, "C"}, {0.230704, 2.773, "B"}, {0.25679, 2.66845, "S"}};
assoc = <|"B" -> 1, "C" -> 2, "S" -> 3|>;

sample2 = sample /. assoc;

ListPlot[Style[{#, #2}, ColorData[97][#3], PointSize[.05]] & @@@ sample2]

or

ListPlot[List/@sample2[[All,{1,2}]], PlotStyle-> ColorData[97] /@ sample2[[All,3]])]

to get

Mathematica graphics

Both of the above are slow for large input data. Using Graphics gives a faster method:

Graphics[{PointSize[.05], Transpose@{ColorData[97]/@sample2[[All,3]], 
 Point/@sample2[[All, {1,2}]]}}, Axes->True]

or

Graphics[{PointSize[.05], Point[sample2[[All, {1,2}]], 
  VertexColors->(ColorData[97] /@ sample2[[All,3]])]}, Axes->True]

Mathematica graphics

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  • $\begingroup$ Thank but this is way too slow, original ListPlot is about 10 seconds, this has already been running for 3 minutes and nothing. $\endgroup$ – leosenko May 2 '17 at 3:02
  • $\begingroup$ @leosenko, it is slow. I think using Graphics would be faster. $\endgroup$ – kglr May 2 '17 at 3:34

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