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I would like to generate a ListPlot with the color for each point in the plot corresponding to a particular value (not associated with the position in the plot). I'd then like to add a legend indicating what the color means.

I'm currently solving the first part of the problem by essentially generating a separate ListPlot for each data point and then assigning a color to that ListPlot. Here's a toy example:

n = 5000;
pos = RandomVariate[NormalDistribution[0, 2], {n, 2}];
altitude = Norm /@ pos;
ListPlot[{#} & /@ pos, 
 PlotStyle -> ((Blend[{{Min[altitude], Yellow}, {Max[altitude], 
         Red}}, #] &) /@ altitude), AspectRatio -> 1]

enter image description here

So my questions are:

(1) Is this the only way to generate such a ListPlot (it does the job but it seems inelegant and I suspect it's inefficient, though that's not a big concern for my application)?

(2) Is there an easy way to generate a legend which indicates the value of the color (i.e., a gradient bar which shows the color scale)?

share|improve this question
up vote 26 down vote accepted

In this case I would use Point for plotting the points. For example

n = 5000;
pos = RandomVariate[NormalDistribution[0, 2], {n, 2}];
altitude = Norm /@ pos;
colorf = Blend[{{Min[altitude], Yellow}, {Max[altitude], Red}}, #] &

pl = Graphics[MapThread[{colorf[#1], Point[#2]} &, {altitude, pos}], 
  Axes -> True, AspectRatio -> 1]

As for plotting legends, that's a reoccurring issue in Mathematica. There is a package called PlotLegends` which you could try but it is not very user friendly and the legends it produces are quite ugly IMHO. I find that it's often faster to just create a legend by hand. For example, this is a function I use for creating legends with contour plots:

plotLegend[{min_, max_}, n_, col_] := 
      Rectangle[{0, #2[[1]] - 1}, {1, #2[[1]]}]}, {Black, 
      Text[NumberForm[N@#1, {4, 2}], {4, #2[[1]] - .5}, {1, 0}]}} &, 
   Rescale[Range[n], {1, n}, {min, max}]],
  Frame -> True, FrameTicks -> None, PlotRangePadding -> .5]

Here, n is the number of subdivisions and col is the colour function. You could combine the legend with the original plot using Grid, e.g.

leg = plotLegend[Through[{Min, Max}[altitude]], 20, colorf];
Grid[{{Show[pl, ImageSize -> {Automatic, 300}], 
  Show[leg, ImageSize -> {Automatic, 250}]}}]

Mathematica graphics

share|improve this answer
The numbers on the color legend aren't really nice. You could use FindDivisions to get a nice set, like so FindDivisions[Through[{Min, Max}[altitude]], 10] – Sjoerd C. de Vries Feb 4 '12 at 18:20
Very helpful! Many thanks to you and the others who posted answers and suggestions!!! – Cassini Feb 5 '12 at 12:42
WHY are the most minor, simple, obvious things so bloody difficult in MMa? I've been using MMa for about 6 months now, and this is my zillionth visit to SE to find out how to do something you'd accomplish in 1 line in Mathematica. Does Wolfram not think about real world use at all??? – Jerry Guern Jan 24 '15 at 7:08

As usual Heike has a fine method, but I can tighten it up.

This will render quite a bit faster, it will use built-in color functions more easily, and it will IMO interactively rescale better.


n = 5000;
pos = RandomReal[NormalDistribution[0, 2], {n, 2}];
altitude = Norm /@ pos;
colorf = Blend[{Yellow, Red}, #] &;

Legend function (modified):

plotLegend[{min_, max_}, n_, col_] :=
  {{col[(# - 1)/(n - 1)], Rectangle[{0, # - 1}, {1, #}]},
   {Black, Text[
      NumberForm[Rescale[#, {1, n}, {min, max}], {4, 2}],
      {3, # - .5}, {1, 0}]}
  } & /@ Range@n,
  Frame -> True, FrameTicks -> None, PlotRangePadding -> .5]


  Point[pos, VertexColors -> colorf /@ Rescale@altitude],
  Inset[plotLegend[{Min@#, Max@#}& @ altitude, 20, colorf],
    Scaled[{1, 1/2}],
    ImageScaled[{-0.1, 1/2}],
    Scaled[{1/4, 0.9}]]
 }, Axes -> True]

Mathematica graphics

It's easy to use other gradients. With colorf = ColorData["Rainbow"]:

Mathematica graphics

share|improve this answer
As with Heike's solution, yours has an ugly set of numbers. No professional typesetter/layouter would use these. Use FindDivision for a better set of numbers. – Sjoerd C. de Vries Feb 5 '12 at 22:14
@Sjoerd I only noticed this comment now. Okay, I'll improve this later. – Mr.Wizard Feb 11 '12 at 0:29
The plotting here is tremendously faster than the accepted answer, FYI. – Guillochon Mar 20 '14 at 15:34
@Guillochon Thanks for noticing. :-) – Mr.Wizard Mar 20 '14 at 17:57
@Sjoerd I see I never did address your criticism. Sorry. This time I won't claim that I will, but perhaps; we'll see. – Mr.Wizard Mar 20 '14 at 17:58

ListLinePlot (or alternately ListPlot with Joined->True) accepts a ColorFunction, which you can use to color your points. The lines can be later converted to points as:

colorFun = Function[{x, y}, Blend[{{Min[altitude], Yellow}, {Max[altitude], Red}}, 
    Norm[{x, y}]]];    
ListLinePlot[pos, ColorFunction -> colorFun, AspectRatio -> 1, 
      ColorFunctionScaling -> False] /. Line -> Point

Then using Heike's plot legends and Sjoerd's suggestion to use FindDivisions, you get:

enter image description here

share|improve this answer
+1 this gets the same faster rendering via VertexColors that my code does, but is arguably neater. I still like Inset for the legend; consider combining them. – Mr.Wizard Feb 4 '12 at 21:29
@Spartacus Thanks. I will, sometime later in the day. I'm off for now. – R. M. Feb 4 '12 at 21:59

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