I have 2 data sets, data1 and data2 with large (in the thousands) amounts of data and I want to compare them in a plane. They aren't modeled by the normal distribution, but I'm using that here for simplicity.

data1 = RandomReal[NormalDistribution[], {10000, 2}];
data2 = data1+1;

I can put these on a list plot and plot them. However, this is visually unpleasing and even hides some of the data.

combPlot = ListPlot[{data1, data2},
Export[Directory[] <> "/figures/test-comb.pdf", combPlot];

The SmoothDensityHistogram command with the following color function shows the data exactly how I want it for each data set.

histOpts = {ColorFunction -> Function[c, GrayLevel[1 - c]], 
   PlotRange -> {{-4, 4}, {-4, 4}}, ImageSize -> Medium};
hist1 = SmoothDensityHistogram[data1, histOpts];
Export[Directory[] <> "/figures/test-hist1.pdf", hist1];
hist2 = SmoothDensityHistogram[data2, histOpts];
Export[Directory[] <> "/figures/test-hist2.pdf", hist2];

How can I combine these into a single plot, preferably with different colors? I had no success using the Show command. Again, the idea is to visually compare the 2 data sets. If there are any other alternatives, I'd like to see them too.

Update: After seeing Diego Zviovich's post, I was able to get it working for my example. However, my actual data sets may significantly differ so the bounds aren't the same:

data1 = RandomReal[NormalDistribution[], {10000, 2}];
data2 = data1+10;

hist1 = SmoothDensityHistogram[data1,
    ColorFunction->Function[c, Hue[215/360, .973, 1, c]], 
    PlotRange->All, ImageSize -> Medium];
hist2 = SmoothDensityHistogram[data2, 
    ColorFunction->Function[c, Hue[311/360, .973, 1, c]], 
    PlotRange->All, ImageSize -> Medium];
combHist = Show[

Export[Directory[] <> "/figures/test-hist1.pdf", hist1];
Export[Directory[] <> "/figures/test-hist2.pdf", hist2];
Export[Directory[] <> "/figures/test-combHist.pdf", combHist];




How do I correct the bounds for this?

Update: Using m_goldberg's suggestion, I'm able to get what I want:

I'll leave this opened for a little for any further discussion, but I'm satisfied with this now. Thanks to everyone who helped!

Further update (sorry): I'm a little unsatisfied with the results for my actual data. The first one appears fine, but the second is overly distorted:

Does anybody have any further suggestions for these?

  • $\begingroup$ I don't fully understand why that works. Is there any weird hack to get Show to act as I expect here? $\endgroup$ Nov 27, 2012 at 2:48
  • $\begingroup$ Hi @Highphi, what about switching the X-Axis to Log X or Ln X? $\endgroup$
    – Zviovich
    Nov 27, 2012 at 20:56
  • $\begingroup$ Just a guess, but perhaps their are few outlying data points that are expanding the plot range unduly. Perhaps you can restrict the plot range with something like `PlotRange -> {{0,45000},{0,100}}. $\endgroup$
    – m_goldberg
    Nov 28, 2012 at 1:00
  • $\begingroup$ If its point clouds, try downsampling your data to every 100th point, for example. Then you can still get the sense of the distribution, but see through one to the other. $\endgroup$
    – MikeY
    Jan 9, 2019 at 15:48

3 Answers 3


I you add PlotStyle -> Opacity[0.5] to histOpts you can to evaluate Show[hist1, hist2] to get

overlaid plots

  • $\begingroup$ Sorry for forgetting about this! $\endgroup$ Mar 14, 2013 at 22:42
  • $\begingroup$ Beautiful plot... $\endgroup$
    – MikeY
    Jan 11, 2019 at 19:01

Would this work for you?

g1 = Rasterize[hist1];
g2 = Rasterize[hist2];

Mathematica graphics

  • $\begingroup$ Yes! See my update for a slight issue I'm having. $\endgroup$ Nov 27, 2012 at 3:35

Another way to go about this is to visualize in 3D. Taking the two data sets above, concatenate "1" to all the elements of the first data set and "-1" to the second data set.

dat1 = Partition[Flatten[Riffle[data1, 1, 2]], 3];
dat2 = Partition[Flatten[Riffle[data2, -1, 2]], 3];

Then you can plot in 3D.

ListPointPlot3D[{dat1, dat2}]

The advantage is that you can interactively rotate and play with the graph to view it from the different angles.


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