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I have a dataset in which each element is a list of 3 numbers eg.

data={{0,0,0},{1,2,0.7},{3,4,2.3},{3,4,3.3}}

I'd like to make a 2 dimensional plot in this way: for each list in the dataset the first 2 elements specify the point coordinate in the 2D plot and the third element specifies a color taken from a color gradient (for example using ColorData[]), so that close numbers have close colors; moreover, I'd like to show a bar legend with colors.

An example that solve the problem is this

mM=MinMax@data[[All,3]]
dataPoints=Tooltip[{Blend[{Red,Blue},Rescale[#[[3]],mM]],Point[{#[[1]],#[[2]]}]},#[[3]]]&/@data
Legended[Graphics[{AbsolutePointSize[10],dataPoints},Frame->True],BarLegend[{Blend[{Red,Blue},#]&,mM},ColorFunctionScaling->True]]

Apart that maybe there is a bug in the frontend, since it says that «ColorFunctionScaling is not an option for BarLegend», can I obtain such a plot without using the Graphics command and related primitives?

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data = {{0, 0, 0}, {1, 2, 0.7}, {3, 4, 2.3}, {3, 4, 3.3}};

mM = MinMax@data[[All, 3]];

cf = Blend[{Red, Blue}, Rescale[#, mM]] &;

dataPoints = Tooltip[{cf[#[[3]]], Point[Most@#]}, #[[3]]] & /@ data;

Legended[Graphics[{AbsolutePointSize[10], dataPoints}, Frame -> True],
  BarLegend[{cf, mM}]]

enter image description here

Note that two of your points are at the same coordinates {3, 4} and only the last will show. You could change your color function to cf = Blend[{Opacity[0.5, Red], Opacity[0.5, Blue]}, Rescale[#, mM]] &; then overlapped points will be darker than singular points. However, in general the colors would then be washed out.

EDIT: Using ListPlot

dataPoints = Tooltip[Style[Most@#, cf[#[[3]]]], #[[3]]] & /@ data;

Legended[
 ListPlot[dataPoints,
  PlotStyle -> AbsolutePointSize[10],
  Frame -> True],
 BarLegend[{cf, mM}]]

enter image description here

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  • $\begingroup$ Well, you've improved my answer but I was looking for an approach different from what I've done. For example, a specific Mathematica function would be a perfect solution. Note that other languages provide such functions. I'll modify my question to be more specific. Thanks anyway :) $\endgroup$ – Giancarlo Nov 4 '18 at 14:45

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