Suppose a data array of the form {{x1,y1,z1},{x2,y2,z2}...}, is it possible to use ListPlot where x,y specify the point position and z (Hue[z]) specify the color of each point?
$\begingroup$
$\endgroup$
3
2 Answers
$\begingroup$
$\endgroup$
1
If you really must use ListPlot and not Graphics as Ubpdqn showed, then here is a hack.
The problem is that ColorFunction for 2D takes only {x,y}. One way is to get the z value yourself from the data. But for this you need the current index.
So simply make an index, set it to zero, and increment it each time the ColorFunction is called. Like this. Notice, must use Joined -> True for this to work.
data = RandomReal[{0, 1}, {40, 3}];
n = 0; (*make sure to set this to zero before each call*)
ListPlot[data[[All, {1, 2}]],PlotStyle->Directive[PointSize[Large]],Joined -> True,
ColorFunction -> Function[{x, y}, Hue[data[[++n, 3]]] ]] /. Line[a__] :> Point[a]
reference 2D ListPlot with separate color function
-
$\begingroup$ Thank you for the solution. How could I use a certain color map like
"TemperatureMapor my own color table:number = Length@data[[All,3]]; colTable = {Table[{Blend[{Blue, Green, Yellow, Red}, x]}, {x, 1/number, 1, 1/number}]}; colTable = Flatten[colTable];$\endgroup$– mrzOct 18, 2017 at 9:01
$\begingroup$
$\endgroup$
You could achieve this using Graphics, for example:
data = RandomReal[{0, 1}, {100, 3}];
Graphics[{Hue[#3], PointSize[0.02], Point[{#1, #2}]} & @@@ data,
Axes -> True]
ColorFunctionsee ref/ListPlot3D under Color by scaled x, y, and z values. there is example there to do this. $\endgroup$