Skip to main content

plotting - How to plot ternary density plots?


How can I get a ternary density plot just like the plots from OriginLab? ternary density plot


ContourPlot and DensityPlot seemingly can accept the [f, {x}, {y}]-style,but cannot accept the [f, {x},{y}, {z}]-style.



Answer



Here's my attempt at an implementation, using ReliefImage[] to give the plots some depth perception:


triangleTicks[arg_List: {5, 4}, tl_: 0.01] := Module[{divs, dQ, sides, st},
dQ = VectorQ[#, IntegerQ] && Length[#] == 2 &;
sides = Partition[{{0, 0}, {1, 0}, {1, Sqrt[3]}/2}, 2, 1, 1];

divs = If[dQ[arg] || (MatrixQ[arg, NumericQ] && First[Dimensions[arg]] == 1),
{arg}, arg];
divs = If[dQ[#],
DeleteCases[MapAt[Function[f, Flatten[ArrayPad[#, -1] & /@ f]],
FindDivisions[{0, 1, 1/Rest[FoldList[Times, 1, #]]}, #],
2], {}], #] & /@
divs[[Mod[Range[3], Length[divs], 1]]];
st = MapThread[Table[Join[Transpose[{1 - d, d}].#1, List /@ d, 2], {d, #2}] &,
{sides, divs}];
MapIndexed[Block[{pt = N[Most[#1]], os},

os = Scaled[RotationTransform[2 π (#2[[1]] - 2)/3][
{tl/#2[[2]], 0}], pt];
If[#2[[2]] == 2, Line[{pt, os}],
{Text[ToString[If[IntegerQ[Last[#1]],
Identity, N][Last[#1]]], os,
{{1, 1}, {-1, -1}, {1, -1}}[[#2[[1]]]]],
Line[{pt, os}]}]] &, st, {3}]]

Options[TernaryReliefPlot] =
{AspectRatio -> Automatic, Background -> None, BaselinePosition -> Automatic,

BaseStyle -> {}, ClippingStyle -> {Black, White}, ColorFunction -> "ThermometerColors",
ColorFunctionScaling -> True, ColorOutput -> Automatic, ContentSelectable -> Automatic,
CoordinatesToolOptions -> Automatic, DisplayFunction :> $DisplayFunction, Epilog -> {},
FormatType :> TraditionalForm, FrameLabel -> None, FrameTicks -> Automatic,
ImageMargins -> 0., ImagePadding -> All, ImageSize -> Automatic,
ImageSizeRaw -> Automatic, LabelStyle -> {}, Method -> Automatic, PlotLabel -> None,
PlotPoints -> Automatic, PlotRange -> All, PlotRegion -> Automatic,
PreserveImageOptions -> Automatic, Prolog -> {}, RotateLabel -> True};

TernaryReliefPlot[f_, opts : OptionsPattern[]] :=

Module[{fl, flt, ft, img, n, rl, sides},
sides = {{0, 0}, {1, 0}, {1, Sqrt[3]}/2};
fl = OptionValue[FrameLabel];
If[fl =!= None,
If[fl === Automatic, fl = ToString /@ Range[3]];
If[Head[fl] =!= List, fl = PadRight[{fl}, 3, ""]];
flt = {fl, ListCorrelate[{{1}, {1}}/2, sides, 1]} ~Join~
If[MatchQ[OptionValue[RotateLabel], True | Automatic],
{{{0, 2.5}, {0, -2.5}, {0, -2.5}},
{{1, 0}, {1, -Sqrt[3]}/2, {1, Sqrt[3]}/2}},

{{{0, 2.5}, {-2.5, 0}, {2.5, 0}}}]];
ft = OptionValue[FrameTicks]; If[ft === Automatic, ft = {5, 4}];
n = OptionValue[PlotPoints]; If[n === Automatic, n = 300];
img = ReliefImage[SparseArray[{j_, k_} /; j >= k :>
f @@ ({j - k, k - 1, n - j}/(n - 1)), {n, n}],
FilterRules[Join[{opts}, Options[TernaryReliefPlot]],
Options[ReliefImage]]];
Graphics[{If[ft =!= None, triangleTicks[ft], {}],
Texture[img], Polygon[sides, VertexTextureCoordinates ->
{{0, 0}, {1, 0}, {0, 1}}],

If[fl =!= None, MapThread[Text, flt], {}]},
Axes -> False, AxesLabel -> None, Frame -> False,
FrameLabel -> None, Method -> Automatic, PlotRange -> All,
FilterRules[Join[{opts}, Options[TernaryReliefPlot]],
Options[Graphics]]]]

Try it out:


TernaryReliefPlot[#3 Sin[10 #1]^2 + #3 (1 - #3) Cos[20 #2]^2 &, 
ColorFunction -> (Hue[0.85 #] &),
FrameLabel -> {Style["p", Large], Style["q", Large], Style["r", Large]},

FrameTicks -> {4, 2}]

a ternary relief plot


It's still missing a few things (e.g. grid lines), but it's a start. I'll try to improve on this when I get the chance.


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...