Skip to main content

graphics - Preventing label crowding in PieChart RadialCallout and RadialCenter


Is it possible to avoid this unreadable situation with label crowding when using PieChart's RadialCallout and RadialCenter methods?


PieChart[tabData109[[All, 2]], 
 SectorOrigin -> {{Pi/2, "Clockwise"}, 1}, 
 ChartStyle -> tabData109[[All, 1]] /. PACE["TAB_COLOR_RULES"],
 LabelingFunction -> (Placed[
     Row[{NumberForm[

        100 # /Plus @@ tabData109[[All, 2]] // N, {3, 1}], "%"}], 
     "RadialCenter"] &),
 ChartLabels -> 
  Placed[tabData109[[All, 1]] /. PACE["TAB_DESCRIPTION_RULES"], 
   "RadialCallout"]]

Gives:


enter image description here



Answer



I had the following thought about the question.




We generate some random crowded test data first:


data = RandomChoice[{20, 15, 8, 7, 6, 5} -> {1, 2, 3, 4, 5, 10}, 50]


{1, 4, 1, 3, 1, 2, 2, 3, 5, 2, 1, 1, 1, 1, 3, 5, 4, 5, 2, 1, 2, 1, 1, 2, 5, 1, 1, 3, 1, 1, 3, 3, 2, 5, 2, 2, 2, 1, 4, 4, 1, 2, 1, 4, 2, 3, 1, 1, 5, 4}



dataLength = Length[data]; 

descriptionData = (FromCharacterCode[RandomInteger[{97, 122}, 

        {RandomInteger[{4, 10}]}]] & ) /@ data

Mathematica graphics


valueData = (NumberForm[#1, {3, 1}] & ) /@ N[(100*data)/Total[data]]; 

labelLst = MapThread[Row[{#1, ":  ", #2, "%"}] & , {descriptionData, valueData}]

Mathematica graphics


Then draw the PieChart using system function:


chartgraph = PieChart[data,

  SectorOrigin -> {{\[Pi]/2, "Clockwise"}, 1},
  LabelingFunction -> (Placed[
      Framed[Style[
        labelLst[[#2[[2]]]],
        Bold, 13],
       Background -> Lighter[Purple, 0.95]],
      "RadialCallout"] &),
  ChartStyle -> EdgeForm[{White, Opacity[0.2]}],
  PlotRange -> All]


Mathematica graphics



Now we'll do some dirty job, modify the underlying data of chartgraph.


First define some functions which are not aesthetic at all, and are very likely not so general for any PieChart. (Their function is adjusting the radial of "RadialCallout" lines.)


Clear[extentFunc]
extentFunc[labeldata_, Radial_] :=
 ReplaceAll[labeldata,
  {{{}, {}}, {{{}, {}},
     {directive1__?(Head[#] =!= LineBox &),
      LineBox[{r0_, R0_}],

      LineBox[{R0_, endpoint_}]},
     {directive2__?(Head[#] =!= LineBox &),
      DiskBox[r0_, diskR_]},
     InsetBox[labeltext_, labPos_, labOPos_]}} :>
   With[{R = Radial/Norm[R0] R0},
    With[{v = R - R0},
     horizonLineLength = Abs[(endpoint - R0)[[1]]];
     {{{}, {}}, {{{}, {}},
       {directive1,
        LineBox[{r0, R}],

        LineBox[{R, endpoint + v}]},
       {directive2, DiskBox[r0, diskR]},
       InsetBox[labeltext, labPos + v, labOPos]}}
     ]]]

Clear[chartExtentFunc]
chartExtentFunc[chartgraph_, Radial_?NumericQ] :=
 ToExpression[ReplacePart[
   ToBoxes[chartgraph],
   {1, 3, 2, 2, 1, 1, 1} -> (

     ReplacePart[#,
        1 -> extentFunc[#[[1]], Radial]
        ] & /@
      ToBoxes[chartgraph][[1, 3, 2, 2, 1, 1, 1]]
     )]]

chartExtentFunc[chartgraph_, Radial_List] :=
 ToExpression[
  With[{num = $ModuleNumber},
    StringReplace[ToString[

      ReplacePart[
       ToBoxes[chartgraph],
       {1, 3, 2, 2, 1, 1, 1} -> (
         MapThread[
          ReplacePart[#1, 1 -> extentFunc[#1[[1]], #2]] &,
          {ToBoxes[chartgraph][[1, 3, 2, 2, 1, 1, 1]],
           Radial}]
         )],
      InputForm],
     "DynamicChart`click$" ~~ (a : DigitCharacter ..) ~~ 

       "$" ~~ (b : DigitCharacter ..) :>
      "DynamicChart`click$" <> a <> "$" <> ToString[num]
     ]] // ToExpression]

Now try them on our chartgraph with random radials:


chartExtentFunc[chartgraph,RandomReal[{2.1, 3},dataLength]]/.Thickness[a_]:>Thickness[.5 a]

Mathematica graphics


It is of course nice to associate radials with correspond polar angles:


\[Theta]Set = \[Pi]/2 - (Accumulate[#] - 1/2 #) &[

data/Total[data] 2 \[Pi]] // N;

2 + If[0 <= # < \[Pi]/8 || \[Pi] - \[Pi]/8 < # < \[Pi] + \[Pi]/8 || 
  2 \[Pi] - \[Pi]/8 < # <= 2 \[Pi], 2.1 Abs[Cos[#]]^12, .3/
 Abs[Sin[#]]] & /@ (\[Pi]/2 - \[Theta]Set);

chartExtentFunc[chartgraph, %]

Mathematica graphics


MapIndexed[

  Piecewise[{{2.4, # == 0}, {3.4, # == 1}, {4.4, # == 2}}] &[
Mod[#2[[1]], 3]] &, (\[Pi]/2 - \[Theta]Set)];

chartExtentFunc[chartgraph, %] /. Thickness[_] :> Thickness[0]

Mathematica graphics



Well the above results are not so nice. So we try to improve it by introducing an optimization function (potential function).


RvariableSet = Table[Symbol["R" <> ToString[i]], {i, dataLength}]


Clear[centerPos]
centerPos[k_] := 
 R[[k]] {Cos[\[Theta][[k]]], Sin[\[Theta][[k]]]} + {L, 0} + {W/2, 0}

Clear[centerPotentialFunc]
centerPotentialFunc[k_, Rmin_, Rmax_] := 
 Exp[-10 (R[[k]] - Rmin)] + Exp[10 (R[[k]] - Rmax)]

Clear[interactionPotentialFunc]
interactionPotentialFunc[i_, j_] := If[i == j, 0,

  With[{d = Sqrt[#.#]/Sqrt[W^2 + H^2] &[centerPos[i] - centerPos[j]]},
   2 Exp[-10 (d - 1.1)]
   ]]

(Here W and H are the max width and height of the label text box separately.)


potentialExpr = 
  Block[{\[Theta] = \[Theta]Set, L = horizonLineLength, W = 1.3, 
H = 0.25, R = RvariableSet},
   Sum[centerPotentialFunc[i, 2.2, 5], {i, 1, dataLength}] + 
Sum[interactionPotentialFunc[i, j], {i, 1, dataLength},

   {j, 1, dataLength}]];

(Here for each i, the upper and lower bound of j can be localized to neighborhood of it to reduce the size of potentialExpr.)


Grad of the total potential (I thinks here I "inject" RvariableSet in an unidiomatic way?):


gradExpr = Module[{CompileTemp},
   CompileTemp[RvariableSet, Evaluate[
      D[potentialExpr, #] & /@ RvariableSet
      ]] /. CompileTemp -> Compile];

Run the kinetics simulation for 300 steps:



initRSet = ConstantArray[3, dataLength];

dt = 1 10^-3;

RSetSet = NestList[Function[paras, Module[{a, v},
     v = paras[[2]];
     a = -gradExpr @@ paras[[1]];
     v = v + 1/2 dt a;
     (If[#[[1]] < 
       2.1, {2.1, -#[[2]]}, {#[[

        1]], .1 #[[2]]}] & /@ ({paras[[1]] + dt v, 
       v}\[Transpose]))\[Transpose]
     ]], {initRSet, ConstantArray[0, dataLength]}, 300];

Manipulate[
 ListPolarPlot[{\[Theta]Set, RSetSet[[k, 1]]}\[Transpose], 
  PlotStyle -> Purple, Joined -> True, 
  Epilog -> {Circle[{0, 0}, 2.1], Circle[{0, 0}, 2]}, PlotRange -> All],
 {k, 1, Length[RSetSet], 1}]


Mathematica graphics


Now try the result on chartgraph:


chartExtentFunc[chartgraph, RSetSet[[-1, 1]]]/.Thickness[a_]:>Thickness[.5 a]

Mathematica graphics


% /. FrameBox[expr_, opt__] :> expr /. Bold -> Plain

Mathematica graphics


So it's kind of better now. (Though still not good enough..)




Thus far, it seems if I choose a proper potential function, I will get a good result. But the final results are not as satisfying as I want. I think there can be more essential improvements, for efficiently and for better result.


Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

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...
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more