Graphics PolarTicks、Polar GridLines&；后台定制

建立在 我想定制R轴和θ轴：

请考虑：

list = {{21, 16}, {16, 14}, {11, 11}, {11, 12}, {13, 15}, {18,17}, {19, 11}, {17, 16}, {16, 19}}

Module[{Countz, maxScale, angleDivisions, dAng}, 
        maxScale = 4;
        angleDivisions = 12;
        dAng = (2 \[Pi])/angleDivisions;
        Countz = BinCounts[ Flatten@Map[ArcTan @@ (# - ScreenCenter) &, list, {1}], 
                            {-Pi, Pi,dAng}];

        SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 
        SectorOrigin -> {-\[Pi]/angleDivisions, "Counterclockwise"}, 
        PolarAxes -> True, PolarGridLines -> {Automatic, {1, 2, 3, 4, 5}},
        PolarAxesOrigin -> {Pi/2, 5},
        PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, i \[Degree]}, 
                       {i, 0, 345, 30}], Range[5]}, 
        ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Red]},
        BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, 
        FontSize -> 12}, ImageSize -> 400, 
        ChartElementFunction -> Function[{range}, 
                                Disk[{0, 0}, range[[2, 2]], -11 Pi/12 + range[[1]]]]]]

2/我想添加一个45度的“θ条”（如图所示）。不过我是用ppt做的。我找不到一个方法来保持桌子以获得主θ轴，并绘制另一个不同颜色的轴

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3/I未能使用Prolog将提供的解决方案应用于Polarplot。是否可以将绘图外圆内的区域涂成灰色，比如说？

您可以使用

Epilog

创建线条和

Prolog

作为背景，例如

Module[{Countz, maxScale, angleDivisions, dAng}, maxScale = 4;
 angleDivisions = 12;
 dAng = (2 \[Pi])/angleDivisions;
 Countz = 
  BinCounts[
   Flatten@Map[ArcTan @@ (# - ScreenCenter) &, list, {1}], {-Pi, Pi, dAng}];

 SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 
  SectorOrigin -> {-\[Pi]/angleDivisions, "Counterclockwise"}, 
  PolarAxes -> True, PolarGridLines -> {Automatic, {1, 2, 3, 4, 5}},
  PolarAxesOrigin -> {Pi/2, 5}, 
  PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, 
      i \[Degree]}, {i, 0, 345, 30}], Range[5]}, 
  ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Red]},
   BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, FontSize -> 12}, 
  ImageSize -> 400, 
  ChartElementFunction -> 
   Function[{range}, 
    Disk[{0, 0}, range[[2, 2]], -11 Pi/12 + range[[1]]]],
  Epilog -> {Red, Thick, Line[{{0, 0}, 5/Sqrt[2] {1, 1}}]},
  Prolog -> {Gray, Disk[{0, 0}, 5]}]]

---

您可以使用

Epilog

创建行，并使用

Prolog

作为背景，例如

Module[{Countz, maxScale, angleDivisions, dAng}, maxScale = 4;
 angleDivisions = 12;
 dAng = (2 \[Pi])/angleDivisions;
 Countz = 
  BinCounts[
   Flatten@Map[ArcTan @@ (# - ScreenCenter) &, list, {1}], {-Pi, Pi, dAng}];

 SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 
  SectorOrigin -> {-\[Pi]/angleDivisions, "Counterclockwise"}, 
  PolarAxes -> True, PolarGridLines -> {Automatic, {1, 2, 3, 4, 5}},
  PolarAxesOrigin -> {Pi/2, 5}, 
  PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, 
      i \[Degree]}, {i, 0, 345, 30}], Range[5]}, 
  ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Red]},
   BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, FontSize -> 12}, 
  ImageSize -> 400, 
  ChartElementFunction -> 
   Function[{range}, 
    Disk[{0, 0}, range[[2, 2]], -11 Pi/12 + range[[1]]]],
  Epilog -> {Red, Thick, Line[{{0, 0}, 5/Sqrt[2] {1, 1}}]},
  Prolog -> {Gray, Disk[{0, 0}, 5]}]]

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Heike，我如何调整线的角度，让

s 37.5度？我不明白这条线在哪个坐标系。非常感谢。@500在结尾：
line[{0,0}，r{Cos@Phi,Sin@Phi}]
其中
r
是半径，
Phi
是半径angle@belisarius非常感谢。我认为它需要{}使之生效。而且我还是很困惑，角度对我来说都是度。phi应采用哪种格式？Heike，对于let
s 37.5度，我如何调整线的角度？我不明白这条线在哪个坐标系。非常感谢。@500在结尾：
line[{0,0}，r{Cos@Phi,Sin@Phi}]
其中
r
是半径，
Phi
是半径angle@belisarius非常感谢。我认为它需要{}使之生效。而且我还是很困惑，角度对我来说都是度。phi应采用哪种格式？