Wolfram mathematica Arctan Binning，从绘图到直方图，技巧 基于Sjoerd，很好的解和延拓，请考虑以下内容： list = {{21, 16}, {16, 14}, {11, 11}, {11, 12}, {13, 15}, {18, 17}, {19, 11}, {17, 16}, {16, 19}} ScreenCenter = {20, 15} ListPolarPlot[{ArcTan[##], EuclideanDistance[##]} & @@@ (# - ScreenCenter & /@ list), PolarAxes -> True, PolarGridLines -> Automatic, Joined -> False, PolarTicks -> {"Degrees", Automatic}, BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, FontSize -> 12}, PlotStyle -> {Red, PointSize -> 0.02}]

如您所见，直方图显示了它应该显示的旋转对称性。 我想尽一切办法把这些事情弄清楚，但没有成功。没有倒车，情况最糟。我试过旋转，但没有成功。我觉得问题出在我的箱子里。 ArcTan输出从-Pi到Pi，而Sjoerd建议我需要从0到2Pi。但是我不知道怎么做。

编辑：问题解决了。多亏了Sjoerd、Belisarius和Heike solutions，我能够在给定图像重心的情况下显示眼睛注视位置的直方图

---

正在检查，但您的第一个情节似乎有缺陷：

list = {{21, 16}, {16, 14}, {11, 11}, {11, 12}, {13, 15}, 
        {18, 17}, {19, 11}, {17, 16}, {16, 19}};
ScreenCenter = {20, 15};

Show[ListPlot[list, PlotStyle -> Directive[PointSize[Medium], Purple]], 
     Graphics[
              {Red, PointSize[Large], Point[ScreenCenter], 
               Circle[ScreenCenter, 10]}], 
AspectRatio -> 1, Axes -> False]

编辑

我没有完全遵循您的代码，但屏幕中心的反射似乎解决了问题：

Module[{Countz, maxScale, angleDivisions, dAng}, 
 Countz = BinCounts[
               {ArcTan[Sequence @@ ##]} & /@ (# + ScreenCenter & /@ -list), 
           {-Pi, Pi, Pi/6}];
 maxScale = 4;
 angleDivisions = 12;
 dAng = (2 Pi)/angleDivisions;

 SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 

  SectorOrigin -> {-Pi/angleDivisions, "Counterclockwise"}, 
  PolarAxes -> True, 
  PolarGridLines -> Automatic, 
  PolarTicks -> {Table[{i \[Degree] + Pi/angleDivisions, 
                        i \[Degree]}, {i, 0, 345, 30}], Automatic}, 
  ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Red]},
   BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, 
    FontSize -> 12}, 
   ImageSize -> 400]]

编辑

在这里，您可能会看到我的代码中的小偏差，这在Heike的回答中得到了解决（投赞成票！）

---

您可以使用

ChartElementFunction

选项精确定位扇区。

ChartElementFunction

的第一个参数的形式为

{{angleMin，angleMax}，{rMin，rMax}

。第一个扇区有边界

{angleMin，angleMax}={-Pi/12，Pi/12}

，第二个扇区有边界

{Pi/12，3pi/12}

，等等。因此，要获得正确的旋转，可以执行以下操作

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, 
  PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, 
      i \[Degree]}, {i, 0, 345, 30}], Automatic}, 
  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]]]]]]

---

@beliarius:当以

ArcTan[x，y]

的形式使用

ArcTan

时，范围是

（-Pi，Pi]

@belisarius，对不起，我现在实在是太离谱了：-（.谢谢你的帮助！@500请注意Heike的解决方案，因为它似乎与betterI的位置相匹配。我在你最初提问的地方给出了这个问题的答案。@Sjoerd也许我错了，但在我看来，你的极坐标直方图并不能代表点密度。也许缺少对称变换。（见你回答中的90-135度范围）我觉得自己很好的那一刻，我回来看到你的解决方案和belisarius的评论，谢谢你，我甚至没有注意到酒吧的位置，这太整洁了。

Module[{Countz, maxScale, angleDivisions, dAng}, 
 Countz = BinCounts[
               {ArcTan[Sequence @@ ##]} & /@ (# + ScreenCenter & /@ -list), 
           {-Pi, Pi, Pi/6}];
 maxScale = 4;
 angleDivisions = 12;
 dAng = (2 Pi)/angleDivisions;

 SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 

  SectorOrigin -> {-Pi/angleDivisions, "Counterclockwise"}, 
  PolarAxes -> True, 
  PolarGridLines -> Automatic, 
  PolarTicks -> {Table[{i \[Degree] + Pi/angleDivisions, 
                        i \[Degree]}, {i, 0, 345, 30}], Automatic}, 
  ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], Red]},
   BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, 
    FontSize -> 12}, 
   ImageSize -> 400]]

Show[Module[{Countz, maxScale, angleDivisions, dAng}, 
  Countz = BinCounts[{ArcTan[
        Sequence @@ ##]} & /@ (# + 
         ScreenCenter & /@ -list), {-\[Pi], \[Pi], \[Pi]/6}];
  maxScale = 4;
  angleDivisions = 12;
  dAng = (2 \[Pi])/angleDivisions;
  SectorChart[{ConstantArray[1, Length[Countz]], Countz}\[Transpose], 
   SectorOrigin -> {-\[Pi]/angleDivisions, "Counterclockwise"}, 
   PolarAxes -> True, PolarGridLines -> Automatic, 
   PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, 
       i \[Degree]}, {i, 0, 345, 30}], Automatic}, 
   ChartStyle -> {Directive[EdgeForm[{Black, Thickness[0.005]}], 
      Red]}, BaseStyle -> {FontFamily -> "Arial", FontWeight -> Bold, 
     FontSize -> 12}, ImageSize -> 400]],
 ListPlot[Plus[# - ScreenCenter] & /@ list/2.5, 
  PlotMarkers -> Image[CrossMatrix[10], ImageSize -> 10]]
 ]

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, 
  PolarTicks -> {Table[{i \[Degree] + \[Pi]/angleDivisions, 
      i \[Degree]}, {i, 0, 345, 30}], Automatic}, 
  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]]]]]]