相关疑难解决方法(0)

我无法理解这一点,这更随机？

rand()

要么

rand() * rand()

我发现它是一个真正的脑筋急转弯,你能帮助我吗？

编辑:

直观地,我知道数学答案将是它们同样随机,但我不禁想到,如果你将两者相乘的话"运行随机数算法"两次,你会创造一些比随机更随机的东西它一次.

基于Sjoerd,使用Mathematica从笛卡尔图到极坐标图的优秀解决方案和扩展,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}]

Module[{Countz, maxScale, angleDivisions, dAng},
        Countz = Reverse[BinCounts[Flatten@Map[ArcTan[#[[1]] - ScreenCenter[[1]], #[[2]] - 
                 ScreenCenter[[2]]] &, …

考虑:

ListPlot[Range[10], 
         Background -> Gray, 
         PlotLabel -> "I don`t want the background here !"]

有没有办法让背景仅适用于实际的绘图区？

不在轴上,不在标签后面.那么基本上就是那个矩形{{0,0},{10,10}}？

编辑:我们可以使用PolarListPlot做同样的事情吗？

使用Mathematica在笛卡尔图上使用Sjoerd解决方案 到极坐标直方图:

dalist = {{21, 22}, {26, 13}, {32, 17}, {31, 11}, {30, 9}, 
          {25,12}, {12, 16}, {18, 20}, {13, 23}, {19, 21}, 
          {14, 16}, {14,22}, {18, 22}, {10, 22}, {17, 23}}

ScreenCenter = {20, 15}

ListPolarPlot[{ArcTan[##],EuclideanDistance[##]} & @@@ (# - ScreenCenter & /@ dalist), 
               PolarAxes -> True, PolarGridLines -> Automatic, Joined -> False, 
               PolarTicks -> {"Degrees", Automatic}, 
               BaseStyle -> …