4

请考虑以下分布:

rs={{400, 0.00929}, {410, 0.0348}, {420, 0.0966}, {430, 0.2}, {440, 0.328}, {450, 0.455}, 
    {460, 0.567}, {470, 0.676}, {480, 0.793}, {490, 0.904}, {500, 0.982}, {510, 0.997}, 
    {520,0.935}, {530, 0.811}, {540, 0.65}, {550, 0.481}, {560, 0.329}, {570,0.208}, 
    {580, 0.121}, {590, 0.0655}, {600, 0.0332}, {610, 0.0159}, {620, 0.00737}, 
    {630, 0.00334}, {640, 0.0015}, {650,0.000677}, {660, 0.000313}, {670, 0.000148}, 
    {680, 0.0000715}, {690,0.0000353}, {700, 0.0000178}}

在此处输入图像描述

如何插入此分布以获得 X 轴上任何位置的点的值?

4

3 回答 3

7

只需使用标准Interpolation功能:

rsInterpolation = Interpolation@rs;
Plot[rsInterpolation@x, {x, 400, 700}]

结果

如果您想拟合特定类别的函数(例如正态分布),请改用FindFit.

于 2011-10-10T19:36:59.737 回答
5

如果你需要好的衍生品,你可以这样做:

interp = Interpolation[rs, InterpolationOrder -> 3, Method -> "Spline"]
Show[Plot[{interp[x], 10 interp'[x]}, {x, Min[First /@ rs], Max[First /@ rs]},
          PlotRange -> Full],
     ListPlot@rs]

在此处输入图像描述

查看使用“样条”方法时导数行为的差异:

interp  = Interpolation[rs, InterpolationOrder -> 3, Method -> "Spline"]
interp1 = Interpolation[rs, InterpolationOrder -> 3]
Show[Plot[{interp1'[x], interp'[x] - .005}, 
          {x, Min[First /@ rs], Max[First /@ rs]}, PlotRange -> Full]]

在此处输入图像描述

于 2011-10-11T04:20:29.930 回答
0

如果是发行版,我认为您应该改用SmoothKernelDistribution

于 2012-04-10T07:09:41.003 回答