如何得到ListPlot图像中的具体点？

Question

tpop = {{10, 211}, {98, 209}, {106, 208}, {113, 206}, {123,
    204}, {131, 199}, {138, 194}, {143, 188}, {150, 181}, {154,
    175}, {161, 167}, {168, 160}, {176, 155}, {182, 150}, {191,
    143}, {201, 139}, {207, 134}, {216, 130}, {227, 124}, {234,
    119}, {242, 113}, {249, 107}, {257, 101}, {264, 95}, {270,
    89}, {277, 81}, {283, 73}, {287, 64}, {290, 56}, {293, 46}, {296,
    34}, {297, 9}};

c = Table[{200, y}, {y, 0, 200, 0.01}];

ListLogLogPlot[{tpop, c}, Joined -> True]

请问如何得到两条线的交点？

Answer 1

使用线性拟合，下面的例子只考虑点集tpop中离直线x=200最近的两点，当然也可以取多个点用线性拟合,改一下n就好了：

(*运行代码前先输入tpop*)
n = 2;
y = Fit[tpop[[Flatten[
       FirstPosition[tpop[[;; , 1]], #] & /@ 
        Nearest[tpop[[;; , 1]], 200, n]]]], {1, x}, x] /. x -> 200;
yourPoint = {200, y}

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也可以使用多个附近的点，使用其它函数拟合（自己选）。比如取4个点或更多，使用3次多项式拟合：

(*运行代码前先输入tpop*)
n = 4;
y = Fit[tpop[[Flatten[
       FirstPosition[tpop[[;; , 1]], #] & /@ 
        Nearest[tpop[[;; , 1]], 200, n]]]], {1, x, x^2, x^3}, x] /. 
   x -> 200;
yourPoint = {200, y}

 

Comments

ListLogLogPlot的Join->True使用的仅仅是线性插值。例如
data = {{1, 1}, {2, 3}, {3, 8}, {4, 15}, {5, 100}};
x0 = Log@2.7;
y0 = Interpolation[Log@N@data, InterpolationOrder -> 1][x0];
ListLogLogPlot[data, Joined -> True, Mesh -> All,
 Epilog -> {Red, PointSize[0.02], Point[{x0, y0}]}]

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插值函数用挺好的。那只用第一种n=2的方法就行了。