3 
 X=c(-    0.000000446,0.000006498,0.00001344,0.00002039,0.00002733,0.00003427,0.00004122,0.00004816,0.00005511,0.00006205,0.000069,0.0000836,0.00009821,0.0001128,0.0001274,0.000142,0.0001566,0.0001713,0.0001859,0.0002005,0.0002151,0.0002297,0.0002443,0.0002589,0.0002735,0.0002881,0.0003027,0.0003173,0.000332,0.0003466,0.0003612,0.0003758,0.0003904,0.000405,0.0004196,0.0004342,0.0004488,0.0004634,0.000478,0.0004926,0.0005073,0.0005219,0.0005365,0.0005511,0.0005657,0.0005803,0.0005949,0.0006095,0.0006241,0.0006387,0.0006533,0.000668,0.0006826,0.0006972,0.0007118,0.0007264,0.000741,0.0007556,0.0007664)
 Y=c(0,61820,73630,83830,93720,103600,109000,112700,116100,119500,120900,122100,123300,124300,125300,126600,127800,129000,130100,131200,132300,133300,134200,135200,136100,136900,137800,138600,139400,140100,140900,141500,142200,142800,143500,144000,144600,145100,145600,146100,146600,147000,147400,147800,148200,148500,148800,149100,149300,149600,149800,149800,149900,149800,149700,149300,148500,148000,147600)
 plot(X,Y)
 linear1=lm(Y~X)
 linear2=segmented(linear1,seg.Z=~X, psi = 0.0005,data=d)
 plot(linear2)
 points(X,Y)
 slope(linear2)

我想要做的是得到这里给出的曲线的双线性表示。首先我想问是否有比我的代码更好的方法？第二个也是重要的事情是我不知道什么函数或命令显示线的交点和每条线的方程。所以问题是如何从这个分段命令中得到线方程。

4

2 回答 2

2

不清楚你所说bilinear的曲线表示是什么意思。要获得线方程，您可以使用intercept和slope这样的：

intercept(linear2)
$X
             Est.
intercept1  22730
intercept2 119200

> slope(linear2)
$X
            Est.   St.Err. t value CI(95%).l CI(95%).u
slope1 2.938e+09 288100000   10.20 2.360e+09 3.515e+09
slope2 4.872e+07   3840000   12.69 4.103e+07 5.642e+07
于 2013-07-28T01:09:09.707 回答
2

这是一个老问题，但我一直在寻找有关使用分段的信息，因为我想我明白你在问什么，所以可以提供一个快速的答案。两条曲线相交的“x”值是断点。使用您的示例：

线性2

调用：segmented.lm(obj = linear1, seg.Z = ~X, psi = 5e-04, data = d)

线性项的有意义系数：

(截取) X U1.X
2.997e+04 2.212e+09 -2.167e+09

估计断点 psi1.X : 4.215e-05

您的模型的线性方程组是：

当 x <= 断点时，y = slope1*x + intercept1

当 x > 断点时，y = slope2*x + intercept2

于 2015-02-25T20:02:21.003 回答