R中的指数回归

问题描述

我有些点看起来像对数曲线.我试图获得的曲线如下所示:y = a * exp(-b * x)+ c

I have some points that look like a logarithmic curve. The curve that I'm trying to obtain look like: y = a * exp(-b*x) + c

我的代码:

x <- c(1.564379666,1.924250092,2.041559879,2.198696382,2.541267447,2.666400433,2.922534874,2.965726615,3.009969443,3.248480245,3.32927682,3.371404563,3.423759668,3.713001284,3.841419166,3.847632349,3.947993339,4.024541136,4.030779671,4.118849343,4.154008445,4.284232251,4.491359108,4.585182188,4.643299476,4.643299476,4.643299476,4.684369939,4.84424144,4.867973977,5.144490521,5.324298915,5.324298915,5.988637637,6.146599422,6.674937463)
y <- c(25600,23800,11990,14900,15400,19000,9850,7500,10000,12500,11400,8950,10900,3600,11500,9990,4000,3500,4000,3000,8000,5500,6000,7900,2800,2800,2800,2950,4990,4999,3500,6001,6000,1100,1200,6000)

df <- data.frame(x, y)
m <- nls(y ~ I(a*exp(-b*x)+c), data=df, start=list(a=max(y), b=0, c=10), trace=T)

输出:

Error en nlsModel(formula, mf, start, wts) :
    singular gradient matrix at initial parameter estimates

我在做什么错了?

推荐答案

我认为b = 0时，nls无法计算相对于a和c的梯度，因为b = 0从等式中去除了x.从不同的b值开始(糟糕，我看到上面的注释).这是示例...

I think that with b = 0, then nls can't calculate the gradient with respect to a and c since b=0 removes x from the equation. Start with a different value of b (oops, I see the comment above). Here's the example...

m <- nls(y ~ I(a*exp(-b*x)+c), data=df, start=list(a=max(y), b=1, c=10), trace=T)
y_est<-predict(m,df$x)
plot(x,y)
lines(x,y_est)
summary(m)

Formula: y ~ I(a * exp(-b * x) + c)

Parameters:
   Estimate Std. Error t value Pr(>|t|)
a 6.519e+04  1.761e+04   3.702 0.000776 ***
b 6.646e-01  1.682e-01   3.952 0.000385 ***
c 1.896e+03  1.834e+03   1.034 0.308688
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2863 on 33 degrees of freedom

Number of iterations to convergence: 5
Achieved convergence tolerance: 5.832e-06

这篇关于R中的指数回归的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！