use*_*441 3 r curve-fitting model-fitting
我有两个变量的散点图,例如:
x<-c(0.108,0.111,0.113,0.116,0.118,0.121,0.123,0.126,0.128,0.131,0.133,0.136)
y<-c(-6.908,-6.620,-5.681,-5.165,-4.690,-4.646,-3.979,-3.755,-3.564,-3.558,-3.272,-3.073)
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我想找到更适合这两个变量之间关系的函数.
准确地说,我想拟合比较三种模式:linear,exponential和logarithmic.
我正在考虑将每个函数拟合到我的值,计算每种情况下的可能性并比较AIC值.
但我真的不知道如何或从哪里开始.任何可能的帮助将非常感激.
非常感谢你提前.
蒂娜.
我会从一个解释性的情节开始,像这样:
x<-c(0.108,0.111,0.113,0.116,0.118,0.121,0.123,0.126,0.128,0.131,0.133,0.136)
y<-c(-6.908,-6.620,-5.681,-5.165,-4.690,-4.646,-3.979,-3.755,-3.564,-3.558,-3.272,-3.073)
dat <- data.frame(y=y,x=x)
library(latticeExtra)
library(grid)
xyplot(y ~ x,data=dat,par.settings = ggplot2like(),
panel = function(x,y,...){
panel.xyplot(x,y,...)
})+
layer(panel.smoother(y ~ x, method = "lm"), style =1)+ ## linear
layer(panel.smoother(y ~ poly(x, 3), method = "lm"), style = 2)+ ## cubic
layer(panel.smoother(y ~ x, span = 0.9),style=3) + ### loeess
layer(panel.smoother(y ~ log(x), method = "lm"), style = 4) ## log
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看起来你需要一个立方模型.
summary(lm(y~poly(x,3),data=dat))
Residual standard error: 0.1966 on 8 degrees of freedom
Multiple R-squared: 0.9831, Adjusted R-squared: 0.9767
F-statistic: 154.8 on 3 and 8 DF, p-value: 2.013e-07
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以下是比较五种模型的示例.由于前两个模型的形式,我们可以lm用来获得良好的起始值.(请注意,y不应比较使用不同变换的模型,因此我们不应该使用lm1和lm2作为比较模型,而只能用于起始值.)现在nls为前两个中的每一个运行一个.在这两个模型之后,我们尝试了不同程度的多项式x.幸运的是lm,nls使用一致的AIC定义(尽管其他R模型拟合函数不一定正确AIC定义),因此我们可以只使用lm多项式.最后,我们绘制前两个模型的数据和拟合.
较低的AIC的更好,所以nls1就是最好的,然后lm3.2继nls2.
lm1 <- lm(1/y ~ x)
nls1 <- nls(y ~ 1/(a + b*x), start = setNames(coef(lm1), c("a", "b")))
AIC(nls1) # -2.390924
lm2 <- lm(1/y ~ log(x))
nls2 <- nls(y ~ 1/(a + b*log(x)), start = setNames(coef(lm2), c("a", "b")))
AIC(nls2) # -1.29101
lm3.1 <- lm(y ~ x)
AIC(lm3.1) # 13.43161
lm3.2 <- lm(y ~ poly(x, 2))
AIC(lm3.2) # -1.525982
lm3.3 <- lm(y ~ poly(x, 3))
AIC(lm3.3) # 0.1498972
plot(y ~ x)
lines(fitted(nls1) ~ x, lty = 1) # solid line
lines(fitted(nls2) ~ x, lty = 2) # dashed line
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添加了一些模型,然后修复它们并更改了符号.另外,为了跟进Ben Bolker的评论,我们可以用AICcmodavg包替换AIC上面的任何地方AICc.