r - 使用 ggplot 绘制非线性回归列表

Question

作为此链接的非线性回归分析的输出图

https://stats.stackexchange.com/questions/209087/non-linear-regression-mixed-model

有了这个数据集:

zz <-(" iso temp diam
 Itiquira   22  5.0
 Itiquira   22  4.7
 Itiquira   22  5.4
 Itiquira   25  5.8
 Itiquira   25  5.4
 Itiquira   25  5.0
 Itiquira   28  4.9
 Itiquira   28  5.2
 Itiquira   28  5.2
 Itiquira   31  4.2
 Itiquira   31  4.0
 Itiquira   31  4.1
 Londrina   22  4.5
 Londrina   22  5.0
 Londrina   22  4.4
 Londrina   25  5.0
 Londrina   25  5.5
 Londrina   25  5.3
 Londrina   28  4.6
 Londrina   28  4.3
 Londrina   28  4.9
 Londrina   31  4.4
 Londrina   31  4.1
 Londrina   31  4.4
    Sinop   22  4.5
    Sinop   22  5.2
    Sinop   22  4.6
    Sinop   25  5.7
    Sinop   25  5.9
    Sinop   25  5.8
    Sinop   28  6.0
    Sinop   28  5.5
    Sinop   28  5.8
    Sinop   31  4.5
    Sinop   31  4.6
    Sinop   31  4.3"
)
df <- read.table(text=zz, header = TRUE)

这个拟合模型，带有四个参数:

thx:最佳温度

thy: 最佳直径

thq:曲率

thc:偏度

library(nlme) 

df <- groupedData(diam ~ temp | iso, data = df, order = FALSE) 

n0 <- nlsList(diam ~ thy * exp(thq * (temp - thx)^2 + thc * (temp - thx)^3),               
      data = df, 
      start = c(thy = 5.5, thq = -0.01, thx = 25, thc = -0.001))

> n0
# Call:
#  Model: diam ~ thy * exp(thq * (temp - thx)^2 + thc * (temp - thx)^3) | iso 

# Coefficients:
              thy          thq      thx           thc
# Itiquira 5.403118 -0.007258245 25.28318 -0.0002075323
# Londrina 5.298662 -0.018291649 24.40439  0.0020454476
# Sinop    5.949080 -0.012501783 26.44975 -0.0002945292

# Degrees of freedom: 36 total; 24 residual
# Residual standard error: 0.2661453

有没有办法在ggplot中绘制拟合值，比如smooth()的特定函数？



我想我找到了...(基于 http://rforbiochemists.blogspot.com.br/2015/06/plotting-two-enzyme-plots-with-ggplot.html)

ip <- ggplot(data=daf,  aes(x=temp, y=diam, colour = iso)) +  
  geom_point() + facet_wrap(~iso)

ip + geom_smooth(method = "nls", 
                method.args = list(formula = y ~ thy * exp(thq * (x-thx)^2 + thc * (x - thx)^3), 
                                   start = list(thy=5.4, thq=-0.01, thx=25, thc=0.0008)),
                se = F, size = 0.5, data = subset(daf, iso=="Itiquira")) +

  geom_smooth(method = "nls", 
              method.args = list(formula = y ~ thy * exp(thq * (x-thx)^2 + thc * (x - thx)^3), 
                                 start = list(thy=5.4, thq=-0.01, thx=25, thc=0.0008)),
              se = F, size = 0.5, data = subset(daf, iso=="Londrina")) +

  geom_smooth(method = "nls", 
              method.args = list(formula = y ~ thy * exp(thq * (x-thx)^2 + thc * (x - thx)^3), 
                                 start = list(thy=5.4, thq=-0.01, thx=25, thc=0.0008)),
              se = F, size = 0.5, data = subset(daf, iso=="Sinop")) 

Answer 1

用稍微有原则的方式回答这个问题 ggplot方法(将输出组合成一个单一的数据帧，其结构与原始数据的结构相匹配)。不幸的是在 nls 上找到置信区间预测并不那么容易(搜索涉及引导或增量方法的解决方案):

tempvec <- seq(22,30,length.out=51)
pp <- predict(n0,newdata=data.frame(temp=tempvec))
## combine predictions with info about species, temp
pdf <- data.frame(iso=names(pp),
                  temp=rep(tempvec,3),
                  diam=pp)

创建图形:

library(ggplot2)
ggplot(df,aes(temp,diam,colour=iso))+
  stat_sum()+
  geom_line(data=pdf)+
  facet_wrap(~iso)+
  theme_bw()+
  scale_size(range=c(1,4))+
  scale_colour_brewer(palette="Dark2")+
  theme(legend.position="none",
        panel.spacing=grid::unit(0,"lines"))