使用 ggplot 绘制非线性回归列表

Question

作为非线性回归分析的输出图 link

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

使用此数据集：

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，就像 smooth() 的特定函数一样？

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

用一种稍微更有原则的 ggplot 方法来回答这个问题（将输出组合成一个结构与原始数据相匹配的数据框）。不幸的是，找到 nls 预测的置信区间并不那么容易（搜索涉及自举或 delta 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)

创建图表：

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"))

使用 ggplot 绘制非线性回归列表

Plotting a list of non linear regressions with ggplot

r

list

ggplot2

nlme