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

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)

而这个适合的模型,白色四个参数:

thx:最佳温度

你的:最佳直径

thq:Curvature

thc:偏斜

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？

我想我发现了......(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

用稍微更有原则性的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单个数据帧）。不幸的是，找到预测的置信区间nls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"))