ggplot2 中的半正态图残差图

Question

我正在尝试为 ggplot2 中的半正态 plot 中的残差制作下面显示的图表。

但是，我遇到了一个错误，标题为 Error: data must be a data frame, or other object coercible by fortify () ，而不是 S3 object with class hnp。

数据可以在下面看到。

library(hnp)
Stones <- c(rep("Stone1",1), rep("Stone2",1), rep("Stone3",1))
treat <- c(rep("T6",9), rep("T7",9), rep("T8",9), 
           rep("T10",9), rep("T11",9),
           rep("T12",9),rep("T14",9),rep("T15",9),
           rep("T16",9))
Samples <- c(rep("SampleI",3), rep("SampleF",3), rep("SampleR",3))
Count <- c(3,3,1,1,2,2,0,0,0,1,2,3,0,0,1,0,0,0,2,0,2,2,0,0,0,0,0,6,
           1,0,2,2,2,0,0,0,4,4,1,3,2,5,0,0,0,3,4,4,5,1,2,0,0,0,
           4,4,6,1,2,1,0,0,0,4,6,4,2,3,1,0,0,0,3,2,2,1,0,5,0,1,0)
Total <- c(rep("6", 81))
dados <- data.frame(treat, Stones, Samples, Count, Total);dados
dados$treat = as.factor(dados$treat)
dados$Samples = as.factor(dados$Samples)
dados$Total = as.numeric(dados$Total)
dados$Prop = dados$Count/dados$Total
dados16 = dados[dados$treat=="T16",];dados16`

model 调整如下所示。

resp16 <-cbind(dados16$Count,dados16$Total - dados16$Count); resp16
m16 <- glm(resp16 ~ Samples,
           data = dados16,
           family = quasibinomial)

Graph16=hnp(m16, xlab = 'Percentil da N(0,1)', ylab = 'Resíduos', 
    main = 'Gráfico Normal de Probabilidades')

错误就在这里。

ggplot(data = Graph16)
Erro: `data` must be a data frame, or other object coercible by `fortify()`, not an S3 object with class hnp

Answer 1

如错误所示，ggplot2 需要一个data.frame object 或一个 object ，为此编写了一个fortify() S3 方法。 The hnp class object doesn't have a fortify method, so probably our best option is to get the data out of the Graph16 object and put it in a dataframe ourselves before plotting. 下面的示例假设Graph16 object 与您的代码生成的相同。

library(ggplot2)

df <- data.frame(
  x = Graph16$x,
  lower = Graph16$lower,
  median = Graph16$median,
  upper = Graph16$upper,
  residuals = Graph16$residuals
)

ggplot(df, aes(x)) +
  geom_ribbon(aes(ymin = lower, ymax = upper),
              alpha = 0.5) +
  geom_point(aes(y = residuals)) +
  geom_line(aes(y = median))