如何使用 predict 或 inv.logit 正确绘制 logistic 函数？

Question

我有一些观察结果，我用这些观察结果来确定基于不同浓度化学物质的死亡率。我根据它们背后的观察数量对这些比率进行了加权，并将它们拟合到 glm（二项式（link=logit））模型中。我一直在尝试在 ggplot 中显示该模型的图，包括原始观察结果（大小 = 重量）、模型拟合线和置信区间，但没有成功。我可以获得一个简单的 plot() 来工作，但是我无法显示我需要的其他图形。有任何想法吗？提前致谢！！！

#data:
C <- data.frame("region" = c("r29","r31","r2325","r25","r2526", "r26"),"conc" = c(755.3189,1689.6680,1781.8450,1902.8830,2052.1133,4248.7832),"nr_dead" = c(1,1,18,44,170,27), "nr_survived" = c(2,3,29,1370,1910,107),"death_rate" = c(0.33333333,0.25000000,0.38297872,0.03111740,0.08173077
,0.20149254))
C$tot_obsv <- (C$nr_survived+C$nr_dead)
#glm model:
C_glm <- glm(cbind(nr_dead, nr_survived) ~ conc, data = C, family = "binomial")
#ggplot line is incorrect:
ggplot(C_glm, aes(C$conc,C$death_rate, size = C$tot_obsv)) + coord_cartesian(ylim = c(0, 0.5)) + theme_bw() + geom_point() + geom_smooth(method = "glm", mapping = aes(weight = C$tot_obsv))

#correct plot of inv.logit = logistic function (1/(1+exp(-x)))
plot(inv.logit(-3.797+0.0005751*(0:6700)))

#using predict function works, but doesn't display confidence interval or nice point sizes:
x_conc <-seq (750, 6700, 1)
y_death_rate <- predict.glm(C_glm, list(conc=x_conc), type="response")
plot(C$conc, C$death_rate, pch = 10, lwd = 3, cex = C$tot_obsv/300, ylim = c(0, 0.5), xlim = c(0,7000), xlab = "conc", ylab = "death rate")
lines(x_conc, y_death_rate, col = "red", lwd = 2)

基本上，我正在尝试使用 ggplot 绘制 glm 预测逻辑曲线、观察权重和置信区间，但只能使用 plot() 使曲线正确显示。

Answer 1

您可以在 predict 函数中包含回归变量的实现值，然后仅使用来自回归变量实现的值的数据添加 geom_point。如果您使用 size = tot_obs 它只会绘制该列不丢失的点，即仅 C.

中的值

tibble(
  x_conc = c(seq(750, 6700, 1), C$conc), 
  y_death_rate = predict.glm(C_glm, list(conc = x_conc), type = "response")
  ) %>% 
  left_join(C, by = c('x_conc' = 'conc')) %>% 
  ggplot(aes(x = x_conc, y = y_death_rate)) +
    geom_line(aes(size = 0.8)) +
    geom_point(aes(y = death_rate, size = tot_obsv))

Answer 2

建立在 @IceCreamToucan's 答案上

tibble(
  x_conc = c(seq(750, 6700, 1), C$conc), 
  y_death_rate = predict.glm(C_glm, list(conc = x_conc), type = "response")
  ) %>% 
  left_join(C, by = c('x_conc' = 'conc')) %>% 
  ggplot(aes(x = x_conc, y = y_death_rate)) +
    #geom_line(aes(size = 0.8)) + commented out as binomial smooth does this
    geom_point(aes(y = death_rate, size = tot_obsv)) + binomial_smooth()

当然我们需要定义函数binomial_smooth 这是取自：https://ggplot2.tidyverse.org/reference/geom_smooth.html

binomial_smooth <- function(...) {
    geom_smooth(method = "glm", method.args = list(family = "binomial"), ...)
}

如何使用 predict 或 inv.logit 正确绘制 logistic 函数？

How can I ggplot a logistic function correctly using predict or inv.logit?

r

predict

ggplot2

glm