如何在 R 中绘制 logistic glm 预测值和置信区间

How to plot logistic glm predicted values and confidence interval in R

我有一个 presence/absence 响应变量的二项式 glm 和一个具有 9 个水平的因子变量,如下所示:

data$y<-factor(data$y,levels=c(0,1),labels=c("absent","present"))
table(data$y,data$site_name)

          Andulay Antulang Basak Dauin Poblacion District 1 Guinsuan Kookoo's Nest Lutoban Pier Lutoban South Malatapay Pier
  absent        4        4     1                          0        3             1            5             5              2
  present       2        2     5                          6        3             5            1             1              4

model <- glm(y~site_name,data=data,binomial)

为了简洁起见,我只是跳过模型推理和验证,我如何在箱线图中绘制每个站点获得 "present" 的概率及其置信区间?我想要的是 Plot predicted probabilities and confidence intervals in R 中显示的内容,但我想用箱线图显示它,因为我的回归变量 site_name 是一个具有 9 个级别的因子,而不是连续变量。

我想我可以按如下方式计算必要的值(但我不能 100% 确定正确性):

将模型系数转换回成功概率的函数:

calc_val <- function(x){return(round(1/(1+1/(exp(x))),3))}

基于模型的预测概率:

prob <- tapply(predict(model,type="response"),data$site_name,function(x){round(mean(x),3)})
means <- as.data.frame(prob)

预测概率的 75% 和 95% 置信区间:

ci <- cbind(confint(model,level=0.9),confint(model,level=0.5))
rownames(ci) <- gsub("site_name","",rownames(ci))
ci <- t(apply(ci,1,calc_val))

将所有内容合二为一table

ci<-cbind(means,ci)
ci
                            prob   5 %  95 %  25 %  75 %   Pr(>|z|) stderr
Andulay                    0.333 0.091 0.663 0.214 0.469 0.42349216  0.192
Antulang                   0.333 0.112 0.888 0.304 0.696 1.00000000  0.192
Basak                      0.833 0.548 0.993 0.802 0.964 0.09916496  0.152
Dauin Poblacion District 1 1.000 0.000    NA 0.000 1.000 0.99097988  0.000
Guinsuan                   0.500 0.223 0.940 0.474 0.819 0.56032414  0.204
Kookoo's Nest              0.833 0.548 0.993 0.802 0.964 0.09916496  0.152
Lutoban Pier               0.167 0.028 0.788 0.130 0.501 0.51171512  0.152
Lutoban South              0.167 0.028 0.788 0.130 0.501 0.51171512  0.152
Malatapay Pier             0.667 0.364 0.972 0.640 0.903 0.25767454  0.192

所以我的问题是双重的:

  1. 概率和置信区间的计算是否正确?
  2. 如何在 bloxplot(盒须图)中绘制它?

EDIT 下面是一些来自 dput 的示例数据(它还修改了上面的 tables 以匹配数据):

# dput(data[c("y", "site_name")])
data <- structure(list(y = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L), .Label = c("absent", "present"), class = "factor"), site_name = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 9L, 9L, 9L, 9L, 9L, 9L, 4L, 4L, 4L, 4L, 4L, 4L, 8L, 8L, 8L, 8L, 8L, 8L, 7L, 7L, 7L, 7L, 7L, 7L, 5L, 5L, 5L, 5L, 5L, 5L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("Andulay", "Antulang", "Basak", "Dauin Poblacion District 1", "Guinsuan", "Kookoo's Nest", "Lutoban Pier", "Lutoban South", "Malatapay Pier"), class = "factor")), .Names = c("y", "site_name"), row.names = c(125L, 123L, 126L, 124L, 128L, 127L, 154L, 159L, 157L, 158L, 156L, 155L, 111L, 114L, 116L, 115L, 112L, 113L, 152L, 151L, 148L, 150L, 153L, 149L, 143L, 146L, 144L, 147L, 142L, 145L, 164L, 165L, 161L, 163L, 160L, 162L, 120L, 122L, 121L, 117L, 118L, 119L, 137L, 136L, 139L, 141L, 140L, 138L, 129L, 134L, 131L, 135L, 133L, 130L), class = "data.frame")
#

这是一个最小公分母、仅基础包的解决方案。

拟合模型:

mm <- glm(y~site_name,data=dd,family=binomial)

用站点名称组成预测框:

pframe <- data.frame(site_name=unique(dd$site_name))

预测(在 logit/linear-predictor 范围内),标准误差

pp <- predict(mm,newdata=pframe,se.fit=TRUE)
linkinv <- family(mm)$linkinv ## inverse-link function

将预测、下限和上限放在一起,并反向转换为概率尺度:

pframe$pred0 <- pp$fit
pframe$pred <- linkinv(pp$fit)
alpha <- 0.95
sc <- abs(qnorm((1-alpha)/2))  ## Normal approx. to likelihood
alpha2 <- 0.5
sc2 <- abs(qnorm((1-alpha2)/2))  ## Normal approx. to likelihood
pframe <- transform(pframe,
                    lwr=linkinv(pred0-sc*pp$se.fit),
                    upr=linkinv(pred0+sc*pp$se.fit),
                    lwr2=linkinv(pred0-sc2*pp$se.fit),
                    upr2=linkinv(pred0+sc2*pp$se.fit))

情节。

with(pframe,
{
    plot(site_name,pred,ylim=c(0,1))
    arrows(as.numeric(site_name),lwr,as.numeric(site_name),upr,
           angle=90,code=3,length=0.1)
})

作为箱线图:

with(pframe,
{
    bxp(list(stats=rbind(lwr,lwr2,pred,upr2,upr),
             n = rep(1,nrow(pframe)),
             conf = NA,
             out = NULL,
             group = NULL,
             names=as.character(site_name)))
})

还有很多其他方法可以做到这一点;我会推荐

library("ggplot2")
ggplot(pframe,aes(site_name,pred))+
     geom_pointrange(aes(ymin=lwr,ymax=upr))+
     geom_linerange(aes(ymin=lwr2,ymax=upr2),lwd=1.5)+
     coord_flip()

另一种解决方案是通过y~site_name-1拟合模型,在这种情况下将为每个站点的概率分配一个单独的参数,并使用profile()/confint()来找到置信区间;这将比上面答案中依赖 parameters/predictions 的采样分布的正态性更准确。