Question

我有一个quasibinomial glm，有两个连续的解释变量（让我们说“LogPesticide”和“LogFood”）和一个互动。我想在不同食物量（例如最小和最大食物价值）下以置信区间计算农药的LC50。如何实现这一目标？

示例：首先，我生成一个数据集。

jira = JIRA('http://jiraurl.com/', basic_auth=('user', 'pass'))

new_issue = jira.create_issue(project='ER', summary='summary', description='desc', issuetype={'name': 'Custom Issue Type'})

new_issue.update(reporter='new_user')

然后我适应了整个glm。模型诊断正常，所有交互术语都很重要。

mydata <- data.frame(
            LogPesticide = rep(log(c(0, 0.1, 0.2, 0.4, 0.8, 1.6) + 0.05), 4),
            LogFood = rep(log(c(1, 2, 4, 8)), each = 6)
          )

set.seed(seed=16) 

growth <- function(x, a = 1, K = 1, r = 1) {            # Logistic growth function. a = position of turning point
  Fx <- (K * exp(r * (x - a))) / (1 + exp(r * (x - a))) # K = carrying capacity
  return(Fx)                                            # r = growth rate (larger r -> narrower curve)
}

y <- rep(NA, length = length(mydata$LogPesticide))
y[mydata$LogFood == log(1)] <- growth(x = mydata$LogPesticide[mydata$LogFood == log(1)], a = log(0.1), K = 1, r = 6)
y[mydata$LogFood == log(2)] <- growth(x = mydata$LogPesticide[mydata$LogFood == log(2)], a = log(0.2), K = 1, r = 4)
y[mydata$LogFood == log(4)] <- growth(x = mydata$LogPesticide[mydata$LogFood == log(4)], a = log(0.4), K = 1, r = 3)
y[mydata$LogFood == log(8)] <- growth(x = mydata$LogPesticide[mydata$LogFood == log(8)], a = log(0.8), K = 1, r = 1)
mydata$Dead <- rbinom(n = length(y), size = 20, prob = y)
mydata$Alive <- 20 - mydata$Dead
mydata$Mortality <- cbind(mydata$Dead, mydata$Alive)

我尝试用MASS包中的dose.p（）估算LC50。如果LogFood是一个因素，当我按照this post中的讨论重新拟合模型时，这将起作用。但是有两个连续的解释变量，你只得到1个截距，2个斜率和斜率的差异（用于交互）。

我可以使用effect（）估算LC50，但不知道如何获取LogPesticide的关联CI。

model <- glm(Mortality ~ LogPesticide * LogFood, family = quasibinomial, data = mydata)
plot(model)
Anova(model)
summary(model)

从dose.p（）的代码我看到必须使用vcov矩阵。 effect（）还提供了一个vcov矩阵，但我无法修改dose.p（）以正确处理该信息。我会很感激任何想法！

Answer 1

复制数据（更新：ggplot2的新版本可能不喜欢带有矩阵的奇怪数据框？）

mydata <- data.frame(
        LogPesticide = rep(log(c(0, 0.1, 0.2, 0.4, 0.8, 1.6) + 0.05), 4),
        LogFood = rep(log(c(1, 2, 4, 8)), each = 6)
      )
set.seed(seed=16) 

growth <- function(x, a = 1, K = 1, r = 1) {
    ## Logistic growth function. a = position of turning point
    ## K = carrying capacity
    ## r = growth rate (larger r -> narrower curve)
    return((K * exp(r * (x - a))) / (1 + exp(r * (x - a))))
}

rlf <- data.frame(LogFood=log(c(1,2,4,8)),
                              a=log(c(0.1,0.2,0.4,0.8)),
                              r=6,4,3,1)
mydata <- merge(mydata,rlf)
mydata <- plyr::mutate(mydata,
               y=growth(LogPesticide,a,K=1,r),
               Dead=rbinom(n=nrow(mydata),size=20,prob=y),
               N=20,
               Alive=N-Dead,
               pmort=Dead/N)


model <- glm(pmort ~ LogPesticide * LogFood, family = quasibinomial,
          data = mydata, weights=N)

为方便起见：

cc <- setNames(coef(model),c("b_int","b_P","b_F","b_PF"))
vv <- vcov(model)
dimnames(vv) <- list(names(cc),names(cc))

基本预测数据框：

pframe <- with(mydata,
         expand.grid(LogPesticide=seq(min(LogPesticide),max(LogPesticide),
                      length=51),
                     LogFood=unique(LogFood)))
pframe$pmort <- predict(model,newdata=pframe,type="response")

现在让我们打破这个。给定水平的（log）食物F和（log）农药P的预测值是

logit(surv) = b_int + b_P*P + b_F*F + b_PF*F*P

因此，F级农药的逻辑曲线是

logit(surv) = (b_int+b_F*F) + (b_P+b_PF*F)*P

我们想知道logit（surv）为0（LC50）的P的值，所以我们需要

0 = (b_int+b_F*F) + (b_P+b_PF*F)*P50
P50 = -(b_int+b_F*F)/(b_P+b_PF*F)

翻译代码：

P50mean <- function(logF) {
    with(as.list(cc), -(b_int+b_F*logF)/(b_P+b_PF*logF))
}
with(mydata,P50mean(c(min=min(LogFood),max=max(LogFood))))


pLC50 <- data.frame(LogFood=unique(mydata$LogFood))
pLC50 <- transform(pLC50,
               pmort=0.5,
               LogPesticide=P50mean(LogFood))

为了获得置信区间，两种最简单的方法是（1）delta方法和（2）后验预测区间（在某些情况下也称为参数Bayes＆＃39;）。（您也可以进行非参数自举。）

Delta方法

我试图手工做到这一点，但意识到它太毛茸茸了（所有四个系数都是强相关的，所有这些相关性都必须在计算中保持跟踪 - 它并不像分子和分母是独立值的通常公式......）

library("emdbook")
deltavar(-(b_int+b_F*2)/(b_P+b_PF*2),meanval=cc,Sigma=vv)
## have to be a bit fancy here with eval/substitute ...
pLC50$var1 <- sapply(pLC50$LogFood,
            function(logF)
                 eval(substitute(
                     deltavar(-(b_int+b_F*logF)/(b_P+b_PF*logF),
                               meanval=cc,Sigma=vv),
                     list(logF=logF))))

人口预测间隔

这假设（稍微弱一点）参数的采样分布是多元正态。

PP <- function(logF,n=1000) {
    b <- MASS::mvrnorm(n,mu=cc,Sigma=vv)
    pred <- with(as.data.frame(b),
         -(b_int+b_F*logF)/(b_P+b_PF*logF))
    return(var(pred))
}
set.seed(101)
pLC50$var2 <- sapply(pLC50$LogFood,PP)

PPI实际上允许我们通过获得预测LC50的分布的分位数来稍微放宽假设......事实证明（见下文）基于PPI的置信区间比Delta更宽一点方法的，但他们并没有相差甚远。

现在描绘整个混乱：

library(ggplot2); theme_set(theme_bw())
gg0 <- ggplot(mydata,aes(LogPesticide,pmort,
              colour=factor(LogFood),
              fill = factor(LogFood))) + geom_point() +
       ## individual fits -- a bit ugly
       ##       geom_smooth(method="glm",aes(weight=N),
       ##           method.args=list(family=binomial),alpha=0.1)+
       geom_line(data=pframe,linetype=2)+
       geom_point(data=pLC50,pch=5,size=4)+
       geom_hline(yintercept=0.5,col="gray")

 gg0 + geom_errorbarh(data=pLC50,lwd=2,alpha=0.5,
                       aes(xmin=LogPesticide-1.96*sqrt(var1),
                           xmax=LogPesticide+1.96*sqrt(var1)),
                       height=0)+
       geom_errorbarh(data=pLC50,
                       aes(xmin=LogPesticide-1.96*sqrt(var2),
                           xmax=LogPesticide+1.96*sqrt(var2)),
                      height=0.02)

来自多重回归glm的LC50 / LD50置信区间与相互作用

1 个答案:

Delta方法

人口预测间隔