r - 具有定量和定性解释变量之间相互作用的多元 Logistic 回归

Question

作为后续 this question ，我拟合了具有定量和定性解释变量之间相互作用的多元 Logistic 回归。 MWE如下:

Type  <- rep(x=LETTERS[1:3], each=5)
Conc  <- rep(x=seq(from=0, to=40, by=10), times=3)
Total <- 50
Kill  <- c(10, 30, 40, 45, 38, 5, 25, 35, 40, 32, 0, 32, 38, 47, 40)

df <- data.frame(Type, Conc, Total, Kill)

fm1 <- 
  glm(
      formula = Kill/Total~Type*Conc
    , family  = binomial(link="logit")
    , data    = df
    , weights = Total
    )

summary(fm1)

Call:
glm(formula = Kill/Total ~ Type * Conc, family = binomial(link = "logit"), 
    data = df, weights = Total)

Deviance Residuals: 
   Min      1Q  Median      3Q     Max  
-4.871  -2.864   1.204   1.706   2.934  

Coefficients:
            Estimate Std. Error z value Pr(>|z|)    
(Intercept) -0.65518    0.23557  -2.781  0.00541 ** 
TypeB       -0.34686    0.33677  -1.030  0.30302    
TypeC       -0.66230    0.35419  -1.870  0.06149 .  
Conc         0.07163    0.01152   6.218 5.04e-10 ***
TypeB:Conc  -0.01013    0.01554  -0.652  0.51457    
TypeC:Conc   0.03337    0.01788   1.866  0.06201 .  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 277.092  on 14  degrees of freedom
Residual deviance:  96.201  on  9  degrees of freedom
AIC: 163.24

Number of Fisher Scoring iterations: 5

anova(object=fm1, test="LRT")

Analysis of Deviance Table

Model: binomial, link: logit

Response: Kill/Total

Terms added sequentially (first to last)


          Df Deviance Resid. Df Resid. Dev Pr(>Chi)    
NULL                         14    277.092             
Type       2    6.196        12    270.895  0.04513 *  
Conc       1  167.684        11    103.211  < 2e-16 ***
Type:Conc  2    7.010         9     96.201  0.03005 *  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1


df$Pred <- predict(object=fm1, data=df, type="response")

df1 <- with(data=df,
               expand.grid(Type=levels(Type)
                           , Conc=seq(from=min(Conc), to=max(Conc), length=51)
                           )
      )
df1$Pred <- predict(object=fm1, newdata=df1, type="response")

library(ggplot2)
ggplot(data=df, mapping=aes(x=Conc, y=Kill/Total, color=Type)) + geom_point() +
  geom_line(data=df1, mapping=aes(x=Conc, y=Pred), linetype=2) +
  geom_hline(yintercept=0.5,col="gray")

我要计算 LD50 , LD90和 LD95与他们的置信区间。由于相互作用很重要，所以我想计算 LD50 , LD90和 LD95每个 Type (A, B, and C) 的置信区间分别地。

LD 代表 lethal dose.它是杀死 X% (LD50 = 50%) 测试群体所需的物质量。

已编辑 Type是代表不同类型药物的定性变量， Conc是代表不同药物浓度的定量变量。

Answer 1

您使用 drc适合逻辑剂量 react 模型的软件包。

先拟合模型

require(drc)
mod <- drm(Kill/Total ~ Conc, 
           curveid = Type, 
           weights = Total, 
           data = df, 
           fct =  L.4(fixed = c(NA, 0, 1, NA)), 
           type = 'binomial')

这里 curveid=指定数据的分组和 fct=指定一个 4 参数逻辑函数，下键和上键的参数固定为 0 和 1。

注意与 glm 的区别可以忽略不计:

df2 <- with(data=df,
            expand.grid(Conc=seq(from=min(Conc), to=max(Conc), length=51),
                        Type=levels(Type)))
df2$Pred <- predict(object=mod, newdata = df2)

这是与 glm 预测差异的直方图

hist(df2$Pred - df1$Pred)

从模型估计有效剂量(和 CI)

这很容易使用 ED()功能:

ED(mod, c(50, 90, 95), interval = 'delta')

Estimated effective doses
(Delta method-based confidence interval(s))

     Estimate Std. Error   Lower  Upper
A:50   9.1468     2.3257  4.5885 13.705
A:90  39.8216     4.3444 31.3068 48.336
A:95  50.2532     5.8773 38.7338 61.773
B:50  16.2936     2.2893 11.8067 20.780
B:90  52.0214     6.0556 40.1527 63.890
B:95  64.1714     8.0068 48.4784 79.864
C:50  12.5477     1.5568  9.4963 15.599
C:90  33.4740     2.7863 28.0129 38.935
C:95  40.5904     3.6006 33.5334 47.648

对于每组，我们使用 CI 获得 ED50、ED90 和 ED95。