定量和定性解释变量之间相互作用的多元逻辑回归 [英] Multiple Logistic Regression with Interaction between Quantitative and Qualitative Explanatory Variables

问题描述

作为对该问题的后续措施，我将多元Logistic回归与定量解释性定性变量和定性解释性变量之间的交互作用进行了拟合. MWE如下:

As a follow up to this question, I fitted the Multiple Logistic Regression with Interaction between Quantitative and Qualitative Explanatory Variables. MWE is given below:

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")

我想用其置信区间计算LD50，LD90和LD95.由于交互作用很重要，因此我想分别计算每个Type (A, B, and C)的LD50，LD90和LD95及其置信区间.

I want to calculate LD50, LD90 and LD95 with their confidence intervals. As the interaction is significant so I want to calculate LD50, LD90 and LD95 with their confidence intervals for each Type (A, B, and C) separately.

LD 代表致命剂量.这是所需物质的量.杀死X％(LD50 = 50％)的测试人群.

LD stands for lethal dose. It is the amount of substance required to kill X% (LD50 = 50%) of the test population.

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.

Edited Type is a qualitative variable representing different types of drugs and Conc is a quantitative variable representing different Concentrations of drugs.

推荐答案

您使用drc软件包来拟合逻辑剂量反应模型.

You use the drc package to fit logistic dose-response models.

首先拟合模型

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

此处curveid=指定数据分组，并且fct=指定4参数逻辑函数，其上下键的参数固定为0和1.

Here curveid=specifies the grouping of the data and fct= specifies a 4 parameter logistic function, with parameters for lower and upper bond fixed at 0 and 1.

请注意，与glm的区别可以忽略不计:

Note the differences to glm are negligible:

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glm预测差异的直方图

Here's a histgramm of the differences to the glm prediction

hist(df2$Pred - df1$Pred)

从模型中估算有效剂量(和CI)

使用ED()函数很容易:

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

对于每个组，我们获得ED50，ED90和带有CI的ED95.

For each group we get ED50, ED90 & ED95 with CI.

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