Posthoc测试使用R中的multcomp嵌套方差分析

Question

我的数据嵌套在组级别。有五种不同的治疗方法。在每种治疗中，将四个参与者分组。它是关于竞争下参与者的捐赠行为（因变量=捐赠，度量，以€为单位）（解释性变量=治疗，顺序）。数据结构是这样的：

Treatment   Session   player.cumulative_donation:
CG             uk4rlbdo         2.5
CG             uk4rlbdo         1.4 
CG             uk4rlbdo         0
CG             uk4rlbdo         1
CG             dg0bqvit         0
CG             dg0bqvit         0
CG             dg0bqvit         0.5
CG             dg0bqvit         0
TG1            g6n3z46r         1
TG1            g6n3z46r         0
TG1            g6n3z46r         0
TG1            g6n3z46r         0.2

基于Rcompanion计算方差分析后，我想使用multcomp函数执行Posthoc测试。

但是，如果我跑步

library(multcomp)

posthoc = glht(model,
               linfct = mcp(Treatment="Tukey"))

我收到了我不理解的错误消息

Error in model.frame.lme(object) : object does not contain any data
Error in factor_contrasts(model) : 
  no ‘model.matrix’ method for ‘model’ found!

模型中存储了数据：

> model
Linear mixed-effects model fit by REML
  Data: NULL 
  Log-restricted-likelihood: -166.8703
  Fixed: Donation ~ Treatment 
 (Intercept) TreatmentTG1 TreatmentTG2 TreatmentTG3 TreatmentTG4 
   0.7492227    1.3343727    0.2981268    1.4943010    0.5274175 

Random effects:
 Formula: ~1 | Session
        (Intercept) Residual
StdDev:   0.1759392 1.651152

Number of Observations: 88
Number of Groups: 27

变量是：

$ player.cumulative_donation: num  2.5 1.4 0 1 0 0 0.5 0 1 0 ...
$ player.treatmentgroup     : chr  "CG" "CG" "CG" "CG" ...
$ Session code              : chr  "uk4rlbdo" "uk4rlbdo" "uk4rlbdo" "uk4rlbdo" ...

编辑：R命令创建模型：

library(nlme)

model = lme(Donation ~ Treatment, random=~1|Session,
            method="REML")

anova.lme(model,
          type="sequential",
          adjustSigma = FALSE)

Answer 1

如果数据在环境中是变量，则回归有效，但是对于下游分析，他们要求将其存储为lme对象内的data.frame：

例如，这很好用

library(nlme)
library(multcomp)
df = data.frame(Treatment=sample(c("TG1","TG2","TG3"),100,replace=TRUE),
                Donation=rnorm(100),
                Session = sample(c("uk4rlbdo","dg0bqvit"),100,replace=TRUE))
model = lme(Donation ~ Treatment, random=~1|Session,data=df)
glht(model,linfct=mcp(Treatment="Tukey"))

而当您将变量放入环境时，我会遇到相同的错误：

Donation = df$Donation
Treatment = df$Treatment
Session =df$Session
model = lme(Donation ~ Treatment, random=~1|Session)
glht(model,linfct=mcp(Treatment="Tukey"))