trying to understand emmeans [General Sta­tis­tics]

posted by mittyri – Russia, 2019-11-02 02:26  – Posting: # 20745
Views: 311

Dear Zizou. dear All,

» the reason of the difference is that sequences were unbalanced (not equal number of subjects in each of the sequences).

looks like that reason is not the dominant reason.

» For unbalanced sequences there might be more questions ...

when the party comes to incomplete cases, everyone has more fun!

I am lazy man too, but some code for your pleasure

library(dplyr)
library(nlme)
library(replicateBE)
library(emmeans)
messupdataset<- function(dataset, treatment = 'R'){
  dataset$logPK <- ifelse(dataset$treatment==treatment, dataset$logPK*runif(1),dataset$logPK)
  return(dataset)
}

emmeanscomparison <- function(dataset, messuprefdata = F){
  dataset$logPK <- log(dataset$PK)
  if(messuprefdata==T){
    dataset <- messupdataset(dataset, "R")
  } 
 
  M=lm(logPK ~ sequence + subject + period + treatment, data = dataset)
  cat(paste0("\nLSMeans ratio by lm(): \nT/R = ", exp(coef(M)[["treatmentT"]])))
 
  newdat <- expand.grid(treatment = levels(dataset$treatment),
                        sequence = levels(dataset$sequence),
                        subject = levels(dataset$subject),
                        period = levels(dataset$period))
  preddata <- cbind(newdat, predict(M, newdat))
  preddatawoNA <- na.omit(left_join(preddata, dataset, by = c("treatment", "sequence", "subject", "period")))
  lsmeansbyhand <-
    preddatawoNA%>%
    group_by(treatment, sequence, period) %>%
    summarize(subjectmean = mean(`predict(M, newdat)`)) %>%
    group_by(treatment) %>%
    summarize(lsmean = exp(mean(subjectmean))) %>%
    mutate(lsmeansratio = lsmean[treatment=='T']/lsmean[treatment=='R']) %>%
    as.data.frame()
 
  cat(paste0("\nLSMeans by hand: \nT = ", lsmeansbyhand$lsmean[lsmeansbyhand$treatment=='T']))
  cat(paste0("\nLSMeans by hand: \nR = ", lsmeansbyhand$lsmean[lsmeansbyhand$treatment=='R']))
  cat(paste0("\nLSMeans ratio by hand: \nT/R = ", lsmeansbyhand$lsmeansratio[1], "\n"))
 
  emmeansdf <- data.frame(emmeans(M, 'treatment'))
  cat(paste0("\nLSMeans by emmeans: \nT = ", exp(emmeansdf$emmean[emmeansdf$treatment=='T'])))
  cat(paste0("\nLSMeans by emmeans: \nR = ", exp(emmeansdf$emmean[emmeansdf$treatment=='R'])))
  cat(paste0("\nLSMeans ratio by emmeans: \nT/R = ", exp(emmeansdf[emmeansdf$treatment=='T',]$emmean - emmeansdf[emmeansdf$treatment=='R',]$emmean), "\n"))
}

# balanced complete
emmeanscomparison(rds11, messuprefdata = F)
seed = 123
emmeanscomparison(rds11, messuprefdata = T)

# unbalanced complete
emmeanscomparison(rds16, messuprefdata = F)
seed = 123
emmeanscomparison(rds16, messuprefdata = T)

# balanced incompete
emmeanscomparison(rds26, messuprefdata = F)
seed = 123
emmeanscomparison(rds26, messuprefdata = T)

# unbalanced incompete
emmeanscomparison(rds14, messuprefdata = F)
seed = 123
emmeanscomparison(rds14, messuprefdata = T)


so I can get the same results for complete cases, but not for INcomplete. Please also take a look at the lsmeans for T when R is messed up for unbalanced complete and unbalanced incompltete. Looks like competeness has a predominant value
so my understanding of emmeans is also INcomplete and Unbalanced:-D

Kind regards,
Mittyri

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