Bioequivalence and Bioavailability Forum

Hi Shuanghe,

❝ # remove subjects who has imcomplete periods. More elegant solution?


❝ library(dplyr)

Didn’t know this package. The syntax within some functions (i.e., the %>%-operator) is – well – un­con­ven­tio­nal.

Code to keep only subjects completing all four periods:

summary(ema[2:4])                # show structure
per.4  <- summary(ema$subj) == 4 # subjs have 4 levels?
incomp <- per.4[per.4 == FALSE]  # incomplete ones
print(sort(names(incomp)))
comp   <- per.4[per.4 == TRUE]   # complete ones
print(sort(names(comp)))
d      <- subset(ema, subj %in% names(comp))
print(d)
summary(d[2:4])
seq1   <- as.numeric(summary(d[, 3])[names(summary(d[, 3]))[1]])
seq2   <- as.numeric(summary(d[, 3])[names(summary(d[, 3]))[2]])
if(seq1 == seq2) {
 txt <- "Balanced: "
 } else {
 txt <- "Unbalanced: "
}
cat(paste0(txt, names(summary(d[, 3]))[1], "=", seq1, ", ",
 names(summary(d[, 3]))[2], "=", seq2, "\n"))

Lines in red can be dropped (checking results only).

Which subjects are incomplete? print(sort(names(incomp))) gives

[1] "11" "20" "24" "31" "42" "67" "69" "71"

summary(d[2:4]) gives

 per      seq      treat
 1:69   RTRT:144   R:138
 2:69   TRTR:132   T:138
 3:69
 4:69

and the goody

Unbalanced: RTRT=144, TRTR=132

The code works for fancy non-numeric subject-IDs and other sequence-codings as well. Try

ema$subj <- as.factor(paste0("Sub #", ema$subj))
levels(ema$seq)[levels(ema$seq)=="RTRT"]  <- 1
levels(ema$seq)[levels(ema$seq)=="TRTR"]  <- 2
levels(ema$treat)[levels(ema$treat)=="T"] <- "A"
levels(ema$treat)[levels(ema$treat)=="R"] <- "B"

and run the code. I got for print(sort(names(incomp)))

[1] "Sub #11" "Sub #20" "Sub #24" "Sub #31" "Sub #42" "Sub #67" "Sub #69"
[8] "Sub #71

and finally for summary(d[2:4])

 per    seq     treat
 1:69   1:144   B:138
 2:69   2:132   A:138
 3:69       
 4:69

Same as above (apart from differing sequences and treatments).

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Edit: Once I saw Detlew’s code below, I can only say about my snippet: Forget it!