How to format describeBy table in R? - r

Question

I have this data set:
Defects.I Defects.D Treatment
1 2 A
1 3 B
And I'm trying to do a descriptive statistics for defects detected and isolated, grouped per treatment.
After searching for a while I found a nice function on the psych library called describeBy().
With the following code:
describeBy(myData[1:2],myData$Treatment)
I got this output:
Treatment A
Mean. Median. Trimed.
Defects.I x x x
Defects.D x x x
Treatment B
Mean. Median. Trimed.
Defects.I x x x
Defects.D x x x
But in reality I was looking for something like
Mean. Median. Trimed.
A B A B A B
Defects.I x x x x x x
Defects.D x x x x x x
Data
myData <- structure(list(Defects.I = c(1L, 1L), Defects.D = 2:3, Treatment = c("A",
"B")), .Names = c("Defects.I", "Defects.D", "Treatment"), class = "data.frame", row.names = c(NA,
-2L))

Answer 1

Since describeBy returns a lists of data frames, we could just cbind them all, but that doesn't get the right order. Instead we can interleave the columns
myData <- structure(list(Defects.I = c(1L, 1L), Defects.D = 2:3,
Treatment = c("A", "B")),
.Names = c("Defects.I", "Defects.D", "Treatment"),
class = "data.frame", row.names = c(NA, -2L))
l <- psych::describeBy(myData[1:2], myData$Treatment)
So interleave using this order
order(sequence(c(ncol(l$A), ncol(l$B))))
# [1] 1 14 2 15 3 16 4 17 5 18 6 19 7 20 8 21 9 22 10 23 11 24 12 25 13 26
rather than what cbind alone would do
c(1:13, 1:13)
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 7 8 9 10 11 12 13
so this
do.call('cbind', l)[, order(sequence(lengths(l)))]
# A.vars B.vars A.n B.n A.mean B.mean A.sd B.sd A.median B.median A.trimmed B.trimmed A.mad B.mad
# Defects.I 1 1 1 1 1 1 NA NA 1 1 1 1 0 0
# Defects.D 2 2 1 1 2 3 NA NA 2 3 2 3 0 0
# A.min B.min A.max B.max A.range B.range A.skew B.skew A.kurtosis B.kurtosis A.se B.se
# Defects.I 1 1 1 1 0 0 NA NA NA NA NA NA
# Defects.D 2 3 2 3 0 0 NA NA NA NA NA NA
or as a function
interleave <- function(l, how = c('cbind', 'rbind')) {
how <- match.arg(how)
if (how %in% 'rbind')
do.call(how, l)[order(sequence(sapply(l, nrow))), ]
else do.call(how, l)[, order(sequence(sapply(l, ncol))), ]
}
interleave(l)
# A.vars B.vars A.n B.n
# Defects.I 1 1 1 1
# Defects.D 2 2 1 1 ...
# ...
interleave(l, 'r')
# vars n mean sd median trimmed mad min max range skew kurtosis se
# A.Defects.I 1 1 1 NA 1 1 0 1 1 0 NA NA NA
# B.Defects.I 1 1 1 NA 1 1 0 1 1 0 NA NA NA
# A.Defects.D 2 1 2 NA 2 2 0 2 2 0 NA NA NA
# B.Defects.D 2 1 3 NA 3 3 0 3 3 0 NA NA NA

Answer 2

You can try the mat = TRUE argument. It's not exactly what you're looking for, but it's closer:
library(psych)
mydata = data.frame(Defects.I = c(1,1), Defects.D = c(2,3), Treatment = c('A','B'))
describeBy(mydata[1:2], mydata$Treatment, mat = TRUE)
gives
item group1 vars n mean sd median trimmed mad min max range skew kurtosis se
Defects.I1 1 A 1 1 1 NA 1 1 0 1 1 0 NA NA NA
Defects.I2 2 B 1 1 1 NA 1 1 0 1 1 0 NA NA NA
Defects.D1 3 A 2 1 2 NA 2 2 0 2 2 0 NA NA NA
Defects.D2 4 B 2 1 3 NA 3 3 0 3 3 0 NA NA NA