Pairwise t test using plyr - r

Question

I would like to use the R package plyr to run a pairwise t test on a really large data frame, but I'm not sure how to do it. I recently learned how to do correlations using plyr, and I really like how you can specify which groups you want to compare and then plyr breaks down the data for you. For example, you could have plyr calculate the correlation between sepal length and sepal width for each species of iris in the iris dataset like this:
Correlations <- ddply(iris, "Species", function(x) cor(x$Sepal.Length, x$Sepal.Width))
I could break the data frame down myself by specifying that the data for the setosa species of iris are in rows 1:50 and so on, but plyr would be less likely than me to mess up and accidentally say rows 1:51, for example.
So how do I do something similar with a paired t test? How can I specify which observations are the pairs? Here's some example data that are similar to what I'm working with, and I'd like the pairs to be the Subject and I'd like to break the data down by Pesticide:
Exposure <- data.frame("Subject" = rep(1:4, 6),
"Season" = rep(c(rep("summer", 4), rep("winter", 4)),3),
"Pesticide" = rep(c("atrazine", "metolachlor", "chlorpyrifos"), each=8),
"Exposure" = sample(1:100, size=24))
Exposure$Subject <- as.factor(Exposure$Subject)
In other words, the question I'd like to evaluate is whether there is a difference in pesticide exposure for each person during the winter versus during the summer, and I'd like to answer that question separately for each of the three pesticides.
Much thanks in advance!
An edit: To clarify, this is how to do an unpaired t test in plyr:
TTests <- dlply(Exposure, "Pesticide", function(x) t.test(x$Exposure ~ x$Season))
And if I add "paired=T" in there, plyr will do a paired t test, but it assumes that I always have the pairs in the same order. While I do have them all in the same order in the example data frame above, I don't in my real data because I sometimes have missing data.

Answer 1

Do you want this?
library(data.table)
# convert to data.table in place
setDT(Exposure)
# make sure data is sorted correctly
setkey(Exposure, Pesticide, Season, Subject)
Exposure[, list(res = list(t.test(Exposure[Season == "summer"],
Exposure[Season == "winter"],
paired = T)))
, by = Pesticide]$res
#[[1]]
#
# Paired t-test
#
#data: Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -4.1295, df = 3, p-value = 0.02576
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -31.871962 -4.128038
#sample estimates:
#mean of the differences
# -18
#
#
#[[2]]
#
# Paired t-test
#
#data: Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -6.458, df = 3, p-value = 0.007532
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -73.89299 -25.10701
#sample estimates:
#mean of the differences
# -49.5
#
#
#[[3]]
#
# Paired t-test
#
#data: Exposure[Season == "summer"] and Exposure[Season == "winter"]
#t = -2.5162, df = 3, p-value = 0.08646
#alternative hypothesis: true difference in means is not equal to 0
#95 percent confidence interval:
# -30.008282 3.508282
#sample estimates:
#mean of the differences
# -13.25

Answer 2

I don't know ddply, but here's how I would do using some base functions.
by(data = Exposure, INDICES = Exposure$Pesticide, FUN = function(x) {
t.test(Exposure ~ Season, data = x)
})
Exposure$Pesticide: atrazine
Welch Two Sample t-test
data: Exposure by Season
t = -0.1468, df = 5.494, p-value = 0.8885
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-49.63477 44.13477
sample estimates:
mean in group summer mean in group winter
60.50 63.25
----------------------------------------------------------------------------------------------
Exposure$Pesticide: chlorpyrifos
Welch Two Sample t-test
data: Exposure by Season
t = -0.8932, df = 4.704, p-value = 0.4151
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-83.58274 41.08274
sample estimates:
mean in group summer mean in group winter
52.25 73.50
----------------------------------------------------------------------------------------------
Exposure$Pesticide: metolachlor
Welch Two Sample t-test
data: Exposure by Season
t = 0.8602, df = 5.561, p-value = 0.4252
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-39.8993 81.8993
sample estimates:
mean in group summer mean in group winter
62.5 41.5