On Dec 10, 2011, at 9:13 AM, R. Michael Weylandt wrote:
Perhaps something like this (untested) -- it's going to depend on the
exact structure of your data so if this doesn't work, please use
dput() to send a plain text representation:
tapply(data, data$animal, function(d) d[, c("A01", "A02")] - d[d$time
== "d0", c("A01", "A02")] )
`tapply` is not documented to work on dataframes as a first argument.
I think you may want to use aggregate or the split/lapply approach.
--
David.
In short, take "data" split it into groups by "data$animal" and then
apply the function which consists of subtracting the "A01" and "A02"
columns by the row that has time == "d0"
Michael
On Fri, Dec 9, 2011 at 1:44 PM, Junyu Lee <junyu0...@gmail.com> wrote:
Hello,
I have a matrix
animal time A01 A02
A d0 -5.4 2.7
A d112 4.6 5.9
A d224 3.9 6.3
B d0 7.1 5.6
B d112 1.5 3.2
B d224 2.9 3.6
C d112 3.7 5.8
C d0 7.1 NA
C d224 4.2 5.7
I have three animal A, B, C. Each animal have two measurements(A01
and A02)
at three different time points (d0, d112 and d224). I'd like to
calculate:
animal A:
A01 at d112 (4.6) - A01 at d0 (-5.4)
A01 at d224 (3.9) - A01 at d0 (-5.4)
A02 at d112 (5.9) - A02 at d0 (2.7)
A02 at d224 (6.3) - A02 at d0 (2.7)
Same for animal B and C
I really appreciate your help.
Junyu
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David Winsemius, MD
West Hartford, CT
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