From: Christoph Lehmann
Dear Sundar, dear Andy
manyt thanks for the length(unique(x)) hint. It solves of course my
problem in a very elegant way. Just of curiosity (or for
potential future
problems): how could I solve it in a way, conceptually
different, namely,
that the computation on 'meas' being dependent on the
variable 'date'?,
means the computation on a variable x in the function passed
to aggregate
is conditional on the value of another variable y? I hope you
understand
what I mean, let's think of an example:
E.g for the example data.frame below, the sum shall be taken over the
variable meas only for all entries with a corresponding 'data' != 2
for this do I have to nest two aggregate statements, or is
there a way
using sapply or similar apply-based commands?
thanks a lot for your kind help.
Cheers!
Christoph
aggregate(data$meas, list(id = data$id), sum)
Christoph Lehmann wrote on 4/15/2005 9:51 AM:
Hi I have a question concerning aggregation
(simple demo code S. below)
I have the data.frame
id meas date
1 a 0.637513747 1
2 a 0.187710063 2
3 a 0.247098459 2
4 a 0.306447690 3
5 b 0.407573577 2
6 b 0.783255085 2
7 b 0.344265082 3
8 b 0.103893068 3
9 c 0.738649586 1
10 c 0.614154037 2
11 c 0.949924371 3
12 c 0.008187858 4
When I want for each id the sum of its meas I do:
aggregate(data$meas, list(id = data$id), sum)
If I want to know the number of meas(ures) for each id I do, eg
aggregate(data$meas, list(id = data$id), length)
NOW: Is there a way to compute the number of meas(ures)
not identical date (e.g using diff()?
so that I get eg:
id x
1 a 3
2 b 2
3 c 4
I am sure it must be possible
thanks for any (even short) hint
cheers
Christoph
--------------
data <- data.frame(c(rep("a", 4), rep("b", 4), rep("c", 4)),
runif(12), c(1, 2, 2, 3, 2, 2, 3, 3,
names(data) <- c("id", "meas", "date")
m <- aggregate(data$meas, list(id = data$id), sum)
names(m) <- c("id", "cum.meas")
How about:
m <- aggregate(data["date"], data["id"],
function(x) length(unique(x)))
--sundar