aggregation question

If I understood you correctly, here's one way:

sumWO2 <- sapply(split(dat, dat$id), function(d) sum(d$meas[d$date !=

2]))

sumWO2

        a         b         c 
0.9439614 0.4481582 1.6967618 

Andy

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) 

for each id 
with

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, 

1, 2, 3, 4))

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

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