grouped colSums without for loops?

lapply(split(d, d$foo), function(x) colSums(x[,-1]))

Consider the following data.frame, d.

d <- data.frame(c("a","b","c","d","b","a"), c(1,4,85,3,4,1), c(7,6,2,4,15,33))
names(d) <- c("foo", "bar", "baz")

To get the colsum of d[[2]] for the rows that have d$foo == "a" I know
I can use

colSums(subset(d, d[[1]] == "a", select = 2))

But what is needed to get a list of colSums for d[[2]] for each factor
of d$foo ? Can it be done without a for loop?

--
Hans Ekbrand (http://sociologi.cjb.net) <hans at sociologi.cjb.net>
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Consider the following data.frame, d.

d <- data.frame(c("a","b","c","d","b","a"), c(1,4,85,3,4,1), c(7,6,2,4,15,33))
names(d) <- c("foo", "bar", "baz")
 
To get the colsum of d[[2]] for the rows that have d$foo == "a" I know
I can use

colSums(subset(d, d[[1]] == "a", select = 2))

But what is needed to get a list of colSums for d[[2]] for each factor
of d$foo ? Can it be done without a for loop?

Is this what you want?

lapply(split(d, d$foo), function(x) colSums(x[,-1]))

* PGP Signed by an unknown key: 03/18/2008 at 06:22:03 AM
On Tue, Mar 18, 2008 at 05:57:59AM -0500, jim holtman wrote:

Is this what you want?

lapply(split(d, d$foo), function(x) colSums(x[,-1]))

Yes! Thank you! the *apply functions seem very powerful, thanks again
for giving me a hint on how to use them.

-- 
Hans Ekbrand (http://sociologi.cjb.net) <hans at sociologi.cjb.net>
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try:
aggregate(d[,2:3], by=list(d$foo), FUN=sum)