converting "by" to a data.frame?
Thanks to Thomas Lumley, Sundar Dorai-Raj, and Don McQueen for their
suggestions. I need the INDICES as part of the output data.frame, which
McQueen's solution provided. I generalized his method as follows:
by.to.data.frame <-
function(x, INDICES, FUN){
# Split data.frame x on x[,INDICES]
# and lapply FUN to each data.frame subset,
# returning a data.frame
#
# Internal functions
get.Index <- function(x, INDICES){
Ind <- as.character(x[,INDICES[1]])
k <- length(INDICES)
if(k > 1)
Ind <- paste(Ind, get.Index(x, INDICES[-1]), sep=":")
Ind
}
FUN2 <- function(data., INDICES, FUN){
vec <- FUN(data.)
Vec <- matrix(vec, nrow=1)
dimnames(Vec) <- list(NULL, names(vec))
cbind(data.[1,INDICES], Vec)
}
# Combine INDICES
Ind <- get.Index(x, INDICES)
# Apply ...: Do the work.
Split <- split(x, Ind)
byFits <- lapply(Split, FUN2, INDICES, FUN)
# Convert to a data.frame
do.call('rbind',byFits)
}
Applying this to my toy problem produces the following:
> by.df <- data.frame(A=rep(c("A1", "A2"), each=3),
+ B=rep(c("B1", "B2"), each=3), x=1:6, y=rep(0:1, length=6))
>
> by.to.data.frame(by.df, c("A", "B"), function(data.)coef(lm(y~x, data.)))
A B (Intercept) x
A1:B1 A1 B1 0.3333333 -1.517960e-16
A2:B2 A2 B2 0.6666667 3.282015e-16
Thanks for the assistance. I can now tackle the real problem that
generated this question.
Best Wishes,
Spencer Graves
########################################
Don MacQueen wrote:
Since I don't have your by.df to test with I may not have it exactly
right, but something along these lines should work:
byFits <- lapply(split(by.df,paste(by.df$A,by.df$B)),
FUN=function(data.) {
tmp <- coef(lm(y~x,data.))
data.frame(A=unique(data.$A),
B=unique(data.$B),
intercept=tmp[1],
slope=tmp[2])
})
byFitsDF <- do.call('rbind',byFits)
That's assuming I've got all the closing parantheses in the right
places, since my email software (Eudora) doesn't do R syntax checking!
This approach can get rather slow if by.df is big, or when the
computations in FUN are extensive (or both).
If by.df$A has mode character (as opposed to being a factor), then
replacing A=unique(data.$A) with A=I(unique(data.$A)) might improve
performance. You want to avoid character to factor conversions when
using an approach like this.
-Don
At 2:54 PM -0700 6/5/03, Spencer Graves wrote:
Dear R-Help:
I want to (a) subset a data.frame by several columns, (b) fit a
model to each subset, and (c) store a vector of results from the fit
in the columns of a data.frame. In the past, I've used "for" loops do
do this. Is there a way to use "by"?
Consider the following example:
> byFits <- by(by.df, list(A=by.df$A, B=by.df$B),
+ function(data.)coef(lm(y~x, data.)))
> byFits
A: A1 B: B1 (Intercept) x 3.333333e-01 -1.517960e-16 ------------------------------------------------------------ A: A2 B: B1 NULL ------------------------------------------------------------ A: A1 B: B2 NULL ------------------------------------------------------------ A: A2 B: B2 (Intercept) x 6.666667e-01 3.282015e-16
#############################
Desired output:
data.frame(A=c("A1","A2"), B=c("B1", "B2"),
.Intercept.=c(1/3, 2/3), x=c(-1.5e-16, 3.3e-16))
What's the simplest way to do this?
Thanks,
Spencer Graves
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