Thanks for your suggestion to use "do.call" in my "get.Index". I discovered that your version actually produces cosmetically different answers in R 1.6.3 and S-Plus 6.1 for Windows. Fortunately, in the context, this difference was unimportant. Since yours is faster, it is clearly superior.
To check my understanding, I generalized my toy example as follows:
> by.df <- data.frame(A=rep(c("A1", "A2"), each=3),
+ B=rep(c("B1", "B2"), each=3), C=rep(c("C1", "C2"), each=3),
+ x=1:6, y=rep(0:1, length=6))With this, your "get.Index" produced the following in R 1.6.2:
> get.Index <- function(x, INDICES) do.call('paste',c(x[INDICES],sep=':'))
> get.Index(by.df, c("A", "B", "C"))
[1] "A1:B1:C1" "A1:B1:C1" "A1:B1:C1" "A2:B2:C2" "A2:B2:C2" "A2:B2:C2"In S-Plus 6.1 for Windows, I got the following:
> get.Index <- function(x, INDICES)
do.call("paste", c(x[INDICES], sep = ":"))
> get.Index(by.df, c("A", "B", "C"))
[1] "1:1:1" "1:1:1" "1:1:1" "2:2:2" "2:2:2" "2:2:2"Fortunately, this difference is unimportant in this context, as "by.to.data.frame" produces the same answer in both cases. Moreover, your answer converts to a single call to "paste", which means that it should be faster. For someone who understands "do.call", your version is also easier to read.
Thanks again for your help. Spencer Graves
######################################
Don MacQueen wrote:
> Glad to hear it was helpful.
>
> You can also use the do.call trick for the paste indices business.
>
> Try
> get.Index <- function(x, INDICES) do.call('paste',c(x[INDICES],sep=':'))
>
> This works because a data frame is actually a list, albeit a special
> kind of list, and do.call() wants a list for its second arg.
>
> -Don
#######################################
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
>>
>> ______________________________________________
>> [EMAIL PROTECTED] mailing list
>> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
>
>
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