[R] converting "by" to a data.frame?

[Spencer Graves]

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
>>
>> ______________________________________________
>> R-help at stat.math.ethz.ch mailing list
>> https://www.stat.math.ethz.ch/mailman/listinfo/r-help
> 
> 
>