[R] aggregate vs tapply; is there a middle ground?

[Peter Dalgaard]

hadley wickham <h.wickham at gmail.com> writes:

> > I faced a similar problem. Here's what I did
> >
> > tmp <-
> > data.frame(A=sample(LETTERS[1:5],10,replace=T),B=sample(letters[1:5],10,replace=T),C=rnorm(10))
> > tmp1 <- with(tmp,aggregate(C,list(A=A,B=B),sum))
> > tmp2 <- expand.grid(A=sort(unique(tmp$A)),B=sort(unique(tmp$B)))
> > merge(tmp2,tmp1,all.x=T)
> >
> > At least fewer than 10 extra lines of code. Anyone with a simpler solution?
> 
> Well, you can almost do this in with the reshape package:
> 
> tmp <-
> data.frame(A=sample(LETTERS[1:5],10,replace=T),B=sample(letters[1:5],10,replace=T),C=rnorm(10))
> a <- recast(tmp, A + B ~ ., sum)
> # see also recast(tmp, A  ~ B, sum)
> add.all.combinations(a, row="A", cols = "B")
> 
> Where add.all.combinations basically does what you outlined above --
> it would be easy enough to generalise to multiple dimensions.

Anything wrong with

> as.data.frame(with(tmp,as.table(tapply(C,list(A=A,B=B),sum))))
   A B       Freq
1  A a         NA
2  B a -0.2524320
3  C a  3.8539264
4  D a         NA
5  A c  0.7227294
6  B c -0.2694669
7  C c  0.4760957
8  D c         NA
9  A e         NA
10 B e  0.1800500
11 C e         NA
12 D e -1.0350928

(except the silly colname, responseName="sum" should fix that).

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