A ddply solution is

dat.out <- ddply(dat, .(time), transform, slope = scale(slope))

but this is not faster than the loop, and slower than the ave() solution:

> system.time(
+ for (i in 1:3) {
+    mat <- dat[dat$time==i, ]
+    outi <- data.frame(mat$time, mat$id, slope=scale(mat$slope))
+    if (i==1) {
+        out <- outi
+    } else {
+        out <- rbind(out, outi)
+    }
+ }
+ )
   user  system elapsed
  0.024   0.000   0.025
>
> system.time(
+ dat.out <- ddply(dat, .(time), transform, slope = scale(slope))
+ )
   user  system elapsed
  0.032   0.000   0.031
>
>
> system.time(
+ cbind(dat, slope = ave(dat$slope, list(dat$time), FUN = scale))
+ )
   user  system elapsed
  0.008   0.000   0.007
>

On Thu, Aug 26, 2010 at 4:33 PM, Bos, Roger <roger.bos@rothschild.com>wrote:

> I created a small example to show something that I do a lot of.  "scale"
> data by month and return a data.frame with the output.  "id" represents
> repeated observations over "time" and I want to scale the "slope"
> variable.  The "out" variable shows the output I want.  My for..loop
> does the job but is probably very slow versus other methods.  ddply
> seems ideal, but despite playing with the baseball examples quite a bit
> I can't figure out how to get it to work with my sample dataset.
>
> TIA for any help, Roger
>
> Here is the sample code:
>
> dat <- data.frame(id=rep(letters[1:5],3),
> time=c(rep(1,5),rep(2,5),rep(3,5)), slope=1:15)
> dat
>
> for (i in 1:3) {
>    mat <- dat[dat$time==i, ]
>    outi <- data.frame(mat$time, mat$id, slope=scale(mat$slope))
>    if (i==1) {
>        out <- outi
>    } else {
>        out <- rbind(out, outi)
>    }
> }
> out
>
> Here is the sample output:
>
> > dat <- data.frame(id=rep(letters[1:5],3),
> time=c(rep(1,5),rep(2,5),rep(3,5)), slope=1:15)
>
> > dat
>   id time slope
> 1   a    1     1
> 2   b    1     2
> 3   c    1     3
> 4   d    1     4
> 5   e    1     5
> 6   a    2     6
> 7   b    2     7
> 8   c    2     8
> 9   d    2     9
> 10  e    2    10
> 11  a    3    11
> 12  b    3    12
> 13  c    3    13
> 14  d    3    14
> 15  e    3    15
>
> > for (i in 1:3) {
> +     mat <- dat[dat$time==i, ]
> +     outi <- data.frame(mat$time, mat$id, slope=scale(mat$slope))
> +     if (i==1) {
> +         out  .... [TRUNCATED]
>
> > out
>   mat.time mat.id      slope
> 1         1      a -1.2649111
> 2         1      b -0.6324555
> 3         1      c  0.0000000
> 4         1      d  0.6324555
> 5         1      e  1.2649111
> 6         2      a -1.2649111
> 7         2      b -0.6324555
> 8         2      c  0.0000000
> 9         2      d  0.6324555
> 10        2      e  1.2649111
> 11        3      a -1.2649111
> 12        3      b -0.6324555
> 13        3      c  0.0000000
> 14        3      d  0.6324555
> 15        3      e  1.2649111
> >
> ***************************************************************
>
> This message is for the named person's use only. It ma...{{dropped:22}}

