searching for data.frame rows / processing of rows

  df <- read.table(stdin(),header=TRUE)

aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)

 system.time(  aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)  )

 system.time(  do.call(rbind, by(df, df[2:1], colMeans, na.rm = TRUE)))

df <- data.frame(Location=rep(1:50,50),

 system.time(  aggregate(df[,3:4],df[,1:2],FUN=mean,na.rm=TRUE)  )

 system.time(  do.call(rbind, by(df, df[2:1], colMeans, na.rm = TRUE)))

Grouping the data frame by the first two columns, apply colMeans
and then rbind the resulting by-structure together:

do.call(rbind, by(DF, DF[2:1], colMeans, na.rm = TRUE))


On 10/5/06, Greg Tarpinian <sasprog474 at yahoo.com> wrote:

R 2.3.1, WinXP:

I have a puzzling problem that I suspect may be solved using
grep or a regular expression but am at a loss how to actually do it...
My data.frame looks like

 Location   Time    X    Y
 --------   ----   ---  ---
 1          0      1.6  9.3
 1          3      4.2  10.4
 1          6      2.7  16.3
 2          0      0.5  2.1
 2          3      NA   3.6
 2          3      5.0  0.06
 2          6      3.4  14.0

and so forth.  I would like to search for duplicate Time values
within a Location and take the numerical average (where possible)
of the elements in X and Y.  These numerical averages should
then be used to create a single row where multiple rows once
existed.  So, I would like to obtain

 2          3      5.0  1.83

for the two Time = 3 rows for Location = 2 and use it to replace
these two rows.  Ideally, avoiding for(i in 1:blah) loops would be
nice because the data.frame has about 10,000 rows that need to
be searched and processed.  My intent is to do some comparing of
SAS to R -- the DATA step processing in SAS is quite fast and
using the RETAIN statement along with the LAG( ) function allows
this sort of thing to be done rapidly.


Thanks in advance,

      Greg