Try this:
evaluation$maxVol <- ave(evaluation$vol, evaluation$name, FUN = max)
or using SQL via sqldf like this:
library(sqldf)
sqldf("select * from evaluation join
(select name, max(vol) from evaluation group by name) using (name)")
On 8/31/07, Calle <[EMAIL PROTECTED]> wrote:
> Hello,
>
> struggling with the very basic needs... :( any help appreciated.
>
> #using the package doBY
> #who drinks how much beer per day and therefor cannot calculate rowise
> maxvals
> evaluation=data.frame(date=c(1,2,3,4,5,6,7,8,9),
> name=c("Michael","Steve","Bob",
> "Michael","Steve","Bob","Michael","Steve","Bob"), vol=c(3,5,4,2,4,5,7,6,7))
> evaluation #
>
> maxval=summaryBy(vol ~ name,data=evaluation,FUN = function(x) { c(ma=max(x))
> } )
> maxval # over all days per person
>
> #function
> getMaxVal=function(x) { maxval$vol.ma[maxval$name==x] }
> getMaxVal("Steve") # testing the function for one name is ok
>
> #we want to add a column, that shows the daily drinkingvolume in relation to
> the persons max-vol.
> evaluation[,"relDrink"]= evaluation$vol/getMaxVal(evaluation$name)
> #
> # this brings the error:
> #
> #Warning message:
> # Korrupter Data Frame: Spalten werden abgeschnitten oder mit NAs
> # aufgefüllt in: format.data.frame(x, digits = digits, na.encode = FALSE)
>
> errortest= evaluation$vol/getMaxVal(evaluation$name)
> errortest
> # this brings:
> # numeric(0)
>
>
> #target was the following:
> #show in each line the daily consumed beer per person and in the next column
>
> #the all time max consumed beer for this person´(or divided by daily vol):
> #
> # date name vol relDrink
> #1 1 Michael 3 7
> #2 2 Steve 5 6
> #3 3 Bob 4 7
> #4 4 Michael 2 7
> #5 5 Steve 4 7
> #6 6 Bob 5 7
> #7 7 Michael 7 7
> #8 8 Steve 6 6
> #9 9 Bob 7 7
>
> # who can help???
>
> [[alternative HTML version deleted]]
>
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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