r - Colmeans in a dataframe by factor variable -

i'm trying mean of variables inside dataframe different factors. have:

    time    geo var1    var2   var3    var4 1   1990    @  1       7      13       19 2   1991    @  2       8      14       20 3   1992    @  3       9      15       21 4   1990    de  4       10     16       22 5   1991    de  5       11     17       23 6   1992    de  6       12     18       24 

and want:

    time    geo var1    var2   var3    var4   m_var2   m_var3 1   1990    @  1       7      13       19    8        14 2   1991    @  2       8      14       20    8        14 3   1992    @  3       9      15       21    8        14 4   1990    de  4       10     16       22    11       17 5   1991    de  5       11     17       23    11       17 6   1992    de  6       12     18       24    11       17 

i've tried few things by() , lapply() think goes direction of ddply

require(plyr) dataset <- data.frame(time=rep(c(1990:1992),2),geo=c(rep("at",3),rep("de",3))       ,var1=as.numeric(c(1:6)),var2=as.numeric(c(7:12)),var3=as.numeric(c(13:18)),       var4=as.numeric(c(19:24)))  newvars <- c("var2","var3") newdata <- dataset[,c("geo",newvars)] 

currently, can choose between 2 errors:

ddply(newdata,newdata[,"geo"],colmeans)  #where r apparently thinks @ variable?  ddply(newdata,"geo",colmeans) #where r worries factor variable not being numeric? 

my lapply attempts got me quite far left me list not dataframe:

lapply(newvars,function(x){        by(dataset[x],dataset[,"geo"],function(x)         rep(colmeans(x,na.rm=t),length(unique(dataset[,"time"]))))        }) 

i think must able merge , filters here: lapply in dataframe on different variables using filters , can't together. appreciated!

one option use data.table. can convert data.frame data.table (setdt(df1)), mean (lapply(.sd, mean)) selected columns ('var2' , 'var3') specifying column index in .sdcols, grouped 'geo'. create new columns assigning output (:=) new column names (paste('m', names(df1)[4:5]))

library(data.table) setdt(df1)[, paste('m', names(df1)[4:5], sep="_") :=lapply(.sd, mean)             ,by = geo, .sdcols=4:5] #     time geo var1 var2 var3 var4 m_var2 m_var3 #1: 1990  @    1    7   13   19      8     14 #2: 1991  @    2    8   14   20      8     14 #3: 1992  @    3    9   15   21      8     14 #4: 1990  de    4   10   16   22     11     17 #5: 1991  de    5   11   17   23     11     17 #6: 1992  de    6   12   18   24     11     17 

note: method more general. can create mean columns 100s of variables without major change in code. ie. if need mean of columns 4:100, change .sdcols=4:100 , in paste('m', names(df1)[4:100].

data

df1 <- structure(list(time = c(1990l, 1991l, 1992l, 1990l, 1991l, 1992l ), geo = c("at", "at", "at", "de", "de", "de"), var1 = 1:6, var2 = 7:12,  var3 = 13:18, var4 = 19:24), .names = c("time", "geo", "var1",  "var2", "var3", "var4"), class = "data.frame", row.names = c("1",  "2", "3", "4", "5", "6"))