You can do this in reshape package as mentioned earlier.
However, if you need a solution with aggregate here it is
a <- with(data, aggregate(cbind(v1,v2), by=list(x,y,z),sum))
names (a) <- c("x","y","z","v1","v2")
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Jun 29, 2010, at 7:56 PM, Yi wrote:
> Great help. It works when the first and the second columns are
> ordered the same way. But aggregate does not work for the following
> case:
> z=c('ab','ah','bc','ah','dv')
> x=substr(z,start=1,stop=1)
> y=substr(z,start=2,stop=2)
> v1=5:9
> v2=7:11
> data=data.frame(x,y,z,v1,v2)
> > data
> x y z v1 v2
> 1 a b ab 5 7
> 2 a h ah 6 8
> 3 b c bc 7 9
> 4 a h ah 8 10
> 5 d v dv 9 11
>
> ##I want to do the aggregate WRT z and sum up v1 and v2. The
> expected output is:
>
> x y z v1 v2
> 1 a b ab 5 7
> 2 a h ah 14 18
> 3 b c bc 7 9
> 4 d v dv 9 11
>
> ### I do this almost manually. As you see here:
>
> newdata=aggregate(data$v1,by=list(data$z),sum)
> newdata2=aggregate(data$v2,by=list(data$z),sum)
> x=substr(newdata$Group.1,start=1,stop=1)
> y=substr(newdata$Group.1,start=2,stop=2)
> data.frame(x,y,newdata$Group.1,newdata$x,newdata2$x)
> new=data.frame(x,y,newdata$Group.1,newdata$x,newdata2$x)
> names(new)=c('x','y','z','v1','v2')
> new
>
> Because I do not think 'aggregate' can not set z as a list and at
> the same time keep x and y for z.
>
> Any tips? I mean my way is too 'silly'.
>
> Thanks all in advance!
>
> Yi
>
> On Mon, Jun 28, 2010 at 7:58 PM, Nikhil Kaza <nikhil.l...@gmail.com>
> wrote:
>
> aggregate(data$third, by=list(data$first), sum)
>
> or
>
> reqiure(reshape)
> cast(melt(data), ~first, sum)
>
>
>
> On Jun 28, 2010, at 9:30 PM, Yi wrote:
>
>
> first=c('u','b','e','k','j','c','u','f','c','e')
> second
> =
> c
> ('usa
> ','Brazil
> ','England','Korea','Japan','China','usa','France','China','England')
> third=1:10
> data=data.frame(first,second,third)
>
>
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