G'Day,
I want to access in a function called from tapply the current factor. In
my example below, all I want to do is to write the current factor on each
histogram. Needless to say my example does not work. I would be grateful
for pointers in the right direction.
Many thanks
Bernie McConnell
Sea Mammal Reserach Unit
cc <- 1:10
ff <- rep(c("a","b"),5)
pp<- function(x,f) {
hist(x, main=as.character(f))
}
tapply(aa, ff, pp, f=ff)
Bernie McConnell wrote:> G'Day, > > I want to access in a function called from tapply the current factor. > In my example below, all I want to do is to write the current factor on > each histogram. Needless to say my example does not work. I would be > grateful for pointers in the right direction. > > Many thanks > Bernie McConnell > Sea Mammal Reserach Unit > > > cc <- 1:10 > ff <- rep(c("a","b"),5) > > pp<- function(x,f) { > hist(x, main=as.character(f)) > } > > > tapply(aa, ff, pp, f=ff) > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://www.stat.math.ethz.ch/mailman/listinfo/r-help >I think it would be easier and more foolproof to use a for loop: cc <- 1:10 ff <- rep(c("a", "b"), 5) par(mfrow = c(1, 2)) for(f in unique(ff)) hist(cc[ff==f], main = as.character(f)) Regards, Sundar
Here's a way (but as Sundar Dorai-Raj suggests, it might be easier to use a
for loop.)
> cc <- 1:10
> ff <- rep(c("a","b"),5)
> # a different function, so that we can see what data and what label it
gets
> data.lab <- function(data,lab) paste(paste(data, collapse="
"),
paste(as.character(lab), collapse=" "), sep="/")
> # ii is indices of each group of ff
> ii <- tapply(cc, ff, c)
> sapply(seq(along=ii), function(j, ii, cc, labs)
data.lab(data=cc[ii[[j]]],lab=labs[j]), ii=ii, cc=cc, labs=names(ii))
[1] "1 3 5 7 9/a" "2 4 6 8 10/b"
>
>
At Thursday 06:13 PM 4/17/2003 +0100, you wrote:>G'Day,
>
>I want to access in a function called from tapply the current factor. In
>my example below, all I want to do is to write the current factor on each
>histogram. Needless to say my example does not work. I would be grateful
>for pointers in the right direction.
>
>Many thanks
>Bernie McConnell
>Sea Mammal Reserach Unit
>
>
>cc <- 1:10
>ff <- rep(c("a","b"),5)
>
>pp<- function(x,f) {
> hist(x, main=as.character(f))
> }
>
>
>tapply(aa, ff, pp, f=ff)
>
>______________________________________________
>R-help at stat.math.ethz.ch mailing list
>https://www.stat.math.ethz.ch/mailman/listinfo/r-help
I've just been puzzling through this for my own use. There are three different R functions which do related things: tapply(), by() and aggregate(). In this context, I think I would use by() rather than tapply(). Define cc, ff as before, and re-define pp <- function(dat, name) hist(dat$x, main=as.character(unique(dat[[name]]))) then by(data.frame(cc,ff), ff, pp, "ff") # should do what you want The slightly odd syntax at the end of pp() allows its second argument "name" to be either a string naming the selection variable (as here) or else an integer specifying it by position within the data frame. If you want to split the data set on the cross product of two factors, ff x gg, the second argument to by() would have to be list(ff,gg). tapply(), by contrast, requires the list() syntax even when there's only one factor to split on. HTH - tom blackwell - u michigan medical school - ann arbor - On Thu, 17 Apr 2003, Bernie McConnell wrote:> G'Day, > > I want to access in a function called from tapply the current factor. In > my example below, all I want to do is to write the current factor on each > histogram. Needless to say my example does not work. I would be grateful > for pointers in the right direction. > > Many thanks > Bernie McConnell > Sea Mammal Reserach Unit > > cc <- 1:10 > ff <- rep(c("a","b"),5) > > pp<- function(x,f) { > hist(x, main=as.character(f)) > } > > tapply(aa, ff, pp, f=ff)