I recently stumbled across replace() and "[<-". They seem to have similar functionality, for example with "[<-" I can do something like this:
> x.tst <- array(1:6, c(2,3))
> s.tst <- array(0, c(2,3))
> s.tst
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 0 0 0
> s.tst[1:3] <- 1
> "[<-"(x.tst, s.tst==1, 0)
[,1] [,2] [,3]
[1,] 0 0 5
[2,] 0 4 6
> x.tst
[,1] [,2] [,3]
[1,] 1 3 5
[2,] 2 4 6
Can somebody help to clarify the difference? What are the strengths of replace vs "[<-" and vis versa?
They're basically exactly the same thing. If you look at the source code of replace, you'll see :
function (x, list, values)
{
x[list] <- values
x
}
<environment: namespace:base>
So replace is nothing else but a wrapper around [<- :
> replace(x.tst, s.tst==1, 0)
[,1] [,2] [,3]
[1,] 0 0 5
[2,] 0 4 6
Using [<- can give you a speedup if you need to do this a million times, as you lose the extra call to the wrapper function. But it's really marginal, so it's a matter of choice. I would say that replace()is a bit more readible
Btw, x.tst[s.tst==1] <- 0 is quite more readible than "[<-"(x.tst, s.tst==1, 0) . No reason to use that construct, unless you want to save the result in a new dataframe.
To clarify, as @Andrie pointed out, both with replace() and "[<-"(x.tst, s.tst==1, 0) you get a copy of the whole x.tst with the relevant values changed. So you can put that in a new object. This is contrary to x.tst[s.tst==1] <- 0, where you change the values in x.tst itself. Mind you, it doesn't save on memory, as R will make internally a copy of x.tst before doing the manipulation.
Timing results :
> system.time(replicate(1e6, replace(x.tst, s.tst==1, 0)))
user system elapsed
12.73 0.03 12.78
> system.time(replicate(1e6, "[<-"(x.tst, s.tst==1, 0)))
user system elapsed
6.42 0.02 6.44
> system.time(replicate(1e6, x.tst[s.tst==1] <- 0))
user system elapsed
5.28 0.02 5.32
来源：https://stackoverflow.com/questions/6346101/replace-vs