Using lapply with changing arguments

Question

R textbooks continue to promote the use of lapply instead of loops. This is easy even for functions with arguments like

lapply(somelist, f, a=1, b=2) 


        

Answer 1

You just need to work out what to lapply() over. Here the names() of the lists suffices, after we rewrite f() to take different arguments:

somelist <- list(USA = 1:10, Europe = 21:30,
                 Switzerland = seq(1, 5, length = 10))
anotherlist <- list(USA = list(a = 1, b = 2), Europe = list(a = 2, b = 4),
                    Switzerland = list(a = 0.5, b = 1))

f <- function(x, some, other) {
    (some[[x]] + other[[x]][["a"]]) * other[[x]][["b"]]
}

lapply(names(somelist), f, some = somelist, other = anotherlist)

Giving:

R> lapply(names(somelist), f, some = somelist, other = anotherlist)
[[1]]
 [1]  4  6  8 10 12 14 16 18 20 22

[[2]]
 [1]  92  96 100 104 108 112 116 120 124 128

[[3]]
 [1] 1.500000 1.944444 2.388889 2.833333 3.277778 3.722222 4.166667 4.611111
 [9] 5.055556 5.500000

Answer 2

Apply over list names rather than list elements. E.g.:

somelist <- list('USA'=rnorm(10), 'Europe'=rnorm(10), 'Switzerland'=rnorm(10))
anotherlist <- list('USA'=5, 'Europe'=10, 'Switzerland'=4)
lapply(names(somelist), function(i) somelist[[i]] / anotherlist[[i]])

EDIT:

You also ask if there is a way "except for a loop" to do this "efficiently". You should note that the apply will not necessarily be more efficient. Efficiency will probably be determined by how quick your inner function is. If you want to operate on each elements of a list, you will need a loop, whether it is hidden in an apply() call or not. Check this question: Is R's apply family more than syntactic sugar?

The example I gave above can be re-written as a for loop, and you can make some naive benchmarks:

fun1 <- function(){
    lapply(names(somelist), function(i) somelist[[i]] / anotherlist[[i]])
}
fun2 <- function(){
    for (i in names(somelist)){
        somelist[[i]] <- somelist[[i]] / anotherlist[[i]] 
    }
    return(somelist)
}
library(rbenchmark)

benchmark(fun1(), fun2(),
          columns=c("test", "replications",
          "elapsed", "relative"),
          order="relative", replications=10000)

The output of the benchmark on my machine was this:

    test replications elapsed relative
1 fun1()        10000   0.145 1.000000
2 fun2()        10000   0.148 1.020690

Although this is not a real work application and the functions are not realistic tasks, you can see that the difference in computation time is quite negligible.