Method chaining with R

Question

Is it possible to chain functions in R?

Sample data:

m <- matrix(c(1:10, 11:20), nrow = 10, ncol = 2)

For example, I would like to replace the following statements below:

step1 <- mean(m)
step2 <- sum(step1)
res <- step2

Or,

res <- sum(mean(m))

With something like this :

res <- m@mean()@sum()

In some cases, that would clarify my code considerably.

EDIT1 This is a dummy example. I randomly picked 'sum' and 'mean'.

Ben has given a first piece of answer using %@% however, it prevents from using extra arguments within functions :

m %@% function1(arg1, arg2) %@% function2(arg1, arg2)

How can I work around that ?

EDIT2 Adding an example

require(xts)
require(PerformanceAnalytics)
xts.ts <- xts(rnorm(231),as.Date(13514:13744,origin="1970-01-01"))
plot(na.omit(lag( rollapply(xts.ts, width=rolling.per-1, FUN= function(x){sqrt(var(x))*sqrt(252)}), k=1)), main = "Dummy Example")

This example seems to work fine with Charles solution :

`%@%` <- function(x, f) eval.parent(as.call(append(as.list(substitute(f)), list(x), 1)))
xts.ts %@% rollapply( width = rolling.per-1, FUN= function(x) x%@%var%@%sqrt * sqrt(252) ) %@% lag( k=1) %@% na.omit %@% plot(main = "Dummy Example")

Less important to my case, but woth mentioning, the following statment fails with Charles's solution :

 xts.ts %@% names <- 'ts name' 

Answer 1

Try the functional package:

library(functional)
squared <- function(x)x*x
Compose(sum, squared)(m)
## [1] 44100
squared(sum(m))
## [1] 44100

EDIT:

Regarding the question in the comments of another response about arguments here is an example of composing with arguments. Curry is also from the functional package:

addn <- function(n, x) x + n
Compose(Curry(addn, 1), squared)(10)
## [1] 121
squared(addn(1, 10))
## [1] 121

EDIT 2:

Regarding question about debugging, debug works if the function is curried. If its not already curried then wrap it in Curry :

# this works since addn is curried
debug(addn)
Compose(Curry(addn, 1), squared)(10)

# to debug squared put it in a Curry -- so this works:
debug(squared)
Compose(Curry(addn, 1), Curry(squared))(10)

Answer 2

I would use the magrittr package. It has a "pipe" operator that takes the result of one function and passes it in as an argument to the next:

m <- matrix(c(1:10, 11:20), nrow = 10, ncol = 2)

m %>% mean %>% sum

Ceci n'est pas un pipe!

Answer 3

Sort of, but I think it's un-idiomatic and maybe fragile/not a good idea. (This is implied, I think, by @RichieCotton's comment above.)

From http://cran.r-project.org/doc/manuals/R-lang.html :

10.3.4 Special operators

R allows user-defined infix operators. These have the form of a string of characters delimited by the ‘%’ character. The string can contain any printable character except ‘%’. The escape sequences for strings do not apply here.

Note that the following operators are predefined

 %% %*% %/% %in% %o% %x%

"%@%" <- function(x,f) {
    f(x)
}

sqr <- function(x) x^2
x <- 1:4

x %@% mean  ## 2.5
x %@% mean %@% sqr  ## 6.25
x %@% (mean %@% sqr)  ## fails

Given m as defined above -- maybe what you had in mind?

 m %@% colMeans %@% sum  ## 21

Notes:

• your example is a bit funny, because mean(x) always returns a scalar (i.e. a length-1 vector), so sum(mean(x)) is always going to be the same as mean(x)
• the infix operators have to be surrounded by %, so you can't have anything as compact as a single symbol (and %% is taken already).
• this sort of chaining is non-associative, which worries me -- it seems that the examples above work, so R is (apparently) evaluating left-to-right, but I don't know that that's guaranteed ...

edit: the question now asks how additional arguments can be incorporated. I don't think the syntax suggested (x %@% fun1(arg1) %@% fun2(arg2)) will work without some serious magic. This is the closest I can get at the moment -- creating a wrapper function that creates a modified version of the original function.

F <- function(f,...) {
    function(x) {
        f(x,...)
    }
}

Testing:

pow <- function(x,b=2) { x^b }
sqr <- function(x) x^2
x <- 1:4

x %@% F(mean,na.rm=TRUE)  ## 2.5
x %@% F(mean,na.rm=TRUE) %@% F(pow,3)  ## 16.25

(Note that I have used F as a function here, which may be dicey in some situations because it overwrites the F==FALSE shortcut)