k13 k13 - 1 year ago 66
R Question

How to efficiently partially apply a function in R?

Suppose I have a function in R that takes multiple arguments, and I'd like to reduce it to a function of fewer arguments by setting some of the arguments to pre-specified values. I'm trying to figure out what is the best way to do this is in R.

For example, suppose I have a function

f <- function(a,b,c,d){a+b+c+d}

I'd like to create or find a function partial that would do the following

partial <- function(f, ...){
#fill in code here
new_f <- partial(f, a=1, c= 2)

would be a function of
and would return

In python I would do

from functools import partial

def f(a,b,c,d):
return a+b+c+d

new_f = partial(f, a=1, c= 2)

I'm actually doing this repeatedly and so I need for this to be as efficient as possible. Can anyone point me to the most efficient way to do this? Right now the best I can do is

partial <- function(f, ...){
z <- list(...)
formals(f) [names(z)] <- z

Can anyone let me know of a faster way or the best way to do this? This is simply too slow.

Answer Source

There are functions in the pryr package that can handle this, namely partial()

f <- function(a, b, c, d) a + b + c + d 
pryr::partial(f, a = 1, c = 2)
# function (...) 
# f(a = 1, c = 2, ...)

So you can use it like this -

new_fun <- pryr::partial(f, a = 1, c = 2)
new_fun(b = 2, d = 5)
# [1] 10
## or if you are daring ...
new_fun(2, 5)
# [1] 10

You could also simply change f()'s formal arguments with

f <- function(a, b, c, d) a + b + c + d 
formals(f)[c("a", "c")] <- list(1, 2)
# function (a = 1, b, c = 2, d) 
# a + b + c + d
f(b = 2, d = 5)
# [1] 10

But with the latter, you must name the b and d arguments in f() to avoid an error when you want to leave a and c as their default values.

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