Question: 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)

```

`new_f` would be a function of `b` and `d` and would return `1+b+2+d`

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
    f
}

```

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

Answer: 
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)
f
# 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.
You have also `Curry` from package `functional`:

```
library(functional)

f <- function(a, b, c, d) a+b+c+d
ff = Curry(f, a=2, c=10)

ff(1,5)
#[1] 18

ff(b=1,d=5)
#[1] 18

```