k13 - 2 months ago 6x

R 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`

`b`

`d`

`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.

Source (Stackoverflow)

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