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Modifying a function programmatically

5 messages · Søren Højsgaard, Pascal Oettli, Gabor Grothendieck

#
Dear list

# I have a function 
 ff  <- function(a,b=2,c=4){a+b+c}
# which I programmatically want to modify to a more specialized function in which a is replaced by 1
 ff1 <- function(b=2,c=4){1+b+c}

# I do as follows:
 vals  <- list(a=1)
 (expr1 <- as.expression(body(ff)))
expression({
    a + b + c
})
 (expr2 <- do.call("substitute", list(expr1[[1]], vals)))
{
    1 + b + c
}
 
 # This "works", 
 eval(expr2, list(b=10,c=12))
[1] 23
 
 # - but I would like a function. Hence I do:
 ll  <- alist(b=, c=, eval(expr2))
 ff2 <- as.function(ll)
 ff2(b=10,c=12)
[1] 23
 
 # BUT I am only half-way where I want to be because the alist(b=, c=, ...) 
 # requires me plugin the remaining formals by hand. I do:
 newformals <-setdiff(names(formals(ff)), names(vals))
 vv         <- vector("list", length(newformals))
 names(vv)  <- newformals
 (hh <- c(vv, expr2))
$b
NULL
$c
NULL
[[3]]
{
    1 + b + c
}

 (ff3 <- as.function(hh))
function (b = NULL, c = NULL) 
{
    1 + b + c
}
 ff3(10,12)
[1] 23
 # Hence I get the function that returns what I want (given the correct input) 
 # but the arguments will have default values that I don't want. 
 # I want to retain the original default values (if any)
 ff1()
[1] 7
 ff3()
numeric(0)

I have two questions: 
1) How to fix the above?
2) Aren't there more elegant alternatives?

Any help would be appreciated.
Regards
S?ren
#
Hi,

Is it what you are looking for?

 > ff  <- function(a,b,c){a+b+c}

 > ff(1,10,12)
[1] 23

 > ff(589,2,4)
[1] 595

HTH,
Pascal


Le 04/02/2013 19:00, S?ren H?jsgaard a ?crit :
#
No, this is not really what I want. But I have (later) realized that what I am looking for can be achieved as shown below. I am just not sure how bullet-proof the solution is. Suggestions are welcome!
Regards
S?ren

------------------------------------------

specialize <- function(ff, vals){
  expr1 <- as.expression(body(ff))
  expr2 <- do.call("substitute", list(expr1[[1]], vals))
  gg  <- formals(ff)
  idx <-match(names(vals), names(gg))
  gg  <- gg[-idx]
  as.function(c(gg, expr2))
  }

 ff  <- function(a,b=2,c=4){a+b+c}
 (ans<-specialize(ff, list(a=1)))
function (b = 2, c = 4) 
{
    1 + b + c
}
<environment: 0x0000000008b51c68>
 ans()
[1] 7
 ans(b=10)
[1] 15

 (ans<-specialize(rnorm, list(n=10)))
function (mean = 0, sd = 1) 
.Internal(rnorm(10, mean, sd))
<environment: 0x0000000008b5a0b8>
 ans()
 [1]  0.1783645319  2.1151344735 -0.6744270178  0.0009751497 -0.1477708270  0.1710790055
 [7] -0.0064331250  0.5935787352 -0.1426050990 -0.1390478788
 ans(sd=.0001)
 [1]  1.386604e-06 -6.616540e-05  6.592326e-05 -1.299638e-04  1.322658e-05 -4.515388e-05
 [7] -2.809639e-05 -5.719492e-05  1.554319e-04 -1.186992e-04




-----Original Message-----
From: Pascal Oettli [mailto:kridox at ymail.com] 
Sent: 4. februar 2013 11:11
To: S?ren H?jsgaard
Cc: r-help at r-project.org
Subject: Re: [R] Modifying a function programmatically

Hi,

Is it what you are looking for?

 > ff  <- function(a,b,c){a+b+c}

 > ff(1,10,12)
[1] 23

 > ff(589,2,4)
[1] 595

HTH,
Pascal


Le 04/02/2013 19:00, S?ren H?jsgaard a ?crit :
#
On Mon, Feb 4, 2013 at 5:00 AM, S?ren H?jsgaard <sorenh at math.aau.dk> wrote:
This a currying operation. Try:

library(functional)
ff1 <- Curry(ff, a = 1)

# test
identical(ff1(2, 4), ff(1, 2, 4)) # TRUE

--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
#
Dear Gabor,

Thanks for pointing me to this; didn't know the Curry function. 

For what it is worth, the two approaches perform somewhat differently in terms of computing time (my "specialize" function is given at the end):

library(functional)
## EXAMPLE 1
 ff  <- function(a,b=2,c=4){a+b+c}
 ff1 <- specialize(ff, vals=list(a=1,bb=123))
 ff2 <- Curry(ff, a = 1)
 rbenchmark::benchmark(ff1(b=10), ff2(b=10), replications=100000)
         test replications elapsed relative user.self sys.self user.child sys.child
1 ff1(b = 10)       100000    0.39    1.000      0.39        0         NA        NA
2 ff2(b = 10)       100000    0.81    2.077      0.79        0         NA        NA
 
## EXAMPLE 2
 gg <- rnorm
 gg1 <- specialize(gg, list(n=10))
 gg2 <- Curry(gg, n=1000)
 rbenchmark::benchmark(gg1(), gg2(), replications=100000)
   test replications elapsed relative user.self sys.self user.child sys.child
1 gg1()       100000    0.53    1.000      0.53     0.00         NA        NA
2 gg2()       100000    9.25   17.453      9.22     0.01         NA        NA 

where

specialize <- function(ff, vals){
	expr1 <- as.expression(body(ff))
	expr2 <- do.call("substitute", list(expr1[[1]], vals))
	gg  <- formals(ff)
	idx <-match(names(vals), names(gg))
	idx <- idx[!is.na(idx)]
	if (length(idx)>0){	gg  <- gg[-idx]}
	as.function(c(gg, expr2))
}

Best regards
S?ren










-----Original Message-----
From: Gabor Grothendieck [mailto:ggrothendieck at gmail.com] 
Sent: 4. februar 2013 17:31
To: S?ren H?jsgaard
Cc: r-help at r-project.org
Subject: Re: [R] Modifying a function programmatically
On Mon, Feb 4, 2013 at 5:00 AM, S?ren H?jsgaard <sorenh at math.aau.dk> wrote:
This a currying operation. Try:

library(functional)
ff1 <- Curry(ff, a = 1)

# test
identical(ff1(2, 4), ff(1, 2, 4)) # TRUE

--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com