R help - Dec 2002 - applying a different function to rows of a dataframe

Here is a simple example of what I would like to do:

Given a data frame foo with variables x and fn. Suppose fn is a vector of
characters which correspond to  names of previously defined  functions
which have only one argument. I would like a vector returned where fn is
applied to x

foo <-  data.frame(x=c(2,5,7), fn = letters[c(6,7,6)])
foo$fn <- as.character(foo$fn)

 "f" <- function(x){17*x}
 "g" <- function(x){3*x}


foo
  x fn
1 2  f
2 5  g
3 7  f

wanted:  a function which returns c(f(2),g(5),f(7)) = c(34, 15, 119)

A simple application of do.call doesn't do what I need (it applies the
function f to each x): 
> do.call(foo$fn,args=list(x = foo$x))
[1]  34  85 119

Whereas the following works but seems like overkill.
> diag(sapply(foo$fn,do.call,args=list(x=foo$x)))
[1]  34  15 119

Another idea that didn't work:

 do.call("evalq",args
=list(paste(foo$fn,"(foo$x)",sep="")))
[1] "f(foo$x)" "g(foo$x)" "f(foo$x)"

Thanks in advance
Anne

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Anne E. York
National Marine Mammal Laboratory
Seattle WA 98115-0070  USA
e-mail: anne.york at noaa.gov
Voice: +1 206-526-4039
Fax: +1 206-526-6615
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Looks to me like you want

sapply(1:nrow(foo), function(i) get(foo$fn[i])(foo$x[i]))

or the equivalent for loop.

On Mon, 16 Dec 2002, Anne York wrote:
> Here is a simple example of what I would like to do:
>
> Given a data frame foo with variables x and fn. Suppose fn is a vector of
> characters which correspond to  names of previously defined  functions
> which have only one argument. I would like a vector returned where fn is
> applied to x
>
> foo <-  data.frame(x=c(2,5,7), fn = letters[c(6,7,6)])
> foo$fn <- as.character(foo$fn)
>
>  "f" <- function(x){17*x}
>  "g" <- function(x){3*x}
>
>
> foo
>   x fn
> 1 2  f
> 2 5  g
> 3 7  f
>
> wanted:  a function which returns c(f(2),g(5),f(7)) = c(34, 15, 119)
>
> A simple application of do.call doesn't do what I need (it applies the
> function f to each x):
>
> > do.call(foo$fn,args=list(x = foo$x))
> [1]  34  85 119
>
> Whereas the following works but seems like overkill.
>
> > diag(sapply(foo$fn,do.call,args=list(x=foo$x)))
> [1]  34  15 119
>
> Another idea that didn't work:
>
>  do.call("evalq",args
=list(paste(foo$fn,"(foo$x)",sep="")))
> [1] "f(foo$x)" "g(foo$x)" "f(foo$x)"
>
> Thanks in advance
> Anne
>
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> Anne E. York
> National Marine Mammal Laboratory
> Seattle WA 98115-0070  USA
> e-mail: anne.york at noaa.gov
> Voice: +1 206-526-4039
> Fax: +1 206-526-6615
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> http://www.stat.math.ethz.ch/mailman/listinfo/r-help
>
-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595