R help - May 2011 - Evaluating a multivariable function XXXX

>
> Hello everyone,
>
> I have the following R code for a multivariable function:
>
>
> > fn2<-function(x,y,z){(y+2*z)/(5*y-x*z)}
> >
> > fn2(-5,-2,3)
> [1] 0.8
>
>
>
> No problems.
>
> ==>
> If, however, I call the function using a vector substitution for the
> arguments, R sees this as 3 separate calls to the function while supplying
> only the first argument:
>
> > in2<-c(-5,-2,3)
> > in2
> [1] -5 -2  3
> >
> > fn2(in2)
> Error in fn2(in2) : argument "y" is missing, with no default
>
> ==>
> How should I call the function using the vector substitution method so that
> R sees that this is a single call to the function that supplies all 3 of
the
> arguments?
>
> Thank you,
>
> Dan
>
>
>
>
>
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On 11-05-07 4:06 AM, Dan Abner wrote:
>>
>> Hello everyone,
>>
>> I have the following R code for a multivariable function:
>>
>>
>>> fn2<-function(x,y,z){(y+2*z)/(5*y-x*z)}
>>>
>>> fn2(-5,-2,3)
>> [1] 0.8
>>
>>
>>
>> No problems.
>>
>> ==>>
>> If, however, I call the function using a vector substitution for the
>> arguments, R sees this as 3 separate calls to the function while
supplying
>> only the first argument:
>>
>>> in2<-c(-5,-2,3)
>>> in2
>> [1] -5 -2  3
>>>
>>> fn2(in2)
>> Error in fn2(in2) : argument "y" is missing, with no default
>>
>> ==>>
>> How should I call the function using the vector substitution method so
that
>> R sees that this is a single call to the function that supplies all 3
of the
>> arguments?
The simplest way is to write a new wrapper function:

fn3 <- function(xyz) fn2(xyz[1], xyz[2], xyz[3])

You can also do it with do.call:

do.call(fn2, as.list(in2))

Duncan Murdoch