[R] ifelse and piecewise function

[David Winsemius]

On Jan 1, 2010, at 7:44 PM, roger king wrote:

> I am a novice user of "R" and  I'm learning with R version  2.8.1,  
> using
> WinEdt_1.8.1, under Widows Vista Home Version.
>
> ## The test function below, from a vector input, returns vector  
> values:
> #   and it contains an "ifelse"statement
> TEST<- function(x) {
> low<- -x^2
> up<-   x^4
> ifelse(x>=0,up,low )
>                   }
> u<- seq(-1,1,0.5)
> TEST(u)
> ###########################################################
> # #The following function, also containing "ifelse", returns correct
> individual values
> #  (confirmed independently from another program).
> #  Vector input, however, generates the following message:
> #  "In if (is.finite(lower)) { :the condition has length > 1 and  
> only the
> first element
> #   will be used"
> #  And, in fact, correct multiple values are not returned.
> ########################################################
> ASEPF<- function(y, theta, sigma, alpha, kappa){
>
> loww<-((theta - y)/(sigma* kappa))^alpha
>
> upp<- (kappa *(y - theta)/sigma)^alpha
>
> k<- 1/(1 + kappa^2)
>
> a<- 1/alpha
>
> integrand<- function(y){y^(a-1)* exp(-y)/gamma(1/alpha)}
>
> GL<- integrate(integrand, lower = loww, upper= Inf)$value
>
> GU<- integrate(integrand, lower = upp,  upper= Inf)$value
> ifelse(y>=0,1 - k*GU, (kappa^2) *k * GL)
>                                               }
> ASEPF(u,0,1,2,0.5)
> Could someone kindly give me some advice on how to
> rewrite the function ASEPF so that it returns a vector of values?

But, but, but, ... it did return a vector of values. You failed to  
recognize them:

ASEPF(u,0,1,2,0.5)
[1] 0.000935539 0.000935539 0.616399902 0.616399902 0.616399902

The above line is the vector of values that you did not post.

Then you also got.....
Warning messages:
1: In if (is.finite(lower)) { :
   the condition has length > 1 and only the first element will be used
2: In if (is.finite(lower)) { :
   the condition has length > 1 and only the first element will be used

You need to figure out why your invocation of integrate is causing  
complaints (only warnings, mind you) from the R interpreter, but the  
problem is not with any failure to return a vector.

-- 
David.




>
> many thanks,
>
> Rik King
>
> rking681 at gmail.com
>
> (ex Mathematics Department,
> University of Western Sydney, Australia)
>
> 	[[alternative HTML version deleted]]
>
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT