I am trying to create a vectorial function in R using the command ifelse as follows
temp <- function(p){
ifelse(p < 0.5 *(1 + 0.5), (1 + 0.5) * qnorm(p/(1 +0.5)), (1 - 0.5) * qnorm((p - 0.5)/(1 - 0.5)))
}
The function works well with scalar values, for example
> temp(0.1)
[1] -2.251629
> temp(0.9)
[1] 0.4208106
but not for vectors:
> temp(c(0.1,0.9))
[1] -2.2516289 0.4208106
Warning message:
In qnorm((p - 0.5)/(1 - 0.5)) : NaNs produced
The weird thing is that it returns the right answer, but indicating a warning.
What am I doing wrong? It seems that the ifelse is evaluating both functions in all the entries of the vector p, which is supposed to be avoided with this command.
ifelse is basically doing this:
p<- c(.1,.9)
a<-(1 + 0.5) * qnorm(p/(1 +0.5))
b<- (1 - 0.5) * qnorm((p - 0.5)/(1 - 0.5))
c<-NULL
c[which(p < 0.5 *(1 + 0.5))] <-a[which(p < 0.5 *(1 + 0.5))]
c[which(!(p < 0.5 *(1 + 0.5)))] <-b[which(!(p < 0.5 *(1 + 0.5)))]
That is, it creates a vector for 'yes' and a vector for 'no'. The 'no' vector it creates throws the warning.
The examples in the documentation allude to this.
x <- c(6:-4)
sqrt(x) #- gives warning
sqrt(ifelse(x >= 0, x, NA)) # no warning
## Note: the following also gives the warning !
ifelse(x >= 0, sqrt(x), NA)
この記事はインターネットから収集されたものであり、転載の際にはソースを示してください。
侵害の場合は、連絡してください[email protected]
コメントを追加