Calling a procedure

Thanks. So, to be safe, always a good idea to give the argument, e.g.,
q=1.96, log.p=FALSE, skipping mean=0 and sd=1 if not needed. Thanks.

pnorm(q=1.96, log.p = FALSE)

On 2020/9/20 ?? 06:36, Rui Barradas wrote:

Hello,

You are making a confusion between

1. the formal argument log.p
2. the variable log.p

In the function body, log.p is a variable that exists in the
function's frame, not the formal argument of pnorm.
The first and the 3rd calls that follow output the same value.

try(x = 1.2, log.p = TRUE)$a
try(x = 1.2, log.p = TRUE)$b
try(x = 1.2, 1)$a

This is because in the function

  a<-pnorm(x,log.p)       # first call

passes log.p as the *second* argument, not as a value for pnorm's
formal argument log.p. Unless when named, the arguments are passed in
the order they appear in the function's definition:

pnorm(q, mean = 0, sd = 1, lower.tail = TRUE, log.p = FALSE)

and that becomes

  a<-pnorm(x,TRUE)       # first call
  a<-pnorm(x,1)          # first call, coerced to numeric.


Let me give another example. In the function that follows the default
is z = FALSE.

In the first call the name z is not the name of the argument, it's the
name of a variable that exists in the .GlobalEnv.

In the second call, z = z assign the formal argument z the value of
the variable z.


f <- function(x, y = 0, z = FALSE){
  a <- x
  b <- y
  d <- z
  list(a = a, b = b, d = d)
}
z <- 2
f(1, z)
f(1, z = z)


Hope this helps,

Rui Barradas

?s 11:11 de 20/09/20, Steven Yen escreveu:

Can someone tell me a proper call to a procedure, in this case,
pnorm. In what follows, I had expected a = b, but they are not equal.
What are wrong with first call and second call? Thank you!

try<-function(x,log.p=FALSE){
a<-pnorm(x,log.p)       # first call
b<-pnorm(x,log.p=log.p) # second call
list(a=a,b=b)
}

try(x=1.2,log.p=TRUE)$a
try(x=1.2,log.p=TRUE)$b

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