Suming vectors in R?

Question

I have this two functions:

Hwave_1 <- function(t) {
  ifelse(0 <= t & t <= 1 / 2, t, ifelse(1 / 2 < t & t <= 1, 1 - t, 0))
}


HaarI <-function(m,k,t){
  ifelse(t > 1 | t < 0, "HaarErrorTimes", 2^((-m)/2)*Hwave_1((2^m)*t-k))
}

I'm trying to create a function as in the image

where the Z_nj=rnorm(1)

I tried to do some naive things like

simul_weiner<-function(t,m){
  for(i in 0:(2^m-1)){result<-sum(rnorm(1)*HaarI(m,i,t))}
  ifelse(n=0,rnorm(1)*HaarI(0,0,t),result+rnorm(1)*HaarI(0,0,t))
}

but it clearly doesnt work because sum() is summing every entry, I dont know if there exist a workaround or another sum function that can help me make the code work.

EDIT: Steps to reproduce

#create times vector
times<-seq(from=0, to=1,by=2^(-5))
#call the function on times with any m you choose, i choose 2 for simplicity
simul_weiner(times,2)

It should output a list with the same number repeating.

Both the `yes=` and `no=` arguments of `ifelse` **must** be the same class, even if `ifelse` does not enforce it. In this case, your second function will always return a `character` vector unless all of the `t` values are inside `(0,1)`. — r2evans, Dec 06 '20 at 01:32
Beyond that, it's difficult to really know since we know nothing of `m` and do not know how you are calling `simul_weiner`. It would help if the question was more *reproducible*, please see https://stackoverflow.com/q/5963269, [mcve], and https://stackoverflow.com/tags/r/info. Thanks! — r2evans, Dec 06 '20 at 01:35
Yea sorry I should've clarified im evaluating the function on (0,1) so there shouldn't be a problem with that. — Sebastian Cor, Dec 06 '20 at 01:35
Regardless, `HaarI` is just a ticking time-bomb as long as `"HaarErrorTimes"` is in it. Alternatives: use `NA_real_`; or start the function with `stopifnot(all(t >= 0 & t <= 1))`. — r2evans, Dec 06 '20 at 01:37

score 1 · Accepted Answer · answered Dec 06 '20 at 02:40

Looking at

  for(i in 0:(2^m-1)){result<-sum(rnorm(1)*HaarI(m,i,t))}

Realize that before the for loop runs, result is undefined. When i is 0 (first pass), it is defined as the singular value resulting from sum. When i is 1 (second pass), the value of result is over-written with the singular value resulting from this call to sum. Etc. When you finish your for loop, result contains the last sum calculated.

Perhaps you should instead use

  result <- sapply(0:(2^m-1), function(i) sum(rnorm(1)*HaarI(m,i,t)))

Next problem:

ifelse(n=0, ...)

That conditional is fundamentally broken. It is syntactically not an error, but it will always always always be false, because n=0 is an assignment, which in this case will invisibly return 0, which is considered FALSE for ifelse. Furthermore, n is undefined, so it should really be producing an error.

If n is supposed to be m, then ifelse is not needed, as m is length 1. The fact that it includes t in it (with a length greater than 1 in your example) suggests that this is not the case.

If n is a variable in the enclosing environment (not in this function), then ... don't do that. It's sloppy, not reproducible, breaks in unpredictable ways, difficult to troubleshoot, and condition enough for me to suspect any code using it.

'n' was supposed to 'm' you're right that 'ifelse' is not needed. I understand what you say, I'll try and get back to you. — Sebastian Cor, Dec 06 '20 at 02:45

Suming vectors in R?

1 Answers1