Calculating truncated polynomials in R

Question

I am trying to write a function based on this formula.

(x − x0)^ r+ = { (x-x0)^r   if x>x0
                    0       otherwise

what I understood from above is;

y= (x-x0)^r unless x<= x0

so for each element of, identify if greater and x0 if so return y that equals the formula.

tp <- function(x,x0,r) {
  y<-list()
  if (x[i] >x0) {
    for (i in seq_along(x)) {
      y<- append((x[i]-x0)^r)
    } else {
      y <- append(0)
    }
    
  }
  return(y)
}

I have tried doing this but I couldn't make it work. Could anyone advise me if I understood the formula right and if so what is the correct way to coed it.

What about: `tp <- function(x, x0, r) ifelse(x > x0, (x-x0) ^ r, 0)`? — harre, Dec 01 '22 at 12:56

score 4 · Answer 1 · answered Dec 01 '22 at 12:58

4

An alternative to the if else statement is to use boolean logic, which I like. But both will work.

tp <- function(x, x0, r) {
#  ifelse(x > x0, (x - x0)^r, 0)
  (x > x0)*(x - x0)^r
}

x0 <- 2
tp(-2:5, x0, r = 2)
#> [1] 0 0 0 0 0 1 4 9

answered Dec 01 '22 at 12:58

DashdotdotDashdotdot

449
1
9

Nice! Due to several issues with `ifelse` specifically (e.g., https://stackoverflow.com/q/6668963/3358272), I prefer to shy away from its use. Granted, it's fine for most things, but ... it's class-stripping and inconsistencies are frustrating. Boolean logic should always be faster, another perk to this. – r2evans Dec 01 '22 at 17:54

score 2 · Answer 2 · answered Dec 01 '22 at 12:54

2

Use ifelse.

tp <- function(x, x0, r) {
  ifelse(x > x0, (x - x0)^r, 0)
}

x0 <- 2
tp(-2:5, x0, r = 2)
#> [1] 0 0 0 0 0 1 4 9

^{Created on 2022-12-01 with reprex v2.0.2}

answered Dec 01 '22 at 12:54

Rui Barradas

70,273
8
34
66

score 1 · Answer 3 · answered Dec 01 '22 at 12:57

Issues with your function:

append needs a list or vector to which the values are appended, so you need to do y <- append(y, ...);
Your for expression should be before (outside) the first (x[i] > x0) conditional; and
Perhaps subjective, but I suggest that y <- numeric(0) makes more sense to me than list. There are certainly times when a list makes sense, but this is not one of them (in my perspective). Again, not wrong, but later processing will likely be more efficient as a vector.

Perhaps something like:

tp <- function(x,x0,r) {
  y <- numeric(0)
  for (i in seq_along(x)) {
    if (x[i] > x0) {
      y <- append(y, (x[i]-x0)^r)
    } else {
      y <- append(y, 0)
    }
  }
  return(y)
}

In R, it is almost always preferred to use vectorized operations instead of for loops. In this case, I think you can easily use the vectorized function that RuiBarradas provided (which is the preferred method for writing this kind of function).

Calculating truncated polynomials in R

3 Answers3