Using ifelse in R when one of the options produces NAs?

user_8384166 0
Viewed 302

I want to vectorize a function that relies on checking a condition and depending on whether this condition is TRUE or FALSE, return the outcome of one of two functions, respectively. The problem is that, when the condition is FALSE, the first function cannot be evaluated. Then, ifelse returns the correct values but it also produces a warning. I would like to produce a function that does not produce warnings.

I have tried ifelse(), but it does not work. I was expecting that this command would skip the evaluation of the first function when the condition is FALSE.

Here is an illustrative piece of R code

p = c(-1,1,-1,1,-1,-1,-1,1)

ifelse(p>0, sqrt(p), p^2)

which returns

[1] 1 1 1 1 1 1 1 1
Warning message:
In sqrt(p) : NaNs produced

As you can see, the outcome is correct but, for some reason, it evaluates the function at the first function when condition is FALSE. Thus, I would like to somehow avoid this issue.

if-statement r function
2 Answers

We can create a numeric vector and then fill the elements based on the condition put forward by 'p'

out <- numeric(length(p))
out[p > 0] <- sqrt(p[p > 0])
out[p <= 0] <- p[p <= 0]^2

With ifelse we need to have all arguments of the same length. According to ?ifelse

ifelse(test, yes, no)

A vector of the same length and attributes (including dimensions and "class") as test and data values from the values of yes or no

What happens is that we do both the calculations on the entire vector and replace the values of 'p' based on the test condition. For sqrt, the negative values definitely gives warning and output as NaN. While the NaN elements don't show up in the output, the warning was already printed. The warning is a friendly one, but can be suppressed with suppressWarnings

user_37322710

Avoidance through ifelse probably isn't possible. My understanding of the ifelse process is

  1. Create a vector of values based on the expression in yes
  2. Create a vector of values based on the expression in no
  3. Use the result of test to decide whether each element comes from yes or no.

If an error will occur in either yes or no, ifelse will fail.

To get around this, you need to only evaluate expressions where they will succeed. (such as in akrun's answer, a variant of which is given here for completeness)

p = c(-1,1,-1,1,-1,-1,-1,1)

condition <- p > 0

result <- numeric(length(p))

result[g1] <- sqrt(p[condition])
result[!g1] <- p[condition]^2
user_10172760
Related