Swift: Generate random number between multiple ranges

Question

I want to generate a random number that can be between

-3...-8 and 3...8

I know I could find a roundabout way of doing this by first doing a random integer between 0 and 1 and then picking the range on that:

let zeroOrOne = CGFloat.random(in: 0...1)
if zeroOrOne == 0 {
    randomNum = CGFloat.random(in: -3...-8)
} else {
    randomNum = CGFloat.random(in: 3...8)
}

but I am wondering if there is a better way of doing this.

If anything, replace `let zeroOrOne = CGFloat.random(in: 0...1)` with `let posOrNeg = Bool.random()`. — rmaddy, Apr 15 '19 at 18:18
You might also use `let randomNum = CGFloat.random(in: 3...8) * (Bool.random() ? 1 : -1)` — Sulthan, Apr 15 '19 at 18:19
`let randomNum: CGFloat = .random(in: .random() ? -8 ... -3 : 3 ... 8)` You can even omit the Bool type — Leo Dabus, Apr 15 '19 at 18:33

score 3 · Answer 1 · answered Apr 15 '19 at 18:23

I am sure you want to use Bool.random.

One way to write it would be

let randomNum = CGFloat.random(in: 3...8) * (Bool.random() ? 1 : -1)

or

var randomNum = CGFloat.random(in: 3...8)
if Bool.random() {
   randomNum.negate()
}

There is no single correct solution.

Peter O. · Accepted Answer · 2019-04-25T18:10:32.373

There is a general solution for generating a random integer within several nonoverlapping ranges.

Find the number of integers in each range; this is that range's weight (e.g., if each range is a closed interval [x, y], this weight is (y - x) + 1).
Choose a range randomly according to the ranges' weights (see this question for how this can be done in Swift).
Generate a random integer in the chosen range: CGFloat.random(in: range).

Swift: Generate random number between multiple ranges

2 Answers2