Generate random numbers with a given distribution

Check out this question:

Swift probability of random number being selected?

The top answer suggests to use a switch statement, which does the job. However, if I have a very large number of cases to consider, the code looks very inelegant; I have a giant switch statement with very similar code in each case repeated over and over again.

Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider? (like ~30)

This is a Swift implementation strongly influenced by the various answers to Generate random numbers with a given (numerical) distribution.

For Swift 4.2/Xcode 10 and later (explanations inline):

func randomNumber(probabilities: [Double]) -> Int {

    // Sum of all probabilities (so that we don't have to require that the sum is 1.0):
    let sum = probabilities.reduce(0, +)
    // Random number in the range 0.0 <= rnd < sum :
    let rnd = Double.random(in: 0.0 ..< sum)
    // Find the first interval of accumulated probabilities into which `rnd` falls:
    var accum = 0.0
    for (i, p) in probabilities.enumerated() {
        accum += p
        if rnd < accum {
            return i
        }
    }
    // This point might be reached due to floating point inaccuracies:
    return (probabilities.count - 1)
}

Examples:

let x = randomNumber(probabilities: [0.2, 0.3, 0.5])

returns 0 with probability 0.2, 1 with probability 0.3, and 2 with probability 0.5.

let x = randomNumber(probabilities: [1.0, 2.0])

return 0 with probability 1/3 and 1 with probability 2/3.

For Swift 3/Xcode 8:

func randomNumber(probabilities: [Double]) -> Int {

    // Sum of all probabilities (so that we don't have to require that the sum is 1.0):
    let sum = probabilities.reduce(0, +)
    // Random number in the range 0.0 <= rnd < sum :
    let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
    // Find the first interval of accumulated probabilities into which `rnd` falls:
    var accum = 0.0
    for (i, p) in probabilities.enumerated() {
        accum += p
        if rnd < accum {
            return i
        }
    }
    // This point might be reached due to floating point inaccuracies:
    return (probabilities.count - 1)
}

For Swift 2/Xcode 7:

func randomNumber(probabilities probabilities: [Double]) -> Int {

    // Sum of all probabilities (so that we don't have to require that the sum is 1.0):
    let sum = probabilities.reduce(0, combine: +)
    // Random number in the range 0.0 <= rnd < sum :
    let rnd = sum * Double(arc4random_uniform(UInt32.max)) / Double(UInt32.max)
    // Find the first interval of accumulated probabilities into which `rnd` falls:
    var accum = 0.0
    for (i, p) in probabilities.enumerate() {
        accum += p
        if rnd < accum {
            return i
        }
    }
    // This point might be reached due to floating point inaccuracies:
    return (probabilities.count - 1)
}

Is there a nicer, cleaner way to pick a random number with a certain probability when you have a large number of probabilities to consider?

Sure. Write a function that generates a number based on a table of probabilities. That's essentially what the switch statement you've pointed to is: a table defined in code. You could do the same thing with data using a table that's defined as a list of probabilities and outcomes:

probability    outcome
-----------    -------
   0.4            1
   0.2            2
   0.1            3
   0.15           4
   0.15           5

Now you can pick a number between 0 and 1 at random. Starting from the top of the list, add up probabilities until you've exceeded the number you picked, and use the corresponding outcome. For example, let's say the number you pick is 0.6527637. Start at the top: 0.4 is smaller, so keep going. 0.6 (0.4 + 0.2) is smaller, so keep going. 0.7 (0.6 + 0.1) is larger, so stop. The outcome is 3.

I've kept the table short here for the sake of clarity, but you can make it as long as you like, and you can define it in a data file so that you don't have to recompile when the list changes.

Note that there's nothing particularly specific to Swift about this method -- you could do the same thing in C or Swift or Lisp.

This seems like a good opportunity for a shameless plug to my small library, swiftstats: https://github.com/r0fls/swiftstats

For example, this would generate 3 random variables from a normal distribution with mean 0 and variance 1:

import SwiftStats
let n = SwiftStats.Distributions.Normal(0, 1.0)
print(n.random())

Supported distributions include: normal, exponential, binomial, etc...

It also supports fitting sample data to a given distribution, using the Maximum Likelihood Estimator for the distribution.

See the project readme for more info.