ios - 生成具有给定分布的随机数

Question

看看这个问题:

Swift probability of random number being selected?

最佳答案建议使用 switch 语句来完成这项工作。但是，如果我要考虑的情况非常多，代码看起来就很不优雅；我有一个巨大的 switch 语句，在每种情况下都一遍又一遍地重复非常相似的代码。

当您需要考虑大量概率时，是否有更好、更简洁的方法来以一定的概率选择随机数？ (大约 30 个)

Answer 1

这是一个 Swift 实现，深受各种Generate random numbers with a given (numerical) distribution的答案.

对于 Swift 4.2/Xcode 10 及更高版本(内联解释):

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)
}

例子:

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

以 0.2 的概率返回 0，以 0.3 的概率返回 1，和 2，概率为 0.5。

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

以 1/3 的概率返回 0，以 2/3 的概率返回 1。

---

对于 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)
}

---

对于 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)
}