生成给定分布的随机数

Question

最重要的答案是建议使用switch语句来完成这项工作。但是，如果我要考虑的情况很多，那么代码看起来就很笨拙。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。

当您要考虑的概率很大时，是否有更好，更干净的方法来选择具有一定概率的随机数？（例如〜30）

Answer 1

这是一个Swift实现，受各种答案的影响很大，这些答案会生成具有给定（数字）分布的随机数

对于 Swift 4.2 / Xcode 10 和更高版本（嵌入式解释）：

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])

返回概率为0.2的0，概率为0.3的1，概率为0.5的2。

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)
}