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生成 1 到 100 之间的唯一随机数

发布于 2024-08-24 14:30:00 字数 63 浏览 4 评论 0原文

如何使用 JavaScript 生成 1 到 100 之间的一些唯一随机数字？

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不知在何时 2024-08-31 14:30:01

将其实现为生成器使其使用起来非常愉快。请注意，此实现与需要首先对整个输入数组进行混洗的实现不同。

此示例函数以惰性方式工作，每次迭代为您提供1个随机项目，最多可达您要求的N个项目。这很好，因为如果您只想从 1000 列表中选择 3 项，则不必先触摸所有 1000 项。

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let ys = xs.slice(0);
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield ys.splice(i,1)[0];
    n--; len--;
  }
}

// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];

// get 3 random items
for (let i of sample(3) (items))
  console.log(i); // f g c

// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
  console.log(i); // 3 8 7

// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]

我选择以不改变输入数组的方式实现示例，但您可以很容易地认为改变实现是有利的。

例如，shuffle 函数可能希望改变原始输入数组。或者您可能希望在不同时间从相同的输入中进行采样，每次都更新输入。

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield xs.splice(i,1)[0];
    n--; len--;
  }
}

// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));

// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');

// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))

console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]

// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]

由于数组输入突变，sample 不再是一个纯函数，但在某些情况下（如上所示）它可能更有意义。

我选择生成器而不是仅返回数组的函数的另一个原因是，您可能希望继续采样，直到出现某些特定条件。

也许我想要 1,000,000 个随机数列表中的第一个素数。

“我应该采样多少个？” – 您不必指定
“我是否必须先找到所有素数，然后选择一个随机素数？” – 不需要。

因为我们使用的是生成器，所以这个任务很简单。

const randomPrimeNumber = listOfNumbers => {
  for (let x of sample(Infinity) (listOfNumbers)) {
    if (isPrime(x))
      return x;
  }
  return NaN;
}

它将一次连续采样 1 个随机数 x，检查它是否为素数，如果是则返回 x是。如果在找到素数之前耗尽数字列表，则返回 NaN。

注意：

此答案最初是在另一个问题上共享的，该问题作为与此问题的重复项而被关闭。因为它与此处提供的其他解决方案非常不同，所以我决定也在这里分享

Implementing this as a generator makes it pretty nice to work with. Note, this implementation differs from ones that require the entire input array to be shuffled first.

This sample function works lazily, giving you 1 random item per iteration up to N items you ask for. This is nice because if you just want 3 items from a list of 1000, you don't have to touch all 1000 items first.

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let ys = xs.slice(0);
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield ys.splice(i,1)[0];
    n--; len--;
  }
}

// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];

// get 3 random items
for (let i of sample(3) (items))
  console.log(i); // f g c

// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
  console.log(i); // 3 8 7

// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]

I chose to implement sample in a way that does not mutate the input array, but you could easily argue that a mutating implementation is favourable.

For example, the shuffle function might wish to mutate the original input array. Or you might wish to sample from the same input at various times, updating the input each time.

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield xs.splice(i,1)[0];
    n--; len--;
  }
}

// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));

// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');

// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))

console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]

// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]

sample is no longer a pure function because of the array input mutation, but in certain circumstances (demonstrated above) it might make more sense.

Another reason I chose a generator instead of a function that just returns an array is because you may want to continue sampling until some specific condition.

Perhaps I want the first prime number from a list of 1,000,000 random numbers.

"How many should I sample?" – you don't have to specify
"Do I have to find all the primes first and then select a random prime?" – Nope.

Because we're working with a generator, this task is trivial

const randomPrimeNumber = listOfNumbers => {
  for (let x of sample(Infinity) (listOfNumbers)) {
    if (isPrime(x))
      return x;
  }
  return NaN;
}

This will continuously sample 1 random number at a time, x, check if it's prime, then return x if it is. If the list of numbers is exhausted before a prime is found, NaN is returned.

Note:

This answer was originally shared on another question that was closed as a duplicate of this one. Because it's very different from the other solutions provided here, I've decided to share it here as well

生成 1 到 100 之间的唯一随机数

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