在特定范围内的 JavaScript 中生成随机整数?

时间:2023-03-15
本文介绍了在特定范围内的 JavaScript 中生成随机整数?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

如何在 JavaScript 中的两个指定变量之间生成随机整数,例如x = 4y = 8 会输出任何 4, 5, 6, 7, 8?

How can I generate random whole numbers between two specified variables in JavaScript, e.g. x = 4 and y = 8 would output any of 4, 5, 6, 7, 8?

推荐答案

Mozilla 开发者网络 页面:

/**
 * Returns a random number between min (inclusive) and max (exclusive)
 */
function getRandomArbitrary(min, max) {
    return Math.random() * (max - min) + min;
}

/**
 * Returns a random integer between min (inclusive) and max (inclusive).
 * The value is no lower than min (or the next integer greater than min
 * if min isn't an integer) and no greater than max (or the next integer
 * lower than max if max isn't an integer).
 * Using Math.round() will give you a non-uniform distribution!
 */
function getRandomInt(min, max) {
    min = Math.ceil(min);
    max = Math.floor(max);
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

<小时>

这是它背后的逻辑.这是一个简单的三法则:


Here's the logic behind it. It's a simple rule of three:

Math.random() 返回一个介于 0(包括)和 1(不包括)之间的 Number.所以我们有一个这样的区间:

Math.random() returns a Number between 0 (inclusive) and 1 (exclusive). So we have an interval like this:

[0 .................................... 1)

现在,我们想要一个介于 min(包括)和 max(不包括)之间的数字:

Now, we'd like a number between min (inclusive) and max (exclusive):

[0 .................................... 1)
[min .................................. max)

我们可以使用 Math.random 来获取 [min, max) 区间内的对应者.但是,首先我们应该通过从第二个间隔中减去 min 来考虑问题:

We can use the Math.random to get the correspondent in the [min, max) interval. But, first we should factor a little bit the problem by subtracting min from the second interval:

[0 .................................... 1)
[min - min ............................ max - min)

这给出了:

[0 .................................... 1)
[0 .................................... max - min)

我们现在可以应用 Math.random 然后计算对应值.让我们选择一个随机数:

We may now apply Math.random and then calculate the correspondent. Let's choose a random number:

                Math.random()
                    |
[0 .................................... 1)
[0 .................................... max - min)
                    |
                    x (what we need)

所以,为了找到 x,我们会这样做:

So, in order to find x, we would do:

x = Math.random() * (max - min);

别忘了把min加回来,这样我们就可以得到一个在[min, max)区间内的数字:

Don't forget to add min back, so that we get a number in the [min, max) interval:

x = Math.random() * (max - min) + min;

这是 MDN 的第一个功能.第二个,返回 minmax 之间的整数,两者都包含.

That was the first function from MDN. The second one, returns an integer between min and max, both inclusive.

现在要获取整数,您可以使用 roundceilfloor.

Now for getting integers, you could use round, ceil or floor.

您可以使用 Math.round(Math.random() * (max - min)) + min,但这会产生非均匀分布.minmax 都只有大约一半的滚动机会:

You could use Math.round(Math.random() * (max - min)) + min, this however gives a non-even distribution. Both, min and max only have approximately half the chance to roll:

min...min+0.5...min+1...min+1.5   ...    max-0.5....max
└───┬───┘└────────┬───────┘└───── ... ─────┘└───┬──┘   ← Math.round()
   min          min+1                          max

max排除在区间之外,滚动的机会比min还要小.

With max excluded from the interval, it has an even less chance to roll than min.

使用 Math.floor(Math.random() * (max - min +1)) + min 你有一个完全均匀的分布.

With Math.floor(Math.random() * (max - min +1)) + min you have a perfectly even distribution.

min.... min+1... min+2 ... max-1... max.... max+1 (is excluded from interval)
|        |        |         |        |        |
└───┬───┘└───┬───┘└─── ... ┘└───┬───┘└───┬───┘   ← Math.floor()
   min     min+1               max-1    max

您不能在该等式中使用 ceil()-1 因为 max 现在滚动的机会略少,但是您也可以滚动(不需要的)min-1 结果.

You can't use ceil() and -1 in that equation because max now had a slightly less chance to roll, but you can roll the (unwanted) min-1 result too.

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