如何在Java中创建随机BigDecimal？ [英] How can I create a random BigDecimal in Java?

问题描述

这个问题：如何生成随机BigInteger 描述了一种实现与BigIntegers的Random.nextInt（int n）相同语义的方法。

This question: How to generate a random BigInteger describes a way to achieve the same semantics as Random.nextInt(int n) for BigIntegers.

我想对BigDecimal和Random.nextDouble做同样的事情（ ）。

I would like to do the same for BigDecimal and Random.nextDouble().

上述问题的一个答案建议创建一个随机的BigInteger，然后用随机比例从中创建一个BigDouble。一个非常快速的实验表明这是一个非常糟糕的主意：）

One answer in the above question suggests creating a random BigInteger and then creating a BigDouble from it with a random scale. A very quick experiment shows this to be a very bad idea :)

我的直觉是使用这种方法需要通过像 n-log10（R），其中n是输出中所需精度的位数，R是随机BigInteger。这应该允许存在正确的位数，以便（例如）1 - > 10 ^ -64和10 ^ 64 - > 1.

My intuition is that using this method would require the integer to be scaled by something like n-log10(R), where n is the number of digits of precision required in the output and R is the random BigInteger. This should allow the correct number of digits to be present so that (for example) 1 -> 10^-64 and 10^64 -> 1.

缩放值还需要正确选择结果，使其落在[0,1]范围内。

The scaling value also needs to be chosen correctly for the result to fall in the range [0,1].

之前是否有人这样做过，并且他们知道是否结果是否正确分布？有没有更好的方法来实现这一目标？

Has anyone done this before, and do they know if the results are correctly distributed? Is there a better way to achieve this?

编辑：感谢@biziclop纠正我对scale参数的理解。以上不是必需的，恒定的比例因子具有预期的效果。

Thanks to @biziclop for correcting my understanding of the scale argument. The above isn't necessary, a constant scale factor has the desired effect.

供以后参考，我的（显然工作代码）是：

For later reference, my (apparently working code) is:

private static BigDecimal newRandomBigDecimal(Random r, int precision) {
    BigInteger n = BigInteger.TEN.pow(precision);
    return new BigDecimal(newRandomBigInteger(n, r), precision);
}

private static BigInteger newRandomBigInteger(BigInteger n, Random rnd) {
    BigInteger r;
    do {
        r = new BigInteger(n.bitLength(), rnd);
    } while (r.compareTo(n) >= 0);

    return r;
}

推荐答案

这肯定很容易..如果我只知道你想要什么。对于范围[0,1]中的均匀分布数和精度N十进制数​​，生成小于10 * N的统一BigInteger，并将其缩小10 * N.

It's surely very easy... if I only knew what you want. For a uniformly distributed number in range [0, 1) and precision N decimal digits generate a uniform BigInteger less than 10*N and scale it down by 10*N.

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