生成具有成对数字的数组 [英] Generate array with pairs of numbers

问题描述

我有以下代码：

Random num = new Random();
            int check = CheckIfOdd(num.Next(1, 1000000));
            int counter = 1;

            while (check <= 0)
            {
                if (check % 2 == 0)
                {
                    check = CheckIfOdd(num.Next(1, 1000000)); ;
                }
                counter++;
            }
            int[] nArray = new int[check];
            int arLength = 0;
            //generate arrays with pairs of numbers, and one number which does not pair.
            for (int i = 0; i < check; i++)
            {
                arLength = nArray.Length;

                if (arLength == i + 1) 
                {
                    nArray[i] = i + 1;
                }
                else
                {
                    nArray[i] = i;
                    nArray[i + 1] = i;
                }
                i++;
            }

确实可以，但是不如我所愿。

which does kinda work, but not as well as i would like.

它应生成一个包含1-1百万个元素的数组，并且其中的数字可以在1-1十亿之间。

It should generate an array with between 1 - 1 million elements, and the numbers within can be between 1 - 1 billion.

它必须在数组中的随机位置（现在不是）对每个数字进行两对处理，然后它应该包含一个没有对的数字...

it has to make two pairs of each number, in random locations in the array ( which it doesn't now ) and then it should contain 1 number which has no pair...

我正在寻找一种更好的方法，因为它不在随机位置，并且不能正确生成1到10亿之间的数字。

I am just looking for a better way of doing it, since it isn't in random locations, and it doesn't generate numbers correctly between 1- 1 billion.

编辑
我被建议这样做：（由oerkelens）

Edit I have been suggested this: (by oerkelens)

var total = new Random().Next(500000) * 2 + 1;
            var nArray = new int[total];
            for (var i = 1; i < total; i += 2)
            {
                nArray[i] = i;
                nArray[i - 1] = i;
            }
            nArray[total - 1] = total;

哪个更好，但代码不是很多，但不会将值随机排列。

Which is better, and not as much code, but it doesn't place the values in random order.

编辑2
这几乎可以满足我的需要，但不能生成正确的金额。
如前所述，它最多可以生成x个元素，且数字在1-y之间

Edit 2 This almost does what i need, but it does not generate the right amount. as stated, it should generate up to x elements, with numbers between 1-y

Random r = new Random();
int[] output = Enumerable.Range(0, 11).Select(x => x / 2).OrderBy(x => r.Next()).ToArray();

通过谜语

推荐答案

尝试以下代码：

int[] output = Enumerable.Range(0, 11).Select(x => x / 2).ToArray();

它会产生一个具有以下值的数组：

It produces an array with these values:

{ 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5 }

您应该能够将其扩展为所需的任意数量的元素。

You should be able to extend this to as many elements you need.

如果希望随机输出顺序，然后尝试以下操作：

If you want the output in a random order then try this:

Random r = new Random();
int[] output = Enumerable.Range(0, 11).Select(x => x / 2).OrderBy(x => r.Next()).ToArray();

在一个运行中，例如，我得到了这个：

In one run, as an example, I got this:

{ 0, 4, 1, 2, 2, 4, 5, 3, 3, 1, 0 }

---

要生成具有单个元素的大量随机对，可以执行以下操作：

---

To produce a large number of random pairs with one single element you can do this:

Random r = new Random();
int pairs = 5; //elements = 2 * pairs + 1;
int max = 100;
int[] output =
    Enumerable
        .Range(0, pairs)
        .Select(x => r.Next(1, max + 1))
        .SelectMany(x => new [] { x, x })
        .StartWith(r.Next(1, max + 1))
        .OrderBy(x => r.Next())
        .ToArray();

但是，这不能保证您不会遇到3、4的碰撞

However, this doesn't guarantee that you don't end up with collisions of 3, 4, or more, number clashes.

这不需要 System.Interactive：

This doesn't require "System.Interactive":

int[] output =
    new [] { r.Next(1, max + 1) }
        .Concat(
            Enumerable
                .Range(0, pairs)
                .Select(x => r.Next(1, max + 1))
                .SelectMany(x => new [] { x, x }))
        .OrderBy(x => r.Next())
        .ToArray();

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