根据区间获取权重的所有排列

发布于 2024-11-26 17:43:17 字数 553 浏览 2 评论 0原文

如果我有给定数量的“系统”

var nbSystems = 2;

和多个“步骤”，

var nbSteps = 3;

我如何生成以下所有排列：

{0.00, 0.00}
{0.33, 0.00}
{0.67, 0.00}
{1.00, 0.00}
{0.00, 0.33}
{0.33, 0.33}
{0.67, 0.33}
{1.00, 0.33}
{0.00, 0.66}
{0.33, 0.66}
{0.67, 0.66}
{1.00, 0.66}
{0.00, 1.00}
{0.33, 1.00}
{0.67, 1.00}
{1.00, 1.00}

我需要每个系统的权重在 0 到 1 之间，并且我想找到所有排列。

您会看到用于系统权重的间隔是根据 nbSteps 计算得出的（间隔 = 1 / nbSteps）。

我想要一个函数，它接受输入 nbSystems 和 nbSteps （或更好？），并且将输出某种具有所有排列的多维数组......

原文

If I have a given number of 'systems'

var nbSystems = 2;

and a number of 'steps'

var nbSteps = 3;

How can I generate all the following permutations :

{0.00, 0.00}
{0.33, 0.00}
{0.67, 0.00}
{1.00, 0.00}
{0.00, 0.33}
{0.33, 0.33}
{0.67, 0.33}
{1.00, 0.33}
{0.00, 0.66}
{0.33, 0.66}
{0.67, 0.66}
{1.00, 0.66}
{0.00, 1.00}
{0.33, 1.00}
{0.67, 1.00}
{1.00, 1.00}

I need a weight between 0 to 1 for each system and I want to find all permutations.

You see that the interval to use for the systems weights is calculated from the nbSteps (interval = 1 / nbSteps).

I would like to have a function that takes inputs nbSystems and nbSteps (or better?) and that will output some sort of multidimensional array with all the permutations...

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枯寂 2024-12-03 17:43:17

这应该可以做到：

static void Main(string[] args)
{
    var nbSystems = 2;
    var nbSteps = 3;

    var steps = GetSteps(0, 1, nbSteps).Select(n => Math.Round(n, 2)).ToArray();
    foreach (var seq in GetCombinations(steps, nbSystems))
        Console.WriteLine(string.Join(", ", seq));
}

private static IEnumerable<decimal> GetSteps(decimal min, decimal max, int count)
{
    var increment = (max - min) / count;
    return Enumerable.Range(0, count + 1).Select(n => min + increment * n);
}

private static IEnumerable<IEnumerable<T>> GetCombinations<T>(
    ICollection<T> choices, int length)
{
    if (length == 0)
    {
        yield return new T[0];
        yield break;
    }

    foreach (var choice in choices)
        foreach (var suffix in GetCombinations(choices, length - 1))
            yield return Enumerable.Concat(new[] { choice }, suffix);
}

如果您关心组合的顺序，请根据需要进行调整。

编辑：哎呀，修正了一个拼写错误 - 我在类型签名中调换了“max”和“min”。

This should do it:

static void Main(string[] args)
{
    var nbSystems = 2;
    var nbSteps = 3;

    var steps = GetSteps(0, 1, nbSteps).Select(n => Math.Round(n, 2)).ToArray();
    foreach (var seq in GetCombinations(steps, nbSystems))
        Console.WriteLine(string.Join(", ", seq));
}

private static IEnumerable<decimal> GetSteps(decimal min, decimal max, int count)
{
    var increment = (max - min) / count;
    return Enumerable.Range(0, count + 1).Select(n => min + increment * n);
}

private static IEnumerable<IEnumerable<T>> GetCombinations<T>(
    ICollection<T> choices, int length)
{
    if (length == 0)
    {
        yield return new T[0];
        yield break;
    }

    foreach (var choice in choices)
        foreach (var suffix in GetCombinations(choices, length - 1))
            yield return Enumerable.Concat(new[] { choice }, suffix);
}

Adjust as necessary if you care about the order of the combinations.

EDIT: Oops, fixed a typo - I transposed "max" and "min" in my type signature.

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痴者 2024-12-03 17:43:17

尝试

double nbSteps = 3.0;

IEnumerable<double> SystemA = Enumerable.Range (0, (int)nbSteps).Select (x => x / nbSteps); 
IEnumerable<double> SystemB = Enumerable.Range (0, (int)nbSteps).Select (x => x / nbSteps); 

var Result = from a in SystemA from b in SystemB select new { a, b };

try

double nbSteps = 3.0;

IEnumerable<double> SystemA = Enumerable.Range (0, (int)nbSteps).Select (x => x / nbSteps); 
IEnumerable<double> SystemB = Enumerable.Range (0, (int)nbSteps).Select (x => x / nbSteps); 

var Result = from a in SystemA from b in SystemB select new { a, b };

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