在 LINQ 中生成 n 个集合的排列

发布于 2024-11-08 06:12:20 字数 666 浏览 4 评论 0原文

我有一个 IEnumerable 数组 (IEnumerable[]),希望生成这些 IEnumerable 中元素的所有可能组合。 它与此问题类似:使用 LINQ 生成排列 但我不知道我事先会有多少个 IEnumerable,因此无法使用所描述的 LINQ 语句。

举个例子: 我的数组

IEnumerable[] array;

例如,

array[0]={0,1,2};
array[1]={a,b};

具有这些元素然后我希望返回这些元素:

{{0,a},{0,b},{1,a},{1,b},{2,a},{2,b}}

但它也可能包含:

array[0]={0,1,2};
array[1]={a,b};
array[2]={w,x,y,z};

然后我需要适当的排列。我没有关于有多少 IEnumerable 的任何信息,也没有关于每个 IEnumerable 拥有多少元素的信息。

预先感谢您的帮助,

拉斯

I have an array of IEnumerable (IEnumerable[]) and would like to generate all possible combinations of the elements in these IEnumerables.
It is similar to this problem: Generating Permutations using LINQ
but I do not know how many IEnumerables I will have beforehand and thus cannot use the described LINQ statement.

To give an example:
My array

IEnumerable[] array;

has for example these elements

array[0]={0,1,2};
array[1]={a,b};

Then I would like these to be returned:

{{0,a},{0,b},{1,a},{1,b},{2,a},{2,b}}

But it might also hold:

array[0]={0,1,2};
array[1]={a,b};
array[2]={w,x,y,z};

Then I would need the appropriate permutations. I do not have any information on how many IEnumerables and no information on how many elements each IEnumerable holds.

Thanks in advance for any help,

Lars

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评论(2

勿忘心安 2024-11-15 06:12:20

看来您想要笛卡尔积。这应该可行,尽管我没有特别仔细地观察边缘情况

public static IEnumerable<T> Drop<T>(IEnumerable<T> list, long numToDrop)
{
    if (numToDrop < 0) { throw new ArgumentException("Number to drop must be non-negative"); }
    long dropped = 0;
    var iter = list.GetEnumerator();
    while (dropped < numToDrop && iter.MoveNext()) { dropped++; }
    while (iter.MoveNext()) { yield return iter.Current; }
}

public static IEnumerable Concat(object head, IEnumerable list)
{
    yield return head;
    foreach (var x in list) { yield return x; }
}

public static IEnumerable<IEnumerable> CrossProduct(IEnumerable<IEnumerable> lists)
{
    if (lists == null || lists.FirstOrDefault() == null) { yield break; }
    var heads = lists.First();
    var tails = CrossProduct(Drop(lists, 1));
    if (tails.FirstOrDefault() == null)
    {
    foreach (var h in heads) { yield return new object[] { h }; }
    }
    else
    {
    foreach (var head in heads)
    {
        foreach (var tail in tails)
        {
        yield return Concat(head, tail);
        }
    }
    }
}

Seems like you want the Cartesian_product. This should work, although I didn't look particularly carefully at edge-cases

public static IEnumerable<T> Drop<T>(IEnumerable<T> list, long numToDrop)
{
    if (numToDrop < 0) { throw new ArgumentException("Number to drop must be non-negative"); }
    long dropped = 0;
    var iter = list.GetEnumerator();
    while (dropped < numToDrop && iter.MoveNext()) { dropped++; }
    while (iter.MoveNext()) { yield return iter.Current; }
}

public static IEnumerable Concat(object head, IEnumerable list)
{
    yield return head;
    foreach (var x in list) { yield return x; }
}

public static IEnumerable<IEnumerable> CrossProduct(IEnumerable<IEnumerable> lists)
{
    if (lists == null || lists.FirstOrDefault() == null) { yield break; }
    var heads = lists.First();
    var tails = CrossProduct(Drop(lists, 1));
    if (tails.FirstOrDefault() == null)
    {
    foreach (var h in heads) { yield return new object[] { h }; }
    }
    else
    {
    foreach (var head in heads)
    {
        foreach (var tail in tails)
        {
        yield return Concat(head, tail);
        }
    }
    }
}
画骨成沙 2024-11-15 06:12:20

尝试将此作为方向(我确定您需要修改,而且我还没有编译它,因此可能存在一些语法错误)

public IEnumerable<string> CompileCominations
    (IEnumberable<string[]> arrays, List<string> combinations)
{
    if(combinations == null) combinations = new List<string>();
    for(int i = arrays.Count() - 1; i >= 0; i--)
    {
       if(combinations.Count >= 1) combinations = 
           Combine2Lists(combinations, arrays[i]);
       else 
       {
           combinations = Combine2Lists(arrays[i], arrays[i -1];
           i--;
       }
    }
    return combinations;
}

private List<string> Combine2Lists
    (IEnumberable<string> list1, IEnumerable<string> list2)
{
    List<string> currentList = new List<string>();
    for(int i = 0; i < list1.Count(); i ++)
    {
        for(int j = 0; j < list2.Count(); j++)
        {
            currentList.Add(
              string.Format("{0},{1}", list1[i], list2[j]);
        }
    }
    return currentList;
}

同样,未编译,但它应该是什么要做的就是继续将“item1, item2”添加到列表中,其中 item1 = 旧的“item1, item2”,item2 = 只是第 n 个数组中的新条目。

因此,字符串数组 [a, b, c] + 字符串数组 [1, 2, 3, 4] 应该产生: {a, 1}, {a, 2}, {a, 3}, {a, 4 }, {b, 1}... 并将字符串数组 [x, y, z] 添加到第一个 to 将产生: {a, 1, x}, {a, 1, y }、{a, 1, z} 等等。

Try this as a direction (you'll need to modify, I'm sure, and I haven't compiled it, so there may be some syntax errors)

public IEnumerable<string> CompileCominations
    (IEnumberable<string[]> arrays, List<string> combinations)
{
    if(combinations == null) combinations = new List<string>();
    for(int i = arrays.Count() - 1; i >= 0; i--)
    {
       if(combinations.Count >= 1) combinations = 
           Combine2Lists(combinations, arrays[i]);
       else 
       {
           combinations = Combine2Lists(arrays[i], arrays[i -1];
           i--;
       }
    }
    return combinations;
}

private List<string> Combine2Lists
    (IEnumberable<string> list1, IEnumerable<string> list2)
{
    List<string> currentList = new List<string>();
    for(int i = 0; i < list1.Count(); i ++)
    {
        for(int j = 0; j < list2.Count(); j++)
        {
            currentList.Add(
              string.Format("{0},{1}", list1[i], list2[j]);
        }
    }
    return currentList;
}

Again, not compiled, but what it should do is keep adding "item1, item2" to a list with a where item1 = the old "item1, item2" and item2 = just the new entry from the nth array.

Thus string array [a, b, c] + string array [1, 2, 3, 4] should yield: {a, 1}, {a, 2}, {a, 3}, {a, 4}, {b, 1}... and adding string array [x, y, z] to the first to would then yield: {a, 1, x}, {a, 1, y}, {a, 1, z} and so forth.

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