Question

考虑这个List<string>

List<string> data = new List<string>();
data.Add("Text1");
data.Add("Text2");
data.Add("Text3");
data.Add("Text4");

我遇到的问题是：如何获得列表子集的每个组合？有点像这样：

#Subset Dimension 4
Text1;Text2;Text3;Text4

#Subset Dimension 3
Text1;Text2;Text3;
Text1;Text2;Text4;
Text1;Text3;Text4;
Text2;Text3;Text4;

#Subset Dimension 2
Text1;Text2;
Text1;Text3;
Text1;Text4;
Text2;Text3;
Text2;Text4;

#Subset Dimension 1
Text1;
Text2;
Text3;
Text4;

我提出了一个很好的解决方案，值得在这里分享。

Answer 1

与Abaco的回答类似，不同的实施......

foreach (var ss in data.SubSets_LB())
{
    Console.WriteLine(String.Join("; ",ss));
}

public static class SO_EXTENSIONS
{
    public static IEnumerable<IEnumerable<T>> SubSets_LB<T>(
      this IEnumerable<T> enumerable)
    {
        List<T> list = enumerable.ToList();
        ulong upper = (ulong)1 << list.Count;

        for (ulong i = 0; i < upper; i++)
        {
            List<T> l = new List<T>(list.Count);
            for (int j = 0; j < sizeof(ulong) * 8; j++)
            {
                if (((ulong)1 << j) >= upper) break;

                if (((i >> j) & 1) == 1)
                {
                    l.Add(list[j]);
                }
            }

            yield return l;
        }
    }
}

Answer 2

我认为，这个问题的答案需要一些性能测试。我会试一试。它是社区维基，随时可以更新它。

void PerfTest()
{
    var list = Enumerable.Range(0, 21).ToList();

    var t1 = GetDurationInMs(list.SubSets_LB);
    var t2 = GetDurationInMs(list.SubSets_Jodrell2);
    var t3 = GetDurationInMs(() => list.CalcCombinations(20));

    Console.WriteLine("{0}\n{1}\n{2}", t1, t2, t3);
}

long GetDurationInMs(Func<IEnumerable<IEnumerable<int>>> fxn)
{
    fxn(); //JIT???
    var count = 0;

    var sw = Stopwatch.StartNew();
    foreach (var ss in fxn())
    {
        count = ss.Sum();
    }
    return sw.ElapsedMilliseconds;
}

输出：

1281
1604 (_Jodrell not _Jodrell2)
6817

Jodrell的更新

我内置了发布模式，即优化。当我通过Visual Studio运行时，我不会在1或2之间得到一致的偏差，但是在重复运行之后LB的答案获胜，我得到了接近类似的答案，

1190
1260
more

但如果我从命令行运行测试工具，而不是通过Visual Studio运行，我会得到更像这样的结果

987
879
still more

Answer 3

修改

我已经接受了性能挑战，接下来是我的合并，它取得了最好的答案。在我的测试中，它似乎具有最佳性能。

public static IEnumerable<IEnumerable<T>> SubSets_Jodrell2<T>( this IEnumerable<T> source) { var list = source.ToList(); var limit = (ulong)(1 << list.Count); for (var i = limit; i > 0; i--) { yield return list.SubSet(i); } } private static IEnumerable<T> SubSet<T>( this IList<T> source, ulong bits) { for (var i = 0; i < source.Count; i++) { if (((bits >> i) & 1) == 1) { yield return source[i]; } } }

同样的想法，几乎与L.B's answer相同，但我自己的解释。

我避免使用内部List和Math.Pow。

public static IEnumerable<IEnumerable<T>> SubSets_Jodrell( this IEnumerable<T> source) { var count = source.Count(); if (count > 64) { throw new OverflowException("Not Supported ..."); } var limit = (ulong)(1 << count) - 2; for (var i = limit; i > 0; i--) { yield return source.SubSet(i); } } private static IEnumerable<T> SubSet<T>( this IEnumerable<T> source, ulong bits) { var check = (ulong)1; foreach (var t in source) { if ((bits & check) > 0) { yield return t; } check <<= 1; } }

你会注意到这些方法在初始集中不能使用超过64个元素，但无论如何它开始需要一段时间。

Answer 4

我为列表开发了一个简单的ExtensionMethod：

    /// <summary>
    /// Obtain all the combinations of the elements contained in a list
    /// </summary>
    /// <param name="subsetDimension">Subset Dimension</param>
    /// <returns>IEnumerable containing all the differents subsets</returns>
    public static IEnumerable<List<T>> CalcCombinations<T>(this List<T> list, int subsetDimension)
    {
        //First of all we will create a binary matrix. The dimension of a single row
        //must be the dimension of list 
        //on which we are working (we need a 0 or a 1 for every single element) so row
        //dimension is to obtain a row-length = list.count we have to
        //populate the matrix with the first 2^list.Count binary numbers
        int rowDimension = Convert.ToInt32(Math.Pow(2, list.Count));

        //Now we start counting! We will fill our matrix with every number from 1 
        //(0 is meaningless) to rowDimension
        //we are creating binary mask, hence the name
        List<int[]> combinationMasks = new List<int[]>();
        for (int i = 1; i < rowDimension; i++)
        {
            //I'll grab the binary rapresentation of the number
            string binaryString = Convert.ToString(i, 2);

            //I'll initialize an array of the apropriate dimension
            int[] mask = new int[list.Count];

            //Now, we have to convert our string in a array of 0 and 1, so first we 
            //obtain an array of int then we have to copy it inside our mask 
            //(which have the appropriate dimension), the Reverse()
            //is used because of the behaviour of CopyTo()
            binaryString.Select(x => x == '0' ? 0 : 1).Reverse().ToArray().CopyTo(mask, 0);

            //Why should we keep masks of a dimension which isn't the one of the subset?
            // We have to filter it then!
            if (mask.Sum() == subsetDimension) combinationMasks.Add(mask);
        }

        //And now we apply the matrix to our list
        foreach (int[] mask in combinationMasks)
        {
            List<T> temporaryList = new List<T>(list);

            //Executes the cycle in reverse order to avoid index out of bound
            for (int iter = mask.Length - 1; iter >= 0; iter--)
            {
                //Whenever a 0 is found the correspondent item is removed from the list
                if (mask[iter] == 0)
                    temporaryList.RemoveAt(iter);
            }
            yield return temporaryList;
        }
    }
}

所以考虑问题中的例子：

# Row Dimension of 4 (list.Count)
Binary Numbers to 2^4

# Binary Matrix
0 0 0 1 => skip
0 0 1 0 => skip
[...]
0 1 1 1 => added // Text2;Text3;Text4
[...]
1 0 1 1 => added // Text1;Text3;Text4
1 1 0 0 => skip
1 1 0 1 => added // Text1;Text2;Text4
1 1 1 0 => added // Text1;Text2;Text3
1 1 1 1 => skip

希望这可以帮助某人：）

如果您需要澄清，或者您希望随时添加答案或评论（哪一个更合适）。

如何获得子集的所有可能组合？

4 个答案: