Question

我的C＃程序中有一个整数列表。但是，我只在运行时知道列表中的项目数。

让我们说，为了简单起见，我的清单是{1,2,3} 现在我需要生成所有可能的组合，如下所示。 {1,2,3} {1,2} {1,3} {2,3} {1} {2} {3}

有人可以帮忙吗？

Answer 1

试试这个：

static void Main(string[] args)
{

    GetCombination(new List<int> { 1, 2, 3 });
}

static void GetCombination(List<int> list)
{
    double count = Math.Pow(2, list.Count);
    for (int i = 1; i <= count - 1; i++)
    {
        string str = Convert.ToString(i, 2).PadLeft(list.Count, '0');
        for (int j = 0; j < str.Length; j++)
        {
            if (str[j] == '1')
            {
                Console.Write(list[j]);
            }
        }
        Console.WriteLine();
    }
}

Answer 2

以下是强类型列表的两个通用解决方案，它们将返回列表成员的所有唯一组合（如果您可以使用更简单的代码解决此问题，我向您致敬）：

T-SQL

Answer 3

这是使用递归的通用解决方案

public static ICollection<ICollection<T>> Permutations<T>(ICollection<T> list) {
    var result = new List<ICollection<T>>();
    if (list.Count == 1) { // If only one possible permutation
        result.Add(list); // Add it and return it
        return result;
    }
    foreach (var element in list) { // For each element in that list
        var remainingList = new List<T>(list);
        remainingList.Remove(element); // Get a list containing everything except of chosen element
        foreach (var permutation in Permutations<T>(remainingList)) { // Get all possible sub-permutations
            permutation.Add(element); // Add that element
            result.Add(permutation);
        }
    }
    return result;
}

我知道这是一篇旧帖子，但有人可能会觉得这很有帮助。

Answer 4

这个答案使用与ojlovecd相同的算法和（对于他的迭代解决方案）jaolho。我要添加的唯一选项是过滤组合中最少数量项目的结果。这可能很有用，例如，如果您只对包含至少两个项目的组合感兴趣。

编辑：根据@ user3610374的要求，添加了最大项目数的过滤器。

编辑2：正如@stannius所建议的那样，算法已被更改，以使其在不需要所有组合的情况下更有效。

  /// <summary>
  /// Method to create lists containing possible combinations of an input list of items. This is 
  /// basically copied from code by user "jaolho" on this thread:
  /// http://stackoverflow.com/questions/7802822/all-possible-combinations-of-a-list-of-values
  /// </summary>
  /// <typeparam name="T">type of the items on the input list</typeparam>
  /// <param name="inputList">list of items</param>
  /// <param name="minimumItems">minimum number of items wanted in the generated combinations, 
  ///                            if zero the empty combination is included,
  ///                            default is one</param>
  /// <param name="maximumItems">maximum number of items wanted in the generated combinations,
  ///                            default is no maximum limit</param>
  /// <returns>list of lists for possible combinations of the input items</returns>
  public static List<List<T>> ItemCombinations<T>(List<T> inputList, int minimumItems = 1, 
                                                  int maximumItems = int.MaxValue)
  {
     int nonEmptyCombinations = (int)Math.Pow(2, inputList.Count) - 1;
     List<List<T>> listOfLists = new List<List<T>>(nonEmptyCombinations + 1);

     // Optimize generation of empty combination, if empty combination is wanted
     if (minimumItems == 0)
        listOfLists.Add(new List<T>());

     if (minimumItems <= 1 && maximumItems >= inputList.Count)
     {
        // Simple case, generate all possible non-empty combinations
        for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
           listOfLists.Add(GenerateCombination(inputList, bitPattern));
     }
     else
     {
        // Not-so-simple case, avoid generating the unwanted combinations
        for (int bitPattern = 1; bitPattern <= nonEmptyCombinations; bitPattern++)
        {
           int bitCount = CountBits(bitPattern);
           if (bitCount >= minimumItems && bitCount <= maximumItems)
              listOfLists.Add(GenerateCombination(inputList, bitPattern));
        }
     }

     return listOfLists;
  }

  /// <summary>
  /// Sub-method of ItemCombinations() method to generate a combination based on a bit pattern.
  /// </summary>
  private static List<T> GenerateCombination<T>(List<T> inputList, int bitPattern)
  {
     List<T> thisCombination = new List<T>(inputList.Count);
     for (int j = 0; j < inputList.Count; j++)
     {
        if ((bitPattern >> j & 1) == 1)
           thisCombination.Add(inputList[j]);
     }
     return thisCombination;
  }

  /// <summary>
  /// Sub-method of ItemCombinations() method to count the bits in a bit pattern. Based on this:
  /// https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan
  /// </summary>
  private static int CountBits(int bitPattern)
  {
     int numberBits = 0;
     while (bitPattern != 0)
     {
        numberBits++;
        bitPattern &= bitPattern - 1;
     }
     return numberBits;
  }

Answer 5

使用Linq和递归的另一种解决方案......

static void Main(string[] args)
    {
        List<List<long>> result = new List<List<long>>();

        List<long> set = new List<long>() { 1, 2, 3, 4 };

        GetCombination<long>(set, result);

        result.Add(set);

        IOrderedEnumerable<List<long>> sorted = result.OrderByDescending(s => s.Count);

        sorted.ToList().ForEach(l => { l.ForEach(l1 => Console.Write(l1 + " ")); Console.WriteLine(); });
    }

    private static void GetCombination<T>(List<T> set, List<List<T>> result)
    {
        for (int i = 0; i < set.Count; i++)
        {
            List<T> temp = new List<T>(set.Where((s, index) => index != i));

            if (temp.Count > 0 && !result.Where(l => l.Count == temp.Count).Any(l => l.SequenceEqual(temp)))
            {
                result.Add(temp);

                GetCombination<T>(temp, result);
            }
        }
    }

Answer 6

首先，给定一组n个元素，可以计算出k个元素的所有组合（nCk）。您必须将k的值从1更改为n才能满足您的要求。

有关生成组合的C＃代码，请参阅此codeproject article。

如果您有兴趣自行开发组合算法，请查看SO question，其中包含大量相关材料的链接。

Answer 7

这是@ojlovecd答案的改进，无需使用字符串。

    static void Main(string[] args)
    {
        GetCombination(new List<int> { 1, 2, 3 });
    }


    private static void GetCombination(List<int> list)
    {
        double count = Math.Pow(2, list.Count);
        for (int i = 1; i <= count - 1; i++)
        {
            for (int j = 0; j < list.Count; j++)
            {
                int b = i & (1 << j);
                if (b > 0)
                {
                    Console.Write(list[j]);
                }
            }
            Console.WriteLine();
        }
    }

Answer 8

protected List<List<T>> AllCombos<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
    {
        List<List<T>> results = new List<List<T>>();
        List<T> workingWith = items.ToList();
        results.Add(workingWith);
        items.ToList().ForEach((x) =>
        {
            results.Add(new List<T>() { x });
        });
        for (int i = 0; i < workingWith.Count(); i++)
        {
            T removed = workingWith[i];
            workingWith.RemoveAt(i);
            List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
            results.AddRange(nextResults);
            workingWith.Insert(i, removed);
        }
        results = results.Where(x => x.Count > 0).ToList();
        for (int i = 0; i < results.Count; i++)
        {
            List<T> list = results[i];
            if (results.Where(x => comparer(x, list)).Count() > 1)
            {
                results.RemoveAt(i);
            }
        }

        return results;
    }

    protected List<List<T>> AllCombos2<T>(Func<List<T>, List<T>, bool> comparer, params T[] items)
    {
        List<List<T>> results = new List<List<T>>();
        List<T> workingWith = items.ToList();
        if (workingWith.Count > 1)
        {
            results.Add(workingWith);
        }
        for (int i = 0; i < workingWith.Count(); i++)
        {
            T removed = workingWith[i];
            workingWith.RemoveAt(i);
            List<List<T>> nextResults = AllCombos2(comparer, workingWith.ToArray());
            results.AddRange(nextResults);
            workingWith.Insert(i, removed);
        }
        results = results.Where(x => x.Count > 0).ToList();
        for (int i = 0; i < results.Count; i++)
        {
            List<T> list = results[i];
            if (results.Where(x => comparer(x, list)).Count() > 1)
            {
                results.RemoveAt(i);
            }
        }

        return results;
    }

这对我有用，它稍微复杂一点，实际上需要一个比较器回调功能，它实际上有2个功能，不同之处在于AllCombos明确地添加了单个项目列表。这是非常原始的，绝对可以减少，但它完成了工作。欢迎任何重构建议。谢谢，

Answer 9

public class CombinationGenerator{
    private readonly Dictionary<int, int> currentIndexesWithLevels = new Dictionary<int, int>();
    private readonly LinkedList<List<int>> _combinationsList = new LinkedList<List<int>>();
    private readonly int _combinationLength;

    public CombinationGenerator(int combinationLength)
    {
        _combinationLength = combinationLength;
    }

    private void InitializeLevelIndexes(List<int> list)
    {
        for (int i = 0; i < _combinationLength; i++)
        {
            currentIndexesWithLevels.Add(i+1, i);
        }
    }

    private void UpdateCurrentIndexesForLevels(int level)
    {
        int index;
        if (level == 1)
        {
            index = currentIndexesWithLevels[level];
            for (int i = level; i < _combinationLength + 1; i++)
            {
                index = index + 1;
                currentIndexesWithLevels[i] = index;
            }
        }
        else
        {
            int previousLevelIndex;
            for (int i = level; i < _combinationLength + 1; i++)
            {
                if (i > level)
                {
                    previousLevelIndex = currentIndexesWithLevels[i - 1];
                    currentIndexesWithLevels[i] = previousLevelIndex + 1;
                }
                else
                {
                    index = currentIndexesWithLevels[level];
                    currentIndexesWithLevels[i] = index + 1;
                }
            }
        }
    }

    public void FindCombinations(List<int> list, int level, Stack<int> stack)
    {
        int currentIndex;
        InitializeLevelIndexes(list);
        while (true)
        {
            currentIndex = currentIndexesWithLevels[level];
            bool levelUp = false;          
            for (int i = currentIndex; i < list.Count; i++)
            {
                if (level < _combinationLength)
                {
                    currentIndex = currentIndexesWithLevels[level];
                    MoveToUpperLevel(ref level, stack, list, currentIndex);
                    levelUp = true;
                    break;
                }
                levelUp = false;
                stack.Push(list[i]);
                if (stack.Count == _combinationLength)
                {
                    AddCombination(stack);
                    stack.Pop();
                }                                                                                 
            }

            if (!levelUp)
            {
                MoveToLowerLevel(ref level, stack, list, ref currentIndex);
                while (currentIndex >= list.Count - 1)
                {
                    if (level == 1)
                    {
                        AdjustStackCountToCurrentLevel(stack, level);
                        currentIndex = currentIndexesWithLevels[level];
                        if (currentIndex >= list.Count - 1)
                        {
                            return;
                        }
                        UpdateCurrentIndexesForLevels(level);
                    }
                    else
                    {
                        MoveToLowerLevel(ref level, stack, list, ref currentIndex);
                    }
              }
          }                               
       }
    }

    private void AddCombination(Stack<int> stack)
    {
        List<int> listNew = new List<int>();
        listNew.AddRange(stack);
        _combinationsList.AddLast(listNew);
    }

    private void MoveToUpperLevel(ref int level, Stack<int> stack, List<int> list, int index)
    {
        stack.Push(list[index]);
        level++;
    }

    private void MoveToLowerLevel(ref int level, Stack<int> stack, List<int> list, ref int currentIndex)
    {
        if (level != 1)
        {
            level--;
        }
        AdjustStackCountToCurrentLevel(stack, level);
        UpdateCurrentIndexesForLevels(level);
        currentIndex = currentIndexesWithLevels[level];
    }

    private void AdjustStackCountToCurrentLevel(Stack<int> stack, int currentLevel)
    {
        while (stack.Count >= currentLevel)
        {
            if (stack.Count != 0)
                stack.Pop();
        }
    }

    public void PrintPermutations()
    {
        int count = _combinationsList.Where(perm => perm.Count() == _combinationLength).Count();
        Console.WriteLine("The number of combinations is " + count);
    }

}

Answer 10

我们可以使用递归来处理涉及字符串或整数的组合/置换问题。

public static void Main(string[] args)
{
    IntegerList = new List<int> { 1, 2, 3, 4 };

    PrintAllCombination(default(int), default(int));
}

public static List<int> IntegerList { get; set; }

public static int Length { get { return IntegerList.Count; } }

public static void PrintAllCombination(int position, int prefix)
{
    for (int i = position; i < Length; i++)
    {
        Console.WriteLine(prefix * 10 + IntegerList[i]);
        PrintAllCombination(i + 1, prefix * 10 + IntegerList[i]);
    }

}

Answer 11

怎么样？

data['consomation_day'] = data.groupby(['day','site'])['consomation'].transform(lambda v: v.mean())

Answer 12

使用C＃7为Linq稍微更通用的版本。这里使用具有两个元素的项目进行过滤。

library(caret) # V6.0-79
library(recipes) # V0.1.2
library(MASS) # V7.3-47
# transform variables using recipes
rec_box <- recipe(~ ., data = as.data.frame(state.x77)) %>% 
  step_BoxCox(., everything()) %>% 
  prep(., training = as.data.frame(state.x77)) %>% 
  bake(., as.data.frame(state.x77)) 

> head(rec_box)
# A tibble: 6 x 8
  Population Income Illiteracy `Life Exp` Murder `HS Grad` Frost  Area
       <dbl>  <dbl>      <dbl>      <dbl>  <dbl>     <dbl> <dbl> <dbl>
1       8.19   138.     0.647   60171653.   6.89      651.   20.  56.0
2       5.90   185.     0.376   61218586.   5.52     1632.  152. 106. 
3       7.70   155.     0.527   66409311.   4.08     1253.   15.  69.4
4       7.65   133.     0.570   66885876.   5.05      609.   65.  56.4
5       9.96   165.     0.0936  71570875.   5.13     1445.   20.  75.5
6       7.84   161.    -0.382   73188251.   3.62     1503.  166.  67.7

# transform variables using preProcess
pre_box <- preProcess(x = as.data.frame(state.x77), method = c('BoxCox')) %>% 
  predict(. ,newdata = as.data.frame(state.x77)) 

> head(pre_box)
    # A tibble: 6 x 8
      Population Income Illiteracy `Life Exp` Murder `HS Grad` Frost  Area
           <dbl>  <dbl>      <dbl>      <dbl>  <dbl>     <dbl> <dbl> <dbl>
    1       8.19   118.     0.642       2383.   6.83      618.   20.  38.7
    2       5.90   157.     0.374       2401.   5.47     1538.  152.  65.7
    3       7.70   133.     0.524       2488.   4.05     1183.   15.  46.3
    4       7.65   114.     0.566       2496.   5.01      579.   65.  38.9
    5       9.96   141.     0.0935      2571.   5.09     1363.   20.  49.7
    6       7.84   138.    -0.383       2596.   3.60     1418.  166.  45.4


## Subtract recipe transformations from MARS::boxcox via caret::preProcess
colMeans(rec_box - pre_box)

> colMeans(rec_box - pre_box)
  Population       Income   Illiteracy     Life Exp       Murder      HS Grad        Frost         Area 
0.000000e+00 2.215800e+01 2.515464e-03 6.803437e+07 2.638715e-02 5.883549e+01 0.000000e+00 1.745788e+01

Answer 13

假设initail集合中的所有项目都是不同，我们可以尝试使用 Linq 进行查询；让我们概括一下解决方案：

代码：

public static IEnumerable<T[]> Permutations<T>(IEnumerable<T> source) {
  if (null == source)
    throw new ArgumentNullException(nameof(source));

  T[] data = source.ToArray();

  return Enumerable
    .Range(0, 1 << (data.Length))
    .Select(index => data
       .Where((v, i) => (index & (1 << i)) != 0)
       .ToArray());
}

演示：

  var data = new char[] { 'A', 'B', 'C' };

  var result = Permutations(data);

  foreach (var item in result)
    Console.WriteLine($"[{string.Join(", ", item)}]);

结果：

[]
[A]
[B]
[A, B]
[C]
[A, C]
[B, C]
[A, B, C]

如果您要排除初始的空数组，请放置.Range(1, (1 << (data.Length)) - 1)而不是.Range(0, 1 << (data.Length))

Answer 14

请找到一种非常简单的解决方案，无需递归并且不占用RAM。

Unique Combinations

Answer 15

这就是我的做法。

public static List<List<int>> GetCombination(List<int> lst, int index, int count)
{
    List<List<int>> combinations = new List<List<int>>();
    List<int> comb;
    if (count == 0 || index == lst.Count)
    {
        return null;
    }
    for (int i = index; i < lst.Count; i++)
    {
        comb = new List<int>();
        comb.Add(lst.ElementAt(i));
        combinations.Add(comb);
        var rest = GetCombination(lst,i + 1, count - 1);
        if (rest != null)
        {
            foreach (var item in rest)
            {
                combinations.Add(comb.Union(item).ToList());
            }
        }
    }
    return combinations;
}

您将其称为：

List<int> lst= new List<int>(new int[]{ 1, 2, 3, 4 });
var combinations = GetCombination(lst, 0, lst.Length)

Answer 16

我只是遇到需要这样做的情况，这就是我想到的：

private static List<string> GetCombinations(List<string> elements)
{
    List<string> combinations = new List<string>();
    combinations.AddRange(elements);
    for (int i = 0; i < elements.Count - 1; i++)
    {
        combinations = (from combination in combinations
                        join element in elements on 1 equals 1
                        let value = string.Join(string.Empty, $"{combination}{element}".OrderBy(c => c).Distinct())
                        select value).Distinct().ToList();
    }

    return combinations;
}

它可能不太有效，并且确实有改进的空间，但是可以完成工作！

List<string> elements = new List<string> { "1", "2", "3" };
List<string> combinations = GetCombinations(elements);

foreach (string combination in combinations)
{
    System.Console.Write(combination);
}

值列表的所有可能组合

16 个答案: