C# 查找一个项目的可能变化

在我的应用程序中，我有配方的概念，每个配方都有可以被另一种成分替代的成分。我想做的是用每种成分创造每种可能性。我将以下结构作为CLR对象：

public class Food
{
    public Food()
    {
        NutritionalValues = new Collection<FoodNutritionalValue>();
    }

    public string Name { get; set; }
    public string Category { get; set; }

    public ICollection<FoodNutritionalValue> NutritionalValues { get; set; }
}

public class FoodNutritionalValue
{
    public string Type { get; set; }
    public decimal Value { get; set; }
}

public class Recipe
{
    public Recipe()
    {
        IngredientGroups = new Collection<IngredientGroup>();
    }

    public string Name { get; set; }
    public ICollection<IngredientGroup> IngredientGroups { get; set; }
}

public class IngredientGroup
{
    public IngredientGroup()
    {
        Ingredients = new Collection<Food>();
    }

    public ICollection<Food> Ingredients { get; set; }
}

我肯定我错过了一些太大的东西。有人知道我如何在运行时生成每个可能性吗？

根据找到的答案，您可能可以使用相同的帮助器方法：

public static List<List<T>> CombinationsOf<T>(List<List<T>> sets)
{    
  var combinations = new List<List<T>>();

  foreach (var value in sets[0])
    combinations.Add(new List<T> { value });

  foreach (var set in sets.Skip(1))
    combinations = AddSet(combinations, set);

  return combinations;
}

private static List<List<T>> AddSet<T>(List<List<T>> combinations, List<T> set)
{
  var result = from value in set
               from combination in combinations
               select new List<T>(combination) { value };

  return result.ToList();
}

输出

results :=

Food8, Food5, Food3 
Food9, Food5, Food3 
Food8, Food10, Food3 
Food9, Food10, Food3 
Food8, Food11, Food3 
Food9, Food11, Food3 
Food8, Food5, Food4 
Food9, Food5, Food4 
Food8, Food10, Food4 
Food9, Food10, Food4 
Food8, Food11, Food4 
Food9, Food11, Food4 

...
var foods = recipes.SelectMany(recipe => 
  recipe.IngredientGroups.Select(ingredient =>
    ingredient.Ingredients.Select(o => o.Name).ToList()));

var results = from combination in CombinationsOf(foods.ToList())
              select string.Join(", ", combination);
...

results :=

Food8, Food5, Food3 
Food9, Food5, Food3 
Food8, Food10, Food3 
Food9, Food10, Food3 
Food8, Food11, Food3 
Food9, Food11, Food3 
Food8, Food5, Food4 
Food9, Food5, Food4 
Food8, Food10, Food4 
Food9, Food10, Food4 
Food8, Food11, Food4 
Food9, Food11, Food4