Question

我试图弄清楚如何使用C＃创建单个连接列表，源自3个单独的列表。例如：

 List 1: Ugly, Pretty
 List 2: Dogs, Carts, Pigs
 List 3: Rock, Suck

输出：

 Ugly Dogs Rock
 Ugly Dogs Suck
 Ugly Cats Rock
 Ugly Cats Suck
 Ugly Pigs Rock
 Ugly Pigs Suck
 Pretty Dogs Rock
 Pretty Dogs Suck
 Pretty Cats Rock
 Pretty Cats Suck
 Pretty Pigs Rock
 Pretty Pigs Suck

我知道它只是嵌套循环，但我无法弄清楚的部分是如何为每个列表使用List-strings。

Answer 1

这不是笛卡儿的产品吗？

var r = from i1 in list1
        from i2 in list2
        from i3 in list3
        select new { i1, i2, i3 };
        // or String.Format("{0} {1} {2}", i1, i2, i3);

Answer 2

var list = from s1 in list1
           from s2 in list2
           from s3 in list3
           select s1 + " " + s2 + " " + s3;

Answer 3

List<string> list1 = new List<string>(){ "Ugly", "Pretty"};
List<string> list2 = new List<string>(){ "Dogs", "Carts", "Pigs"};
List<string> list3 = new List<string>(){ "Rock", "Suck"};

var result = from s1 in list1
             from s2 in list2
             from s3 in list3
             select new[] { s1, s2, s3 };

foreach (var item in result)
{
    Console.WriteLine(String.Join(",", item));
}

如果您正在寻找更通用的解决方案，不仅仅是3个列表，您可以尝试Eric Lippert的解决方案

foreach (var item in new[] { list1, list2, list3 }.CartesianProduct())
{
    Console.WriteLine(String.Join(",", item));
}

public static partial class MyExtensions
{
    // Eric Lippert’s Blog
    // Computing a Cartesian Product with LINQ
    // http://blogs.msdn.com/b/ericlippert/archive/2010/06/28/computing-a-cartesian-product-with-linq.aspx
    public static IEnumerable<IEnumerable<T>> CartesianProduct<T>(this IEnumerable<IEnumerable<T>> sequences)
    {
        // base case: 
        IEnumerable<IEnumerable<T>> result = new[] { Enumerable.Empty<T>() };
        foreach (var sequence in sequences)
        {
            var s = sequence; // don't close over the loop variable 
            // recursive case: use SelectMany to build the new product out of the old one 
            result =
                from seq in result
                from item in s
                select seq.Concat(new[] { item });
        }
        return result;
    }
}

连接3个单词列表

3 个答案: