使用嵌套for循环将Lambda转换为Linq语句

Question

有没有办法重写GetTransformedCollection下面的方法,以便它使用Linq语句而不是表达式？我目前正试图绕过"一个带有语句体的lambda表达式无法转换为表达式树"错误.

public class Obj1
{
    public int Id { get; set; }
    public string[] Names { get; set; }
    public string[] Tags { get; set; }
}

public class EntCollections
{
    private List<Obj1> _results;

    [SetUp]
    public void SetUp()
    {
        _results = new List<Obj1>
        {
            new Obj1 {Id = 1, Names = new[] {"n1"}, Tags = new[] {"abc", "def"}},
            new Obj1 {Id = 2, Names = new[] {"n2", "n3"}, Tags = new[] {"ghi"}},
            new Obj1 {Id = 3, Names = new[] {"n1", "n3"}, Tags = new[] {"def", "xyz"}}
        };
    }

    private static Dictionary<string, List<string>>
        GetTransformedCollection(IEnumerable<Obj1> results)
    {
        var list = new Dictionary<string, List<string>>();

        foreach (var result in results)
        {
            foreach (var id in result.Names)
            {
                if (list.ContainsKey(id))
                {
                    list[id].AddRange(result.Tags);
                }
                else
                {
                    list.Add(id, result.Tags.ToList());
                }
            }
        }

        return list;
    }

    [Test]
    public void Test()
    {
        var list = GetTransformedCollection(_results);

        Assert.That(list["n1"], Is.EquivalentTo(new [] { "abc", "def", "def", "xyz" }));
        Assert.That(list["n2"], Is.EquivalentTo(new [] { "ghi" }));
        Assert.That(list["n3"], Is.EquivalentTo(new [] { "ghi", "def", "xyz" }));
    }

Ps我并不太担心结果类型是一个字典,这只是将它表达为返回类型的简单方式.

Answer 1

我个人会使用一个ILookup,这是一个很好的选择,只要你有一个Dictionary<T1, List<T2>>,并建立了ToLookup():

// Flatten the objects (lazily) to create a sequence of valid name/tag pairs
var pairs = from result in results
            from name in result.Names
            from tag in result.Tags
            select new { name, tag };

// Build a lookup from name to all tags with that name
var lookup = pairs.ToLookup(pair => pair.name, pair => pair.tag);