Question

我有两个课程＆＃34; Word.cs＆＃34;和＃34; TrainerFile.cs＆＃34;。 TrainerFile包含5个不同List<Word>的集合，Word包含2个字符串Lang1和Lang2。我需要的是TrainerFile中所有单词的List或enum，这样我就可以随机列出所述TrainerFile的单词。我不知道如何访问所有单词。这是两个班级：

public class Word
{
    public string Lang1 { get; set; }
    public string Lang2 { get; set; }

    public Word()
    {
        this.Lang1 = "";
        this.Lang2 = "";
    }

    public Word(string lang1, string lang2)
    {
        this.Lang1 = lang1;
        this.Lang2 = lang2;
    }
}

public class TrainerFile
{
    public List<Word> FolderInitial { get; set; }
    public List<Word> Folder1 { get; set; }
    public List<Word> Folder2 { get; set; }
    public List<Word> Folder3 { get; set; }
    public List<Word> FolderFinal { get; set; }

    public TrainerFile()
    {
        this.FolderInitial = new List<Word>();
        this.Folder1 = new List<Word>();
        this.Folder2 = new List<Word>();
        this.Folder3 = new List<Word>();
        this.FolderFinal = new List<Word>();
    }

    public TrainerFile(List<Word> listInitial, List<Word> list1, List<Word> list2, List<Word> list3, List<Word> listFinal)
    {
        this.FolderInitial = listInitial;
        this.Folder1 = list1;
        this.Folder2 = list2;
        this.Folder3 = list3;
        this.FolderFinal = listFinal;
    }
}

Answer 1

使用linq＆＃39; s Concat：

var result = FolderInitial.Concat(Folder1)
                          .Concat(Folder2)
                          .Concat(Folder3)
                          .Concat(FolderFinal);

如果您想要一个不同的列表，那么何时不使用Concat Union并覆盖Word类的Equals和GetHashCode方法，以便按类的属性进行比较。
另一种方法是提供IEqualityComparer this overload of Union

此外，由于C＃6.0，您可以使用默认构造函数：

public class TrainerFile
{
    public List<Word> FolderInitial { get; set; } = new List<Word>();
    public List<Word> Folder1 { get; set; } = new List<Word>();
    public List<Word> Folder2 { get; set; } = new List<Word>();
    public List<Word> Folder3 { get; set; } = new List<Word>();
    public List<Word> FolderFinal { get; set; } = new List<Word>();
}

Answer 2

看起来像Dictionary<string, List<Word>>

的情况

Dictionary<string, List<Word>> trainerFile = new Dictionary<string, List<Word>>();
trainerFile.Add("FolderInitial", new List<Word>());
trainerFile.Add("Folder1", new List<Word>());
trainerFile.Add("Folder2", new List<Word>());
trainerFile.Add("Folder3", new List<Word>());
trainerFile.Add("FolderFinal", new List<Word>());

然后像这样访问：

List<string> initial = trainerFile["FolderInitial"];
trainerFile["Folder2"].Add(new Word());
Word word = trainerFile["Folder2"][0];

要访问每个单词，

IEnumerable<Word> all = trainerFile.Values.SelectMany(i => i);

从类中创建一个IEnumerable

2 个答案: