如何使用三个键创建和填充嵌套字典

时间:2015-01-06 19:14:12

标签: c# dictionary nested multikey

我有一个独特的双重对应三个字符串的变体。我想填充字典或类似的内容,我可以调用类似dict[key1][key2][key3]的内容并获取值。

我尝试过很多像

这样的事情
    Dictionary<string, Dictionary<string, double>> dict = new Dictionary<string, Dictionary<string, double>> {
        { "Foo", {"Bar", 1.2 } },
        { "Foo", {"Test", 3.4 } }
    };

这给了我语法错误和错误,例如“错误4命名空间不能直接包含字段或方法等成员”

    Dictionary<double, Tuple<string, string>> dict = {
          {1.23, "Blah", "Foo"}
    };

这给了我一些错误,例如“错误1只能使用数组初始值设定项表达式来分配给数组类型。请尝试使用新的表达式。”

并且

    object dict = new Dictionary<string, Dictionary<string, Dictionary<string, string>>>();

    dict["k1"] = new Dictionary<string, Dictionary<string, string>>();
    dict["k1"]["k2"] = new Dictionary<string, string>();
    dict["k1"]["k2"]["k3"] = 3.5;

这给了我语法错误和错误,例如类,结构或接口成员声明中的“错误2无效标记”“k1”'

我应该怎么做?提前谢谢。

![在此处输入图片说明] [1]

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编辑:尝试使用Jonesy的代码:

namespace WindowsFormsApplication1
{
    public partial class Form1 : Form
    {
        string[] grades = { "Grade 1", "Grade 5", "Grade 8", "ASTM A325", "316 Stainless", "Monel", "Brighton Best 1960" };
        string[] sizes = { "#1", "2", "3", "4", "5", "6", "8", "10", "12", "1/4", "5/16", "3/8", "7/16", "1/2", "9/16", "5/8", "3/4", "7/8", "1", "1-1/8", "1-1/4", "1-3/8", "1-1/2" };

        var dict = new Dictionary<string, Dictionary<string, Dictionary<string, double>>>();
        dict["k1"] = new Dictionary<string, Dictionary<string, double>>();
        dict["k1"]["k2"] = new Dictionary<string, double>();
        dict["k1"]["k2"]["k3"] = 3.5;


        public Form1()
        {
            InitializeComponent();
        } 

2 个答案:

答案 0 :(得分:3)

你的最后一次尝试已经结束了,你想要:

var dict = new Dictionary<string, Dictionary<string, Dictionary<string, double>>>();
dict["k1"] = new Dictionary<string, Dictionary<string, double>>();
dict["k1"]["k2"] = new Dictionary<string, double>();
dict["k1"]["k2"]["k3"] = 3.5;

您希望var代替object

(或Dictionary<string, Dictionary<string, Dictionary<string, double>>>如果你喜欢滚动)

,您的最后一个字符串应为double

答案 1 :(得分:0)

据我所知,您有数据并希望在其中执行查找。你为什么不能只为这个目的使用一些数据库?

但如果您真的想要对所有值进行硬编码,则可以。只是不要手动初始化字典,进行简化 - 在运行时解析数据。

像这样的东西。 (我想,你是c#编程的新手,所以为了方便起见,我创建了新的控制台应用程序并复制粘贴了所有代码)

public class Program
{
    // harcode all data as string
    const string RawData =
        "k11,k12,k13=3.4;" +
        "k21,k22,k23=4.42;" +
        "k31,k32,k33=5.91;" +
        "k41,k42,k43=8.14;" +
        "k51,k52,k53=4.13;" +
        "k61,k62,k63=5.4";

    static void Main(string[] args)
    {
        // create dictionary from hardcoded string
        var data = ParseData();

        // use Tuple as key for data lookup
        var value = data[Tuple.Create("k11", "k12", "k13")];

        // check, that value equals expected one
        Debug.Assert(value == 3.4);
    }

    private static IDictionary<Tuple<string, string, string>, double> ParseData()
    {
        var parsedData =
            RawData
                .Split(';')
                .Select(ParseRow)
                .ToDictionary(x => x.Item1, x => x.Item2);

        return parsedData;
    }

    private static Tuple<Tuple<string, string, string>, double> ParseRow(string row)
    {
        var parts = row.Split('=');
        var coefficients = ParseCoefficients(parts[0]);
        var value = Double.Parse(parts[1], CultureInfo.InvariantCulture);

        return Tuple.Create(coefficients, value);
    }

    private static Tuple<string, string, string> ParseCoefficients(string row)
    {
        var coeffs = row.Split(',');
        var result = Tuple.Create(coeffs[0], coeffs[1], coeffs[2]);

        return result;
    }
}

作为另一种简化,您可以使用自定义类作为字典键而不是嵌套字典。编写自己的(注意,它应该覆盖平等成员EqualsGetHashCode),或者使用基类库中的东西。 Tuple<string, string, string>是完美的。