c# - Concat两个具有相同关键记录的字典

我想合并两个字典。尝试此操作时，我收到有关重复键的错误。如何合并具有相同键的两个字典？

class Program
{
    static void Main(string[] args)
    {
        Dictionary<Class1.Deneme, List<string>> dict1 = new Dictionary<Class1.Deneme, List<string>>();
        Dictionary<Class1.Deneme, List<string>> dict2 = new Dictionary<Class1.Deneme, List<string>>();
        Dictionary<Class1.Deneme, List<string>> dict3 = new Dictionary<Class1.Deneme, List<string>>();

        List<string> list1 = new List<string>() { "a", "b" };
        List<string> list2 = new List<string>() { "c", "d" };
        List<string> list3 = new List<string>() { "e", "f" };
        List<string> list4 = new List<string>() { "g", "h" };

        dict1.Add(Class1.Deneme.Buyer, list1);
        dict2.Add(Class1.Deneme.Buyer, list2);
        dict3 = dict1.Concat(dict2).ToDictionary(kvp => kvp.Key, kvp => kvp.Value);

        Console.Read();
    }
}

public class Class1
{
    public enum Deneme
    {
        Customer=1,
        Buyer=2,
        Supplier=3
    }
}

最佳答案

怎么样

static IDictionary<TKey, TValue> Merge<TKey, TValue>(
    this IDictionary<TKey, TValue> left,
    IDictionary<TKey, TValue> right,
    Func<TKey, TValue, TValue, TValue> valueMerger)
{
    var result = new Dictionary<TKey, TValue>();
    foreach(var pair in left.Concat(right))
    {
        if (result.ContainsKey(pair.Key))
        {
            result[pair.Key] = valueMerger(
                                   pair.Key,
                                   pair.Value,
                                   result[pair.Key]);
            continue;
        }

        result.Add(pair.Key, pair.Value);
    }

    return result;
}

因此，您需要提供一个委托来处理键重复的情况。我建议您以您的示例为例，

var dictionary1 = new Dictionary<Class1.Deneme, List<string>>();
var dictionary2 = new Dictionary<Class1.Deneme, List<string>>();

var merged = dictionary1.Merge(
                 dictionary2,
                 (key, left, right) => left.Concat(right));

还是您宁愿抛出异常？

   var merged = dictionary1.Merge(
                 dictionary2,
                 (key, left, right) => throw new ArgumentException("right", ...);