Question

我在dict中有dict<string, list<string>>,个3个键，第一个键有2个值，secodn 3个值，第三个键有3个值。如果我从每个值集中得到一个值，那么我将得到2 * 3 * 3 = 18集的组合如何在c＃中编码？

谢谢

编辑抱歉没说清楚

我想要这样的东西说我有这样的决定

 {"1",new List<String>(){"a", "b"}}, 
    {"2",new List<String>(){"c", "d", "e"}}, 
 {"3", new List<string>() {"f", "g"}

我想要这样的输出 acf，acg，adf，adg，aef，aeg bcf，bcg，bdf，bdg，bef，beg

Answer 1

Linq：

var dict = new Dictionary<String, List<String>>() { 
    {"1",new List<String>(){"a", "b"}},
    {"2",new List<String>(){"c", "d", "e"}},
    {"3",new List<String>(){"f", "g", "h"}},
};
var combis = from kv in dict
             from val1 in kv.Value
             from val2 in kv.Value
             select string.Format("{0}{1}", val1, val2);
foreach (var combi in combis)
    Console.WriteLine(combi);

演示：http://ideone.com/nm7mY

Answer 2

我认为你的意思是这个？

Dictionary<string, int> dict = new Dictionary<string, int>
{
    { "Hello World", 1 },
    { "HelloWorld", 1 },
    { "Hello  World", 1 },
};

foreach (var item in dict) // var is of type KeyValuePair<string, int>
    Console.WriteLine(item.Key + ", " + item.Value);

Answer 3

        Dictionary<string, List<int>> storage = new Dictionary<string, List<int>>();
        storage.Add("key1", new List<int>() { 2, 7 });
        storage.Add("key2", new List<int>() { 8, 4, 1});
        storage.Add("key3", new List<int>() { 3, 9, 3 });
        foreach (string key in storage.Keys)
        {
            //access to single storage...
            List<int> subStorage = (List<int>)storage[key];
            foreach (int item in subStorage)
            {
                //access to single value inside storage...
            }
        }

Answer 4

如果我试图读取或编辑列表中的值，我会尝试以下内容：

Dictionary<int, List<string>> dict = new Dictionary<int, List<string>>();
var arrayOfValues = dict.Values.ToArray();

for (int i = 0; i < arrayOfValues.Length; i++)
{
    for (int j = 0; j < arrayOfValues[i].Count; j++)
    {
        //read/edit arrayOfValues[i][j];
    }
}

你不需要递归，因为你知道“树”的部门。

Answer 5

快速＆amp;脏，但你可以抛光这种方法。结果列表包含预期结果：

用法：

var dict = new Dictionary<String, List<String>>() { 
    {"1",new List<String>(){"a", "b"}},
    {"2",new List<String>(){"c", "d", "e"}},
    {"3",new List<String>(){"f", "g"}},
};

var collections = dict.Select(kvp => kvp.Value).ToArray();            
var result = new List<string>(); 
GetNextProduct(collections, 0, String.Empty, result);

产生结果的方法：

private static void GetNextProduct(IEnumerable<string>[] collections, int collectionIndex, string currentProduct, IList<string> results)
{
    var currentList = collections[collectionIndex];
    bool isLast = collections.Length == collectionIndex + 1;
    foreach (var s in currentList)
    {
        if (isLast) results.Add(currentProduct + s);
        else GetNextProduct(collections, collectionIndex + 1, currentProduct + s, results);
    }
}

从键值列表中获取组合

5 个答案: