Question

有一个类似的问题无法回答我的问题。 -> Count number of element in List>

我有一个包含子列表的列表：

List<string> sublist1 = new List<string>() { "a", "b" };
List<string> sublist2 = new List<string>() { "a", "b" };
List<string> sublist3 = new List<string>() { "a", "c" };

现在我要计算每个列表的出现次数。

a, b --> 2
a, c --> 1

我使用了LINQ的distinct()，但得到了输出：

a, b --> 1
a, b --> 1
a, c --> 1

我认为哈希码是不同的。是否有distinct()的替代项，它正在查看列表值？我想在LINQ中解决这个问题。

编辑：列表项的顺序必须相同！

Answer 1

要使用GroupBy()进行此操作，您需要一个合适的IEqualityComparer<List<string>>来比较字符串列表。没有内置的实现，因此您必须自己动手：

public sealed class StringListEqualityComparer : IEqualityComparer<List<string>>
{
    public bool Equals(List<string> x, List<string> y)
    {
        if (ReferenceEquals(x, y))
            return true;

        if (x == null || y == null)
            return false;

        return x.SequenceEqual(y);
    }

    public int GetHashCode(List<string> strings)
    {
        int hash = 17;

        foreach (var s in strings)
        {
            unchecked
            {
                hash = hash * 23 + s?.GetHashCode() ?? 0;
            }
        }

        return hash;
    }
}

一旦掌握了这些，就可以将其与GroupBy()一起使用，如下所示：

public static void Main()
{
    var sublist1 = new List<string>{ "a", "b" };
    var sublist2 = new List<string>{ "a", "b" };
    var sublist3 = new List<string>{ "a", "c" };

    var listOfLists = new List<List<string>> {sublist1, sublist2, sublist3};

    var groups = listOfLists.GroupBy(item => item, new StringListEqualityComparer());

    foreach (var group in groups)
    {
        Console.WriteLine($"Group: {string.Join(", ", group.Key)}, Count: {group.Count()}");
    }
}

Answer 2

     public JsonResult CountList(){
        List<List<string>> d = new List<List<string>>(); //SuperList
            d.Add(new List<string> { "a", "b" }); //List 1
            d.Add(new List<string> { "a", "b" }); // List 2
            d.Add(new List<string> { "a", "c" }); // List 3
            d.Add(new List<string> { "a", "c", "z" }); //List 4
            var listCount = from items in d
                            group items by items.Aggregate((a,b)=>a+""+b) into groups
                            select new { groups.Key, Count = groups.Count() };
        return new JsonResult(listCount);
     }

这将提供以下结果作为Post Man或Advanced REST Client中的输出

[{
"key": "ab",
"count": 2
},
  {
"key": "ac",
"count": 1
},
  {
"key": "acz",
"count": 1
}],

Answer 3

我认为这会有所帮助

 var list = new List<List<string>>() { sublist1, sublist2, sublist3};
 var result = list.GroupBy(x => string.Join(",",x)).ToDictionary(x => x.Key.Split(',').ToList(), x => x.Count());

Answer 4

您可以尝试以下代码：-

            List<string> sublist1 = new List<string>() { "a", "b" };   
            List<string> sublist2 = new List<string>() { "a", "b" };   
            List<string> sublist3 = new List<string>() { "a", "c" };   
            List<List<string>> listOfLists = new List<List<string>> { sublist1, sublist2, sublist3 };    
            Dictionary<string, int> counterDictionary = new Dictionary<string, int>();                  
                foreach (List<string> strList in listOfLists)       
                {   
                    string concat = strList.Aggregate((s1, s2) => s1 + ", " + s2);   
                    if (!counterDictionary.ContainsKey(concat))   
                        counterDictionary.Add(concat, 1);   
                    else   
                        counterDictionary[concat] = counterDictionary[concat] + 1;   
                }   
                foreach (KeyValuePair<string, int> keyValue in counterDictionary)   
                {   
                       Console.WriteLine(keyValue.Key + "=>" + keyValue.Value);   
                }

Answer 5

我想我将通过以下方式解决此问题：

var equallists = list1.SequenceEqual(list2);

因此，我将不同的列表和具有SequenceEquals（）的列表进行比较，并对它们进行计数。

欢迎使用更好的解决方案。：）

计算List <List <T >>中与订单相关的List <T>的出现次数

5 个答案: