Question

说我有以下列表：

List 1:
{{"John", "Doe", "Tall", "Old"},
{"John", "Doe", "Short", "Old"},
{"Jane", "Doe", "Tall", "Young"},
{"Jane", "Doe", "Short", "Old"}}

我想在列表中搜索{“John”，“Doe”，“Short”，“Old”}。

搜索此嵌套列表条目的最佳方法是什么，并确保我不会获得{“John”，“Doe”，“Tall”，“Old”}？

如果嵌套列表只包含一个string项而不是四个，我会使用LINQ来展平列表并搜索结果List<string>.，即：

List<string> newList = oldList.SelectMany(x => x).Distinct().ToList();
newList.Contains("string");

对于每个嵌套列表包含多个字符串项的列表，我能做些类似的事情吗？

Answer 1

那么列表必须包含该顺序中的所有字符串？然后你可以使用Enumerable.SequenceEqual。如果订单不重要，请将Enumerable.All与Contains一起使用，以便：

var names = new[]{"John", "Doe", "Short", "Old"};
List<List<string>> result = list
    .Where(l => l.SequenceEqual(names)).ToList();

或

result = list
    .Where(l => l.All(name => names.Contains(name))).ToList();

输出：

foreach(List<string> l  in result)
    Console.WriteLine(string.Join(",", l));  // John,Doe,Short,Old

Demo

旁注：如果您将搜索的集合转换为HashSet<T>，则可以提高第二种方法的效率：

var names = new HashSet<string>(new[]{"John", "Doe", "Short", "Old"});
result = list.Where(l => l.All(names.Contains)).ToList();

正如Servy所提到的，第二种方法并不能阻止你获得包含所有项目的列表，而且还有更多。您可以添加Count检查以确保它。

Answer 2

您可以使用允许传递自定义相等比较器（IEqualityComparer<T>）的Contains方法的重载版本。

Answer 3

“扁平化”清单的一种方法如下：

var people = list1.
    Select(lst => new {
        First  = lst[0]
    ,   Last   = lst[1]
    ,   Height = lst[2]
    ,   Age    = lst[3]
    });

现在您可以按如下方式搜索收容：

bool hasShortOldJohnDoe = people
    .Contains(p => p.First=="John"
                && p.Last=="Doe"
                && p.Height == "Short"
                && p.Age=="Old");

Answer 4

您可以尝试以下操作：

List<List<string>> mainList = new List<List<string>>

{
    new List<string>(){"John", "Doe", "Tall", "Old"},
    new List<string>(){"John", "Doe", "Short", "Old"},
    new List<string>(){"Jane", "Doe", "Tall", "Young"},
    new List<string>(){"Jane", "Doe", "Short", "Old"},
};
List<string> searchList = new List<string>() { "John", "Doe", "Short", "Old" };

var temp = mainList[0].Except(searchList).Count();
List<List<string>> result  = mainList
                                .Where(r => r.Except(searchList).Count() == 0)
                                .ToList();

它会在result中为您提供一个项目。

或者：

var  result = mainList
                  .Where(r => !r.Except(searchList).Any());

Answer 5

以下是两种方式：

 [Test]
    public void Using_String_Join()
    {
        var l = new List<List<string>>
        {
            new List<string> {"John", "Doe", "Tall", "Old"},
            new List<string> {"John", "Doe", "Short", "Old"},
            new List<string> {"Jane", "Doe", "Tall", "Young"},
            new List<string> {"Jane", "Doe", "Short", "Old"}
        };

        var l2 = new List<string> {"John", "Doe", "Short", "Old"};

        Assert.That(l.Count(inner => string.Join(",", inner).Equals(string.Join(",", l2))), Is.EqualTo(1));
    }

    [Test]
    public void Using_SequenceEqual()
    {
        var l = new List<List<string>>
        {
            new List<string> {"John", "Doe", "Tall", "Old"},
            new List<string> {"John", "Doe", "Short", "Old"},
            new List<string> {"Jane", "Doe", "Tall", "Young"},
            new List<string> {"Jane", "Doe", "Short", "Old"}
        };

        var l2 = new List<string> {"John", "Doe", "Short", "Old"};

        Assert.That(l.Count(inner => inner.SequenceEqual(l2)), Is.EqualTo(1));
    }

搜索List <list <string>＆gt;？</list <string>的最佳方法是什么？

5 个答案: