假设我有以下列表:
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{"约翰","母鹿","短","旧"}。
搜索此嵌套列表条目并确保我不会得到{"John","Doe","Tall","Old"}的最佳方法是什么?
如果嵌套列表只包含一个string
项而不是四个,我将使用 LINQ 平展列表并搜索生成的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,是否可以执行类似操作?
所以列表必须按该顺序包含所有字符串?然后你可以使用 Enumerable.SequenceEqual
.如果顺序不重要,请使用Enumerable.All
Contains
,因此:
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
演示
旁注:如果将要搜索的集合转换为HashSet<T>
,则可以使第二种方法更有效:
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,第二种方法不会阻止您获取包含所有项目的列表,但还可以获取更多项目。您可以添加Count
检查以确保它。
您可以使用Contains
方法的重载版本,该方法允许传递自定义相等比较器(IEqualityComparer<T>
)。
"扁平化"列表的一种方法是:
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包含:
bool hasShortOldJohnDoe = people
.Contains(p => p.First=="John"
&& p.Last=="Doe"
&& p.Height == "Short"
&& p.Age=="Old");
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();
它会给你一个项目 result
.
或:
var result = mainList
.Where(r => !r.Except(searchList).Any());
这里有几种方法:
[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));
}