LINQ：如何连接 ObjectCollection 中的嵌套列表

发布于 2024-09-10 07:46:20 字数 1054 浏览 6 评论 0原文

是否可以在不使用 foreach 循环的情况下为以下问题定义 LINQ 语句？

public class GroupObject
{
    public String name;
    public List<String> letters;

    public void test()
    {
        List<GroupObject> myGroups = new List<GroupObject> {
            new GroupObject {
                name="test1",
                letters=new List<String>{"x","y","z"}
            },
            new GroupObject {
                name="test2",
                letters=new List<String>{"l","m","n"}
            },
            new GroupObject {
                name="test3",
                letters=new List<String>{"m","x","z"}
            }
        };

        // LINQ problem 1: how to get a list of all 9 letters
        // List<string> allLetters = (from ...).ToList();

        // LINQ problem 2: how to get a list of all 6 DISTINCT letters
        // List<string> allDistictLetters = (from ...).ToList();
    }
}

在写这个问题的时候，我找到了解决问题的方法。我仍然会发布（并回答）这个问题，因为这对我来说是一个真正的麻烦。而且我在这里没有找到合适的现有问题。

再见，尤文

原文

is it possible to define a LINQ statement for the following problems WITHOUT using a foreach loop?

public class GroupObject
{
    public String name;
    public List<String> letters;

    public void test()
    {
        List<GroupObject> myGroups = new List<GroupObject> {
            new GroupObject {
                name="test1",
                letters=new List<String>{"x","y","z"}
            },
            new GroupObject {
                name="test2",
                letters=new List<String>{"l","m","n"}
            },
            new GroupObject {
                name="test3",
                letters=new List<String>{"m","x","z"}
            }
        };

        // LINQ problem 1: how to get a list of all 9 letters
        // List<string> allLetters = (from ...).ToList();

        // LINQ problem 2: how to get a list of all 6 DISTINCT letters
        // List<string> allDistictLetters = (from ...).ToList();
    }
}

While writing this question, I found a solution to the problems. I will still post (and answer) the question, since this one was a real bugger for me. And I did not find a suitable existing question here.

Ciao,
Juve

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半岛未凉 2024-09-17 07:46:20

我相信这个问题可以通过SelectMany和Distinct来解决！

var allLetters = myGroups.SelectMany(go => go.letters);
var allUniqueLetters = allLetters.Distinct();

第二个变量名称具有误导性。除其他字母外，它将返回一个 "m" 实例，该实例在原始集合中不是唯一的;)

I believe this problem can be solved by SelectMany and Distinct!

var allLetters = myGroups.SelectMany(go => go.letters);
var allUniqueLetters = allLetters.Distinct();

The second variable name is misleading tho. It will return, among other letters, one instance of "m", which was not unique in the original collection ;)

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浸婚纱 2024-09-17 07:46:20

如上所述，我自己找到了答案：

List<string> allLetters = (from g in myGroups from l in g.letters select l).ToList();
List<string> allDistinctLetters = allLetters.Distinct().ToList();

我希望这对某人有帮助。

你好，尤文图斯

As stated above, I found the answer myself:

List<string> allLetters = (from g in myGroups from l in g.letters select l).ToList();
List<string> allDistinctLetters = allLetters.Distinct().ToList();

I hope this helpful for somebody.

Ciao, Juve

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