将夫妇 foreachers 转换为 linq

Question

所以我有这个代码，问题是，在我的项目中，它必须用 linq 表达式编写，我不知道如何做到这一点，因为有两个列表。有任何想法吗？

public void GetLecturersWorkloadStatistics(List<Student> studentList, List<Lecturer> lecturerList)

{

    foreach (Lecturer lecturer in lecturerList)

    {

        foreach (Student student in studentList)

        {

            if (lecturer.ModuleName == student.ModuleName &&

                lecturer.LastName == student.LecturerLastName &&

                lecturer.FirstName == student.LecturerFirstName)

            {

                lecturer.Credits = lecturer.Credits + lecturer.ModuleValueInCredits;

            }

        }

    }

}

如上所述，我已经用 linq 完成了内部操作，但我不知道下一步该做什么。

public void GetLecturersWorkloadStatistics(List<Student> studentList, List<Lecturer> lecturerList)

{

    foreach (Lecturer lecturer in lecturerList)

    {

        studentList = (from student in studentList

            where lecturer.ModuleName == student.ModuleName &&

                  lecturer.LastName == student.LecturerLastName &&

                  lecturer.FirstName == student.LecturerFirstName

            select student).ToList();

    }

}

Answer 1

更改您的方法主体以使用 a 的组合Enumerable.Join并Enumerable.All使其在一行中发生<!-- language: c# -->lecturerList.Join(studentList, lecturer => new {lecturer.ModuleName, LecturerLastName = lecturer.LastName, LecturerFirstName = lecturer.FirstName}, student=> new {student.ModuleName, student.LecturerLastName, student.LecturerFirstName}, (lecturer, student) => lecturer).All(lecturer=> {lecturer.Credits = lecturer.Credits + lecturer.ModuleValueInCredits; return true;}

Answer 2

我的 C# 有点生疏，但这里有一种使用扩展方法 LINQ 表示法的方法：lecturerList&nbsp; &nbsp; .SelectMany(lecturer => studentList, Tuple.Create)&nbsp; &nbsp; .Where((lecturer, student) =>&nbsp; &nbsp; &nbsp; &nbsp; lecturer.ModuleName == student.ModuleName &&&nbsp; &nbsp; &nbsp; &nbsp; lecturer.LastName&nbsp; &nbsp;== student.LecturerLastName &&&nbsp; &nbsp; &nbsp; &nbsp; lecturer.FirstName&nbsp; == student.LecturerFirstName)&nbsp; &nbsp; .Select((l, s) => l)&nbsp; &nbsp; .ToList()&nbsp; &nbsp; .ForEach(lecturer => lecturer.Credits = lecturer.Credits + lecturer.ModuleValueInCredits);就像@user2023861 一样，您基本上是在创建所有可能元组的列表（“笛卡尔积”）：(0, 0), (1, 0), (2, 0), ..., (lecturers.Length() - 1, students.Length() - 1)然后应用Where过滤器。然后将其压缩回仅讲师列表。另一个想法是使用Join：lecturerList&nbsp; &nbsp; .Join(studentList,&nbsp; &nbsp; &nbsp; &nbsp; lecturer => Tuple.Create(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lecturer.ModuleName,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lecturer.LastName,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lecturer.FirstName),&nbsp; &nbsp; &nbsp; &nbsp; student => Tuple.Create(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; student.ModuleName,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; student.LecturerLastName,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; student.LecturerFirstName),&nbsp; &nbsp; &nbsp; &nbsp; (lecturer, student) => lecturer);&nbsp; &nbsp; .ToList()&nbsp; &nbsp; .ForEach(lecturer => lecturer.Credits = lecturer.Credits + lecturer.ModuleValueInCredits);如果您首先对数据进行排序，则可以避免这种O(N^2)行为，并N通过利用提前退出来使大数据更快。大多数排序算法都小于，O(N^2)并且通过提前退出，您可以获得类似O(N log N).当然，这仅在您的数据集确实很大时才值得。:)

Answer 3

您正在执行嵌套 foreach 以便对于列表 1 的每个元素您考虑列表 2 的所有元素，这需要在 LINQ 中使用笛卡尔积，这很容易使用 LINQ 语法：var LecturersToUpdate = from lecturer in lecturerList&nbsp; &nbsp; &nbsp; &nbsp;from student in studentList&nbsp;&nbsp; &nbsp; &nbsp; &nbsp;// 2 from clauses one after the other will give the cartesian product of the lists&nbsp; &nbsp; &nbsp; &nbsp;where lecturer.ModuleName == student.ModuleName &&&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lecturer.LastName == student.LecturerLastName &&&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lecturer.FirstName == student.LecturerFirstName&nbsp;&nbsp; &nbsp; &nbsp; // After there where we only get the pair of lecturer/student in the cartesian that match your condition, we stop the query there with a select, we could select both the lecturer and student in a new object but here i'm only grabing the lecturer as that's all you're using in the logic of your foreach&nbsp; &nbsp; &nbsp;select lecturer;foreach(var lect in LecturersToUpdate){&nbsp; &nbsp; lecturer.Credits = lecturer.Credits + lecturer.ModuleValueInCredits;}您的更新版本在正确的轨道上，您只需要额外的 from 子句即可获取所有对。