使用LINQ标准化数据

时间:2010-02-27 20:50:07

标签: linq normalization

假设我们有一些非规范化数据,如下所示:

List<string[]> dataSource = new List<string[]>();
string [] row1 = {"grandParentTitle1", "parentTitle1", "childTitle1"}; 
string [] row2 = {"grandParentTitle1", "parentTitle1", "childTitle2"};
string [] row3 = {"grandParentTitle1", "parentTitle2", "childTitle3"};
string [] row4 = {"grandParentTitle1", "parentTitle2", "childTitle4"};
dataSource.Add(row1);

我需要将其标准化,例如得到IEnumerable&lt;小孩&gt;填充Child.Parent和Child.Parent.GrandParent。

势在必行的方式或多或少都很明确。 Linq会缩短吗?

在一个查询中更好,对于更多实体,这应该是可扩展的。

我试过像分开创建IEnumerable&lt; GrandParent&gt ;,然后IEnumerable&lt;父母&gt;分配等。

PLease能否以功能性的方式实现这一点?

3 个答案:

答案 0 :(得分:1)

您可以使用group by完成您想要的操作。不幸的是我对C#LINQ语法的了解是有限的,所以我只能告诉你调用扩展方法GroupBy的方式。

var normalized = dataSource
    .GroupBy(source => source[0], (grandParent, grandParentChilds) => new { GrandParent = grandParent, Parents = grandParentChilds
        .GroupBy(source => source[1], (parent, parentChilds) => new { Parent = parent, Children = from source in parentChilds select source[2]}) });

foreach (var grandParent in normalized)
{
    Console.WriteLine("GrandParent: {0}", grandParent.GrandParent);
    foreach (var parent in grandParent.Parents)
    {
        Console.WriteLine("\tParent: {0}", parent.Parent);
        foreach (string child in parent.Children)
            Console.WriteLine("\t\tChild: {0}", child);
    }
}

答案 1 :(得分:0)

Linq确实与此相反。即。如果你将它标准化,你可以很容易地说

from g in grandParents
from p in g.Parents
from c in p.Children
select new { GrandParentName = g.Name, ParentName = p.Name, ChildName = c.Name };

做你要问的事情比较棘手。像这样的东西

var grandparents = (from g in dataSource
                    select new GrandParent {
                        Title = g[0],
                        Parents = (from p in dataSource
                                   where p[0] == g[0]
                                   select new Parent {
                                      Title = p[1],
                                      Children = from c in dataSource
                                                 where p[1] == c[1]
                                                 select new
                                                            {
                                                                Title = c[2]
                                                            }
                                   }).Distinct(new ParentTitleComparer())
                    }).Distinct(new GrandParentTitleComparer());

我不相信这比命令式版本更好。

答案 2 :(得分:0)

执行此操作的最基本方法是使用匿名变量:

from ds0 in dataSource group ds0 by ds0[0] into grandparents
select new
{
    Grandparent = grandparents.Key,
    Parents =
        from ds1 in grandparents group ds1 by ds1[1] into parents
        select new
        {
            Parent = parents.Key, 
            Children = from ds2 in parents select ds2[2]
        }
};

如果你想用具体的类做这个,我建议创建一个带有构造函数的Person类,该构造函数使用IEnumerable<Person>代表正在构造的Person的子代。然后你可以这样做:

from ds0 in dataSource
group ds0 by ds0[0] into grandparents
select new Person(grandparents.Key,
    from ds1 in grandparents
    group ds1 by ds1[1] into parents
    select new Person(parents.Key,
        from ds2 in parents
        select new Person(ds2[2])));

这些解决方案中的任何一个都适合您吗?

如果您想要不同的GrandParentParent&amp; Child类型,那么您应该能够修改最后一个示例以适应。

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