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多值比较器

.net

4

steve · 技术社区 · 17 年前

是否可以使用iComparer在对象中使用两个值对列表进行排序?

我有一个自定义比较器类,它根据value1进行排序。但是,对value1和value2进行排序的最佳方法是什么?

3 回复 | 直到 17 年前

1

13

mqp 17 年前

你的 IComparer 类应该处理它。例如:

public class ThingComparer : IComparer
{
    public int Compare(object x, object y)
    {
        // null- and type-checking omitted for clarity

        // sort by A, B, and then C

        if (x.A != y.A) return x.A.CompareTo(y.A);
        if (x.B != y.B) return x.B.CompareTo(y.B);
        return x.C.CompareTo(y.C);
    }
}

2

0

Fredrik Mörk 17 年前

如果实现自己的比较器,则可以执行任何排序:

List<Customer> customers = GetCustomers();
customers.Sort(delegate(Customer x, Customer y)
{
    if (x.Name != y.Name)
    {
        return x.Name.CompareTo(y.Name);
    }

    return x.Location.CompareTo(y.Location);
});

现在,上面的代码不是IComparer类,但是比较方法是相同的。

3

0

Matt Hinze 17 年前

public class ScratchComparer : IComparer<Scratch>
{
    public int Compare(Scratch x, Scratch y)
    {
        return x.Foo.CompareTo(y.Foo).CompareTo(0.CompareTo(x.Bar.CompareTo(y.Bar)));
    }
}

[TestFixture]
public class Scratch
{
    public virtual int Foo { get; set; }
    public virtual int Bar { get; set; }

    [Test]
    public void Should_sort()
    {
        var scratches = new[]
        {
            new Scratch {Foo = 1, Bar = 1},
            new Scratch {Foo = 2, Bar = 1},
            new Scratch {Foo = 1, Bar = 1},
            new Scratch {Foo = 1, Bar = 2}
        };

        // IComparer
        Array.Sort(scratches, new ScratchComparer());

        scratches[0].Foo.ShouldEqual(1);
        scratches[0].Bar.ShouldEqual(1);

        scratches[1].Foo.ShouldEqual(1);
        scratches[1].Bar.ShouldEqual(1);

        scratches[2].Foo.ShouldEqual(1);
        scratches[2].Bar.ShouldEqual(2);

        scratches[3].Foo.ShouldEqual(2);
        scratches[3].Bar.ShouldEqual(1);

        // better
        Scratch[] ordered = scratches.OrderBy(x => x.Foo).ThenBy(x => x.Bar).ToArray();

        ordered[0].Foo.ShouldEqual(1);
        ordered[0].Bar.ShouldEqual(1);

        ordered[1].Foo.ShouldEqual(1);
        ordered[1].Bar.ShouldEqual(1);

        ordered[2].Foo.ShouldEqual(1);
        ordered[2].Bar.ShouldEqual(2);

        ordered[3].Foo.ShouldEqual(2);
        ordered[3].Bar.ShouldEqual(1);
    }
}