Question

鉴于下面的测试用例，我该如何：

根据匹配的IList<TestObject>索引对Id进行排序在IList<int>列表中。
不匹配的值将移至列表末尾并按其原始索引排序。在这种情况下，由于索引列表中不存在3和4，因此我们希望看到list[3] == 3和list[4] == 4。
虽然我知道这可以通过linq实现，但我需要使用原始列表而不是创建一个新列表（由于列表的存储方式）。
源列表必须是IList（我不能使用List<T>）

以下是测试：

    public class TestObject
    {
        public int Id { get; set; }
    }

    [Test]
    public void Can_reorder_using_index_list()
    {
        IList<TestObject> list = new List<TestObject>
        {
            new TestObject { Id = 1 },
            new TestObject { Id = 2 },
            new TestObject { Id = 3 },
            new TestObject { Id = 4 },
            new TestObject { Id = 5 }
        };

        IList<int> indexList = new[] { 10, 5, 1, 9, 2 };

        // TODO sort

        Assert.That(list[0].Id, Is.EqualTo(5));
        Assert.That(list[1].Id, Is.EqualTo(1));
        Assert.That(list[2].Id, Is.EqualTo(2));
        Assert.That(list[3].Id, Is.EqualTo(3));
        Assert.That(list[4].Id, Is.EqualTo(4));
    }

更新

根据要求，这是我尝试过的，但1）它只适用于List<T>和2）我不确定它是最有效的方式：

       var clone = list.ToList();
        list.Sort((x, y) =>
        {
            var xIndex = indexList.IndexOf(x.Id);
            var yIndex = indexList.IndexOf(y.Id);

            if (xIndex == -1)
            {
                xIndex = list.Count + clone.IndexOf(x);
            }
            if (yIndex == -1)
            {
                yIndex = list.Count + clone.IndexOf(y);
            }

            return xIndex.CompareTo(yIndex);
        });

更新2：

感谢@leppie，@ jamiec，@ maych wheat - 这是工作代码：

    public class TestObjectComparer : Comparer<TestObject>
    {
        private readonly IList<int> indexList;
        private readonly Func<TestObject, int> currentIndexFunc;
        private readonly int listCount;

        public TestObjectComparer(IList<int> indexList, Func<TestObject, int> currentIndexFunc, int listCount)
        {
            this.indexList = indexList;
            this.currentIndexFunc = currentIndexFunc;
            this.listCount = listCount;
        }

        public override int Compare(TestObject x, TestObject y)
        {
            var xIndex = indexList.IndexOf(x.Id);
            var yIndex = indexList.IndexOf(y.Id);

            if (xIndex == -1)
            {
                xIndex = listCount + currentIndexFunc(x);
            }
            if (yIndex == -1)
            {
                yIndex = listCount + currentIndexFunc(y);
            }

            return xIndex.CompareTo(yIndex);
        }
    }

    [Test]
    public void Can_reorder_using_index_list()
    {
        IList<TestObject> list = new List<TestObject>
        {
            new TestObject { Id = 1 },
            new TestObject { Id = 2 },
            new TestObject { Id = 3 },
            new TestObject { Id = 4 },
            new TestObject { Id = 5 }
        };

        IList<int> indexList = new[] { 10, 5, 1, 9, 2, 4 };

        ArrayList.Adapter((IList)list).Sort(new TestObjectComparer(indexList, x => list.IndexOf(x), list.Count));

        Assert.That(list[0].Id, Is.EqualTo(5));
        Assert.That(list[1].Id, Is.EqualTo(1));
        Assert.That(list[2].Id, Is.EqualTo(2));
        Assert.That(list[3].Id, Is.EqualTo(3));
        Assert.That(list[4].Id, Is.EqualTo(4));
    }

Answer 1

您可以尝试以下操作：

ArrayList.Adapter(yourilist).Sort();

<强>更新

通用比较器：

class Comparer<T> : IComparer<T>, IComparer
{
  internal Func<T, T, int> pred;

  public int Compare(T x, T y)
  {
    return pred(x, y);  
  }

  public int Compare(object x, object y)
  {
    return Compare((T)x, (T)y);
  }
}

static class ComparerExtensions
{
  static IComparer Create<T>(Func<T, T, int> pred)
  {
    return new Comparer<T> { pred = pred };
  }

  public static void Sort<T>(this ArrayList l, Func<T, T, int> pred)
  {
    l.Sort(Create(pred));
  }
}

<强>用法：

ArrayList.Adapter(list).Sort<int>((x,y) => x - y);

Answer 2

正在考虑这一点，事实上如前所述，你需要ArrayList.Adapter，但是你会注意到它需要一个非通用的IList，所以需要一些强制转换：

ArrayList.Adapter((IList)list)

您还需要编写一个比较器，其中包含进行排序的逻辑。请原谅，但是：

public class WeirdComparer : IComparer,IComparer<TestObject>
{
    private IList<int> order;
    public WeirdComparer(IList<int> order)
    {
        this.order = order;
    }
    public int Compare(object x, object y)
    {
        return Compare((TestObject) x, (TestObject) y);
    }

    public int Compare(TestObject x, TestObject y)
    {
        if(order.Contains(x.Id))
        {
            if(order.Contains(y.Id))
            {
                return order.IndexOf(x.Id).CompareTo(order.IndexOf(y.Id));    
            }
            return -1;
        }
        else
        {
            if (order.Contains(y.Id))
            {
                return 1;
            }
            return x.Id.CompareTo(y.Id);
        }
    }
}

编辑：在上面的comparerr中添加了实现

然后用法如下：

IList<int> indexList = new[] { 10, 5, 1, 9, 2 };
ArrayList.Adapter((IList)list).Sort(new WeirdComparer(indexList));

顺便说一句，this thread解释了一种很好的方法，可以将其转换为扩展方法，使您的代码更易于重复使用，更容易阅读IMO。

Answer 3

这是比较器的通用版本。 IEntity<TId>只是一个简单的界面，其属性“Id”的类型为TId：

public class IndexComparer<T, TId> : Comparer<T> where T : IEntity<TId> where TId : IComparable
{
    private readonly IList<TId> order;
    private readonly int listCount;
    private readonly Func<T, int> currentIndexFunc;

    public IndexComparer(Func<T, int> currentIndexFunc, IList<TId> order, int listCount) {
        this.order = order;
        this.listCount = listCount;
        this.currentIndexFunc = currentIndexFunc;
    }

    public override int Compare(T x, T y)
    {
        var xIndex = order.IndexOf(x.Id);
        var yIndex = order.IndexOf(y.Id);

        if (xIndex == -1)
        {
            xIndex = listCount + currentIndexFunc(x);
        }
        if (yIndex == -1)
        {
            yIndex = listCount + currentIndexFunc(y);
        }

        return xIndex.CompareTo(yIndex);
    }
}

工作测试：

[TestFixture]
public class OrderingTests
{
    public class TestObject : IEntity<int>
    {
        public int Id { get; set; }
    }

    [Test]
    public void Can_reorder_using_index_list()
    {
        IList<TestObject> list = new List<TestObject>
        {
            new TestObject { Id = 1 },
            new TestObject { Id = 2 },
            new TestObject { Id = 3 },
            new TestObject { Id = 4 },
            new TestObject { Id = 5 }
        };

        IList<int> indexList = new[] { 10, 5, 1, 9, 2, 4 };
        ArrayList.Adapter((IList)list)
            .Sort(new IndexComparer<TestObject, int>(x => list.IndexOf(x), indexList, list.Count));

        Assert.That(list[0].Id, Is.EqualTo(5));
        Assert.That(list[1].Id, Is.EqualTo(1));
        Assert.That(list[2].Id, Is.EqualTo(2));
        Assert.That(list[3].Id, Is.EqualTo(4));
        Assert.That(list[4].Id, Is.EqualTo(3));
    }
}

这符合我原始问题中列出的要求。不匹配的元素移动到列表的末尾，然后按原始索引排序。

有效地对IList <t>进行排序而不复制源列表</t>

3 个答案: