在不复制源列表的情况下有效排序IList <T>
鉴于下面的测试案例，我该如何：
IList<int>列表中匹配Id的索引对IList<TestObject>排序。 list[3] == 3和list[4] == 4 。 IList （我不能使用List<T> ） 这是测试：
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：
按照要求，这是我所做的尝试，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，@mitch小麦 - 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));
}
一直在寻找这一点，事实上如前所述，你将需要ArrayList.Adapter，但是你会注意到它需要一个非泛型的IList，因此需要一些投射：
ArrayList.Adapter((IList)list)
你还需要编写一个比较器，其中包含你的排序逻辑。 请原谅这个名字，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：添加到上面的比较器的实现
那么用法如下：
IList<int> indexList = new[] { 10, 5, 1, 9, 2 };
ArrayList.Adapter((IList)list).Sort(new WeirdComparer(indexList));
顺便说一句，这个线程解释了一个很好的方法来将它变成一个扩展方法，这将使你的代码更加可重用并且更易于读取IMO。
您可以尝试以下方法：
ArrayList.Adapter(yourilist).Sort();
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));
}
}
用法：
ArrayList.Adapter(list).Sort<int>((x,y) => x - y);
这是比较器的通用版本。 IEntity<TId>只是一个简单的接口，它具有TId类型的属性“Id”：
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。 不匹配的元素将移动到列表的末尾，然后按其原始索引排序。
链接地址: http://www.djcxy.com/p/53609.html上一篇: Efficiently sort an IList<T> without copying the source list