将自己的 IComparer<T> 与 Linq OrderBy 一起使用

Question

我有一个通用的

List<MyClass>

其中MyClass 有一个属性InvoiceNumber，其中包含以下值：

200906/1
200906/2
..
200906/10
200906/11
200906/12

我的列表绑定到一个

BindingList<T>

支持linq排序：

protected override void ApplySortCore(
           PropertyDescriptor property, ListSortDirection direction)
{

    _sortProperty = property;
    _sortDirection = direction;

    var items = this.Items;

    switch (direction)
    {
        case ListSortDirection.Ascending:
            items = items.OrderByDescending(x => property.GetValue(x)).ToList();
            break;
        case ListSortDirection.Descending:
            items = items.OrderByDescending(x => property.GetValue(x)).ToList();
            break;
    }

    this.Items = items;

}

然而，默认的比较器排序（如假设的那样）如下：

200906/1
200906/10
200906/11
200906/12
200906/2

在这种情况下很糟糕。

现在我想用我自己的IComparer<T> 来处理这个。它看起来像这样：

public class MyComparer : IComparer<Object>
{

    public int Compare(Object stringA, Object stringB)
    {
        String[] valueA = stringA.ToString().Split('/');
        String[] valueB = stringB.ToString().Split('/');

        if(valueA .Length != 2 || valueB .Length != 2)
             return String.Compare(stringA.ToString(), stringB.ToString());

        if (valueA[0] == valueB[0]) 
        {
          return String.Compare(valueA[1], valueB[1]);
        }
        else
        {
          return String.Compare(valueA[0], valueB[0]);
        }

    }

}

并将ApplySortCore 代码更改为使用此IComparer：

case ListSortDirection.Ascending:
    MyComparer comparer = new MyComparer();
    items = items.OrderByDescending(
              x => property.GetValue(x), comparer).ToList();
    break;

当我调试我的代码时，我看到MyComparer.Compare(object, object) 被多次调用并为比较方法返回正确的值（-1、0、1）。

但我的列表仍然以“错误”的方式排序。我错过了什么吗？我不知道。

Answer 1

您的比较器在我看来是错误的。您仍然只是按默认文本顺序排序。当然，您想解析这两个数字并以此为基础进行排序：

public int Compare(Object stringA, Object stringB)
{
    string[] valueA = stringA.ToString().Split('/');
    string[] valueB = stringB.ToString().Split('/');

    if (valueA.Length != 2 || valueB.Length != 2)
    {
        stringA.ToString().CompareTo(stringB.ToString());
    }

    // Note: do error checking and consider i18n issues too :)
    if (valueA[0] == valueB[0]) 
    {
        return int.Parse(valueA[1]).CompareTo(int.Parse(valueB[1]));
    }
    else
    {
        return int.Parse(valueA[0]).CompareTo(int.Parse(valueB[0]));
    }
}

（请注意，这与您的问题不相符，即您已经调试并验证了 Compare 正在返回正确的值 - 但我担心我怀疑这方面的人为错误。）

此外，Sven 的权利 - 更改 items 的值根本不会更改您的绑定列表。您应该添加：

this.Items = items;

在方法的底部。

Answer 2

我遇到了一般自然排序的问题，把解决方法写在这里：

Natural Sort Compare with Linq OrderBy()

public class NaturalSortComparer<T> : IComparer<string>, IDisposable
{
    private bool isAscending;

    public NaturalSortComparer(bool inAscendingOrder = true)
    {
        this.isAscending = inAscendingOrder;
    }

    #region IComparer<string> Members

    public int Compare(string x, string y)
    {
        throw new NotImplementedException();
    }

    #endregion

    #region IComparer<string> Members

    int IComparer<string>.Compare(string x, string y)
    {
        if (x == y)
            return 0;

        string[] x1, y1;

        if (!table.TryGetValue(x, out x1))
        {
            x1 = Regex.Split(x.Replace(" ", ""), "([0-9]+)");
            table.Add(x, x1);
        }

        if (!table.TryGetValue(y, out y1))
        {
            y1 = Regex.Split(y.Replace(" ", ""), "([0-9]+)");
            table.Add(y, y1);
        }

        int returnVal;

        for (int i = 0; i < x1.Length && i < y1.Length; i++)
        {
            if (x1[i] != y1[i])
            {
                returnVal = PartCompare(x1[i], y1[i]);
                return isAscending ? returnVal : -returnVal;
            }
        }

        if (y1.Length > x1.Length)
        {
            returnVal = 1;
        }
        else if (x1.Length > y1.Length)
        { 
            returnVal = -1; 
        }
        else
        {
            returnVal = 0;
        }

        return isAscending ? returnVal : -returnVal;
    }

    private static int PartCompare(string left, string right)
    {
        int x, y;
        if (!int.TryParse(left, out x))
            return left.CompareTo(right);

        if (!int.TryParse(right, out y))
            return left.CompareTo(right);

        return x.CompareTo(y);
    }

    #endregion

    private Dictionary<string, string[]> table = new Dictionary<string, string[]>();

    public void Dispose()
    {
        table.Clear();
        table = null;
    }
}

Answer 3

我们不能这样做吗：

public class MyComparer : IComparer<string>
{

    public int Compare(string stringA, string stringB)
    {
        string small = stringA;
        string big = stringB;
        if (stringA.Length > stringB.Length)
        {
            small = stringB;
            big = stringA;
        }
        else if (stringA.Length < stringB.Length)
        {
            small = stringA;
            big = stringB;
        }
        for (int j = 0; j < small.Length; j++)
        {
            if (Convert.ToInt32(small[j]) > Convert.ToInt32(big[j])) return -1;
            if (Convert.ToInt32(small[j]) < Convert.ToInt32(big[j])) return 1;
        }

        //big is indeed bigger
        if (big.Length > small.Length) return 1;

        //finally they are smae
        return 0;
    }
}

用法：

string[] inputStrings = {"_abc*&","#almnp","abc" };
//string[] inputStrings = { "#", "_", "_a", "@", "_" };
MyComparer computer = new MyComparer();
var kola = inputStrings.OrderBy(x => x, new MyComparer()).ToArray();

这与：

    Array.Sort(inputStrings, StringComparer.Ordinal);

Answer 4

您可以使用 Alphanum 算法：

  (...)
    items.OrderBy(x => property.GetValue(x), new AlphanumComparator())
  (...)

AlphanumComparator

/*
 * The Alphanum Algorithm is an improved sorting algorithm for strings
 * containing numbers.  Instead of sorting numbers in ASCII order like
 * a standard sort, this algorithm sorts numbers in numeric order.
 *
 * The Alphanum Algorithm is discussed at http://www.DaveKoelle.com
 *
 * Based on the Java implementation of Dave Koelle's Alphanum algorithm.
 * Contributed by Jonathan Ruckwood <jonathan.ruckwood@gmail.com>
 *
 * Adapted by Dominik Hurnaus <dominik.hurnaus@gmail.com> to
 *   - correctly sort words where one word starts with another word
 *   - have slightly better performance
 *
 * Released under the MIT License - https://opensource.org/licenses/MIT
 *
 * Permission is hereby granted, free of charge, to any person obtaining
 * a copy of this software and associated documentation files (the "Software"),
 * to deal in the Software without restriction, including without limitation
 * the rights to use, copy, modify, merge, publish, distribute, sublicense,
 * and/or sell copies of the Software, and to permit persons to whom the
 * Software is furnished to do so, subject to the following conditions:
 *
 * The above copyright notice and this permission notice shall be included
 * in all copies or substantial portions of the Software.
 *
 * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
 * EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
 * MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
 * IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM,
 * DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
 * OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE
 * USE OR OTHER DEALINGS IN THE SOFTWARE.
 *
 */
using System;
using System.Collections;
using System.Text;

/*
 * Please compare against the latest Java version at http://www.DaveKoelle.com
 * to see the most recent modifications
 */
namespace AlphanumComparator
{
    public class AlphanumComparator : IComparer
    {
        private enum ChunkType {Alphanumeric, Numeric};
        private bool InChunk(char ch, char otherCh)
        {
            ChunkType type = ChunkType.Alphanumeric;

            if (char.IsDigit(otherCh))
            {
                type = ChunkType.Numeric;
            }

            if ((type == ChunkType.Alphanumeric && char.IsDigit(ch))
                || (type == ChunkType.Numeric && !char.IsDigit(ch)))
            {
                return false;
            }

            return true;
        }

        public int Compare(object x, object y)
        {
            String s1 = x as string;
            String s2 = y as string;
            if (s1 == null || s2 == null)
            {
                return 0;
            }

            int thisMarker = 0, thisNumericChunk = 0;
            int thatMarker = 0, thatNumericChunk = 0;

            while ((thisMarker < s1.Length) || (thatMarker < s2.Length))
            {
                if (thisMarker >= s1.Length)
                {
                    return -1;
                }
                else if (thatMarker >= s2.Length)
                {
                    return 1;
                }
                char thisCh = s1[thisMarker];
                char thatCh = s2[thatMarker];

                StringBuilder thisChunk = new StringBuilder();
                StringBuilder thatChunk = new StringBuilder();

                while ((thisMarker < s1.Length) && (thisChunk.Length==0 ||InChunk(thisCh, thisChunk[0])))
                {
                    thisChunk.Append(thisCh);
                    thisMarker++;

                    if (thisMarker < s1.Length)
                    {
                        thisCh = s1[thisMarker];
                    }
                }

                while ((thatMarker < s2.Length) && (thatChunk.Length==0 ||InChunk(thatCh, thatChunk[0])))
                {
                    thatChunk.Append(thatCh);
                    thatMarker++;

                    if (thatMarker < s2.Length)
                    {
                        thatCh = s2[thatMarker];
                    }
                }

                int result = 0;
                // If both chunks contain numeric characters, sort them numerically
                if (char.IsDigit(thisChunk[0]) && char.IsDigit(thatChunk[0]))
                {
                    thisNumericChunk = Convert.ToInt32(thisChunk.ToString());
                    thatNumericChunk = Convert.ToInt32(thatChunk.ToString());

                    if (thisNumericChunk < thatNumericChunk)
                    {
                        result = -1;
                    }

                    if (thisNumericChunk > thatNumericChunk)
                    {
                        result = 1;
                    }
                }
                else
                {
                    result = thisChunk.ToString().CompareTo(thatChunk.ToString());
                }

                if (result != 0)
                {
                    return result;
                }
            }

            return 0;
        }
    }
}

Answer 5

已排序列表仅绑定到局部变量 items，而不绑定到绑定列表的 Items 属性，因此它保持未排序。

[编辑] 基本上，您只是在丢弃排序工作的结果；-)