C#中的自然排序顺序


Answers:


148

最简单的操作是在Windows中P / Invoke内置函数,并将其用作比较函数IComparer

[DllImport("shlwapi.dll", CharSet = CharSet.Unicode)]
private static extern int StrCmpLogicalW(string psz1, string psz2);

迈克尔·卡普兰(Michael Kaplan)举例说明了此功能的工作原理,以及为使Vista更加直观而进行的更改。此功能的好处是,它将具有与其运行的Windows版本相同的行为,但这确实意味着它在Windows版本之间有所不同,因此您需要考虑这是否对您有问题。

因此,完整的实现应类似于:

[SuppressUnmanagedCodeSecurity]
internal static class SafeNativeMethods
{
    [DllImport("shlwapi.dll", CharSet = CharSet.Unicode)]
    public static extern int StrCmpLogicalW(string psz1, string psz2);
}

public sealed class NaturalStringComparer : IComparer<string>
{
    public int Compare(string a, string b)
    {
        return SafeNativeMethods.StrCmpLogicalW(a, b);
    }
}

public sealed class NaturalFileInfoNameComparer : IComparer<FileInfo>
{
    public int Compare(FileInfo a, FileInfo b)
    {
        return SafeNativeMethods.StrCmpLogicalW(a.Name, b.Name);
    }
}

8
好答案。警告:这对于Win2000不起作用,因为那几个人仍然在该操作系统上运行东西。另一方面,Kaplan的博客和MSDN文档之间有足够的提示来创建类似的功能。
克里斯·查拉巴鲁克

9
这不是便携式的,仅适用于Win32,但不适用于Linux / MacOS / Silverlight / Windows Phone / Metro
linquize 2012年

20
@linquize-他说.NET不是Mono,因此Linux / OSX并不是真正的问题。发布此答案时,Windows Phone / Metro在2008年不存在。以及您多久在Silverlight中执行一次文件操作?因此,对于OP以及大多数其他人来说,这是一个合适的答案。无论如何,您都可以自由提供更好的答案;这就是这个网站的运作方式。
格雷格·比奇

6
这并不意味着原始答案是错误的。我只是添加了具有最新信息的其他信息
linquize 2012年

2
仅供参考,如果您继承自Comparer<T>而不是实现IComparer<T>,则会获得IComparer(非泛型)接口的内置实现,该接口调用您的泛型方法,供在使用该泛型方法的API中使用。基本上也是免费的:只需删除“ I”并更改public int Compare(...)为即可public override int Compare(...)。与IEqualityComparer<T>和相同EqualityComparer<T>
Joe Amenta 2015年

75

只是以为我会添加(使用我能找到的最简洁的解决方案):

public static IOrderedEnumerable<T> OrderByAlphaNumeric<T>(this IEnumerable<T> source, Func<T, string> selector)
{
    int max = source
        .SelectMany(i => Regex.Matches(selector(i), @"\d+").Cast<Match>().Select(m => (int?)m.Value.Length))
        .Max() ?? 0;

    return source.OrderBy(i => Regex.Replace(selector(i), @"\d+", m => m.Value.PadLeft(max, '0')));
}

上面的代码将字符串中的任何数字填充到所有字符串中所有数字的最大长度,并使用结果字符串进行排序。

强制转换为(int?)是为了允许不包含任何数字的字符串集合(.Max()在空的可枚举抛出InvalidOperationException)。


1
+1它不仅是最简洁的,而且是我见过的最快的。除了可接受的答案外,由于机器依赖性,我无法使用该答案。它在大约35秒内对400万个值进行了排序。
Gene S

4
这既美丽又无法阅读。我认为Linq的优势将(至少)意味着最佳的平均性能和最佳情况,因此,我认为我会继续考虑。尽管缺乏明确性。非常感谢@Matthew Horsley
Ian Grainger

1
这非常好,但是某些十进制数字存在一个错误,我的示例是对k8.11与k8.2进行排序。为了解决这个问题,我实现了以下正则表达式:\ d +([[..] \ d)?
devzero

2
填充此代码m.Value.PadLeft(max,'0')
devzero

3
我认为您可以使用.DefaultIfEmpty().Max()而不是强制转换为int?。同样值得做的source.ToList()是避免重新枚举可枚举。
Teejay

30

现有的实现都没有一个看起来很棒,因此我编写了自己的实现。结果几乎与现代版本的Windows资源管理器(Windows 7/8)所使用的排序相同。我看到的唯一区别是1)尽管Windows曾经使用过(例如XP)处理任意长度的数字,但现在仅限于19位数字-我的是无限的,2)Windows提供的某些Unicode数字集结果不一致-我的作品很好(尽管它不会对代理对中的数字进行数字比较; Windows也不会),并且3)如果它们出现在不同的部分(例如“e-1é”与“ é1e-“-数字前后的部分有变音符号和标点符号的重量差异)。

public static int CompareNatural(string strA, string strB) {
    return CompareNatural(strA, strB, CultureInfo.CurrentCulture, CompareOptions.IgnoreCase);
}

public static int CompareNatural(string strA, string strB, CultureInfo culture, CompareOptions options) {
    CompareInfo cmp = culture.CompareInfo;
    int iA = 0;
    int iB = 0;
    int softResult = 0;
    int softResultWeight = 0;
    while (iA < strA.Length && iB < strB.Length) {
        bool isDigitA = Char.IsDigit(strA[iA]);
        bool isDigitB = Char.IsDigit(strB[iB]);
        if (isDigitA != isDigitB) {
            return cmp.Compare(strA, iA, strB, iB, options);
        }
        else if (!isDigitA && !isDigitB) {
            int jA = iA + 1;
            int jB = iB + 1;
            while (jA < strA.Length && !Char.IsDigit(strA[jA])) jA++;
            while (jB < strB.Length && !Char.IsDigit(strB[jB])) jB++;
            int cmpResult = cmp.Compare(strA, iA, jA - iA, strB, iB, jB - iB, options);
            if (cmpResult != 0) {
                // Certain strings may be considered different due to "soft" differences that are
                // ignored if more significant differences follow, e.g. a hyphen only affects the
                // comparison if no other differences follow
                string sectionA = strA.Substring(iA, jA - iA);
                string sectionB = strB.Substring(iB, jB - iB);
                if (cmp.Compare(sectionA + "1", sectionB + "2", options) ==
                    cmp.Compare(sectionA + "2", sectionB + "1", options))
                {
                    return cmp.Compare(strA, iA, strB, iB, options);
                }
                else if (softResultWeight < 1) {
                    softResult = cmpResult;
                    softResultWeight = 1;
                }
            }
            iA = jA;
            iB = jB;
        }
        else {
            char zeroA = (char)(strA[iA] - (int)Char.GetNumericValue(strA[iA]));
            char zeroB = (char)(strB[iB] - (int)Char.GetNumericValue(strB[iB]));
            int jA = iA;
            int jB = iB;
            while (jA < strA.Length && strA[jA] == zeroA) jA++;
            while (jB < strB.Length && strB[jB] == zeroB) jB++;
            int resultIfSameLength = 0;
            do {
                isDigitA = jA < strA.Length && Char.IsDigit(strA[jA]);
                isDigitB = jB < strB.Length && Char.IsDigit(strB[jB]);
                int numA = isDigitA ? (int)Char.GetNumericValue(strA[jA]) : 0;
                int numB = isDigitB ? (int)Char.GetNumericValue(strB[jB]) : 0;
                if (isDigitA && (char)(strA[jA] - numA) != zeroA) isDigitA = false;
                if (isDigitB && (char)(strB[jB] - numB) != zeroB) isDigitB = false;
                if (isDigitA && isDigitB) {
                    if (numA != numB && resultIfSameLength == 0) {
                        resultIfSameLength = numA < numB ? -1 : 1;
                    }
                    jA++;
                    jB++;
                }
            }
            while (isDigitA && isDigitB);
            if (isDigitA != isDigitB) {
                // One number has more digits than the other (ignoring leading zeros) - the longer
                // number must be larger
                return isDigitA ? 1 : -1;
            }
            else if (resultIfSameLength != 0) {
                // Both numbers are the same length (ignoring leading zeros) and at least one of
                // the digits differed - the first difference determines the result
                return resultIfSameLength;
            }
            int lA = jA - iA;
            int lB = jB - iB;
            if (lA != lB) {
                // Both numbers are equivalent but one has more leading zeros
                return lA > lB ? -1 : 1;
            }
            else if (zeroA != zeroB && softResultWeight < 2) {
                softResult = cmp.Compare(strA, iA, 1, strB, iB, 1, options);
                softResultWeight = 2;
            }
            iA = jA;
            iB = jB;
        }
    }
    if (iA < strA.Length || iB < strB.Length) {
        return iA < strA.Length ? 1 : -1;
    }
    else if (softResult != 0) {
        return softResult;
    }
    return 0;
}

签名与Comparison<string>委托人匹配:

string[] files = Directory.GetFiles(@"C:\");
Array.Sort(files, CompareNatural);

这是一个包装器类,可用作IComparer<string>

public class CustomComparer<T> : IComparer<T> {
    private Comparison<T> _comparison;

    public CustomComparer(Comparison<T> comparison) {
        _comparison = comparison;
    }

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

例:

string[] files = Directory.EnumerateFiles(@"C:\")
    .OrderBy(f => f, new CustomComparer<string>(CompareNatural))
    .ToArray();

这是我用于测试的一组好的文件名:

Func<string, string> expand = (s) => { int o; while ((o = s.IndexOf('\\')) != -1) { int p = o + 1;
    int z = 1; while (s[p] == '0') { z++; p++; } int c = Int32.Parse(s.Substring(p, z));
    s = s.Substring(0, o) + new string(s[o - 1], c) + s.Substring(p + z); } return s; };
string encodedFileNames =
    "KDEqLW4xMiotbjEzKjAwMDFcMDY2KjAwMlwwMTcqMDA5XDAxNyowMlwwMTcqMDlcMDE3KjEhKjEtISox" +
    "LWEqMS4yNT8xLjI1KjEuNT8xLjUqMSoxXDAxNyoxXDAxOCoxXDAxOSoxXDA2NioxXDA2NyoxYSoyXDAx" +
    "NyoyXDAxOCo5XDAxNyo5XDAxOCo5XDA2Nio9MSphMDAxdGVzdDAxKmEwMDF0ZXN0aW5nYTBcMzEqYTAw" +
    "Mj9hMDAyIGE/YTAwMiBhKmEwMDIqYTAwMmE/YTAwMmEqYTAxdGVzdGluZ2EwMDEqYTAxdnNmcyphMSph" +
    "MWEqYTF6KmEyKmIwMDAzcTYqYjAwM3E0KmIwM3E1KmMtZSpjZCpjZipmIDEqZipnP2cgMT9oLW4qaG8t" +
    "bipJKmljZS1jcmVhbT9pY2VjcmVhbT9pY2VjcmVhbS0/ajBcNDE/ajAwMWE/ajAxP2shKmsnKmstKmsx" +
    "KmthKmxpc3QqbTAwMDNhMDA1YSptMDAzYTAwMDVhKm0wMDNhMDA1Km0wMDNhMDA1YSpuMTIqbjEzKm8t" +
    "bjAxMypvLW4xMipvLW40P28tbjQhP28tbjR6P28tbjlhLWI1Km8tbjlhYjUqb24wMTMqb24xMipvbjQ/" +
    "b240IT9vbjR6P29uOWEtYjUqb245YWI1Km/CrW4wMTMqb8KtbjEyKnAwMCpwMDEqcDAxwr0hKnAwMcK9" +
    "KnAwMcK9YSpwMDHCvcK+KnAwMipwMMK9KnEtbjAxMypxLW4xMipxbjAxMypxbjEyKnItMDAhKnItMDAh" +
    "NSpyLTAwIe+8lSpyLTAwYSpyLe+8kFwxIS01KnIt77yQXDEhLe+8lSpyLe+8kFwxISpyLe+8kFwxITUq" +
    "ci3vvJBcMSHvvJUqci3vvJBcMWEqci3vvJBcMyE1KnIwMCEqcjAwLTUqcjAwLjUqcjAwNSpyMDBhKnIw" +
    "NSpyMDYqcjQqcjUqctmg2aYqctmkKnLZpSpy27Dbtipy27Qqctu1KnLfgN+GKnLfhCpy34UqcuClpuCl" +
    "rCpy4KWqKnLgpasqcuCnpuCnrCpy4KeqKnLgp6sqcuCppuCprCpy4KmqKnLgqasqcuCrpuCrrCpy4Kuq" +
    "KnLgq6sqcuCtpuCtrCpy4K2qKnLgrasqcuCvpuCvrCpy4K+qKnLgr6sqcuCxpuCxrCpy4LGqKnLgsasq" +
    "cuCzpuCzrCpy4LOqKnLgs6sqcuC1puC1rCpy4LWqKnLgtasqcuC5kOC5lipy4LmUKnLguZUqcuC7kOC7" +
    "lipy4LuUKnLgu5UqcuC8oOC8pipy4LykKnLgvKUqcuGBgOGBhipy4YGEKnLhgYUqcuGCkOGClipy4YKU" +
    "KnLhgpUqcuGfoOGfpipy4Z+kKnLhn6UqcuGgkOGglipy4aCUKnLhoJUqcuGlhuGljCpy4aWKKnLhpYsq" +
    "cuGnkOGnlipy4aeUKnLhp5UqcuGtkOGtlipy4a2UKnLhrZUqcuGusOGutipy4a60KnLhrrUqcuGxgOGx" +
    "hipy4bGEKnLhsYUqcuGxkOGxlipy4bGUKnLhsZUqcuqYoFwx6pilKnLqmKDqmKUqcuqYoOqYpipy6pik" +
    "KnLqmKUqcuqjkOqjlipy6qOUKnLqo5UqcuqkgOqkhipy6qSEKnLqpIUqcuqpkOqplipy6qmUKnLqqZUq" +
    "cvCQkqAqcvCQkqUqcvCdn5gqcvCdn50qcu+8kFwxISpy77yQXDEt77yVKnLvvJBcMS7vvJUqcu+8kFwx" +
    "YSpy77yQXDHqmKUqcu+8kFwx77yO77yVKnLvvJBcMe+8lSpy77yQ77yVKnLvvJDvvJYqcu+8lCpy77yV" +
    "KnNpKnPEsSp0ZXN02aIqdGVzdNmi2aAqdGVzdNmjKnVBZS0qdWFlKnViZS0qdUJlKnVjZS0xw6kqdWNl" +
    "McOpLSp1Y2Uxw6kqdWPDqS0xZSp1Y8OpMWUtKnVjw6kxZSp3ZWlhMSp3ZWlhMip3ZWlzczEqd2Vpc3My" +
    "KndlaXoxKndlaXoyKndlacOfMSp3ZWnDnzIqeSBhMyp5IGE0KnknYTMqeSdhNCp5K2EzKnkrYTQqeS1h" +
    "Myp5LWE0KnlhMyp5YTQqej96IDA1MD96IDIxP3ohMjE/ejIwP3oyMj96YTIxP3rCqTIxP1sxKl8xKsKt" +
    "bjEyKsKtbjEzKsSwKg==";
string[] fileNames = Encoding.UTF8.GetString(Convert.FromBase64String(encodedFileNames))
    .Replace("*", ".txt?").Split(new[] { "?" }, StringSplitOptions.RemoveEmptyEntries)
    .Select(n => expand(n)).ToArray();

数字部分需要按部分进行比较,即“ abc12b”应小于“ abc123”。
SOUser

您可以尝试以下数据:public string [] filenames = {“ -abc12.txt”,“ abc12.txt”,“ 1abc_2.txt”,“ a0000012.txt”,“ a0000012c.txt”,“ a000012.txt” ,“ a000012b.txt”,“ a012.txt”,“ a0000102.txt”,“ abc1_2.txt”,“ abc12 .txt”,“ abc12b.txt”,“ abc123.txt”,“ abccde.txt”,“ b0000.txt”,“ b00001.txt”,“ b0001.txt”,“ b001.txt”,“ c0000.txt”,“ c0000c.txt”,“ c00001.txt”,“ c000b.txt”,“ d0”。 20.2b.txt”,“ d0.1000c.txt”,“ d0.2000y.txt”,“ d0.20000.2b.txt”,“
SOUser

@XichenLi感谢您的良好测试案例。如果让Windows资源管理器对这些文件进行排序,则将根据所使用的Windows版本获得不同的结果。我的代码对这些名称的排序方式与Server 2003(可能是XP)相同,但与Windows 8不同。如果有机会,我将尝试弄清楚Windows 8的工作方式并更新我的代码。
JD

2
有错误。索引超出范围
linquize 2013年

3
很好的解决方案!当我在大约10,000个文件的正常情况下对它进行基准测试时,它比Matthew的regex示例要快,并且性能与StrCmpLogicalW()差不多。上面的代码中有一个小错误:“ while(strA [jA] == zeroA)jA ++;” 和“ while(strB [jB] == zeroB)jB ++;” 应该是“ while(jA <strA.Length && strA [jA] == zeroA)jA ++;”;和“ while(jB <strB.Length && strB [jB] == zeroB)jB ++;”。否则,仅包含零的字符串将引发异常。
kuroki 2015年

22

用于linq orderby的纯C#解决方案:

http://zootfroot.blogspot.com/2009/09/natural-sort-compare-with-linq-orderby.html

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;
    }
}

2
该代码最终来自codeproject.com/KB/recipes/NaturalComparer.aspx(不是面向LINQ的)。
mhenry1384

2
该博客文章将Justin Jones(codeproject.com/KB/string/NaturalSortComparer.aspx归功于 IComparer,而不是Pascal Ganaye。
詹姆斯·麦考马克

1
值得注意的是,此解决方案将忽略与Windows相同的空格,并且不如下面的Matthew Horsley的代码好。因此,您可能会得到例如'string01''string 01''string 02''string02'(看起来很丑)。如果删除空格,它会将字符串向后排序,即'string01'在'string 01'之前,这可能会或可能不会。
2014年

这适用于地址,例如“ 1 Smith Rd”,“ 10 Smith Rd”,“ 2 Smith Rd”等-自然排序。是! 好一个!
Piotr Kula 2015年

顺便说一句,我注意到(而且该链接页面上的注释似乎也表明)Type参数<T>是完全不必要的。
jv-dev

18

Matthews Horsleys的答案是最快的方法,不会根据程序在哪个版本的Windows上更改行为。但是,通过一次创建正则表达式并使用RegexOptions.Compiled,它甚至可以更快。我还添加了插入字符串比较器的选项,因此可以根据需要忽略大小写,并提高了可读性。

    public static IEnumerable<T> OrderByNatural<T>(this IEnumerable<T> items, Func<T, string> selector, StringComparer stringComparer = null)
    {
        var regex = new Regex(@"\d+", RegexOptions.Compiled);

        int maxDigits = items
                      .SelectMany(i => regex.Matches(selector(i)).Cast<Match>().Select(digitChunk => (int?)digitChunk.Value.Length))
                      .Max() ?? 0;

        return items.OrderBy(i => regex.Replace(selector(i), match => match.Value.PadLeft(maxDigits, '0')), stringComparer ?? StringComparer.CurrentCulture);
    }

使用者

var sortedEmployees = employees.OrderByNatural(emp => emp.Name);

排序100,000个字符串需要450毫秒,而默认.net字符串比较则需要300毫秒-相当快!



16

我的解决方案:

void Main()
{
    new[] {"a4","a3","a2","a10","b5","b4","b400","1","C1d","c1d2"}.OrderBy(x => x, new NaturalStringComparer()).Dump();
}

public class NaturalStringComparer : IComparer<string>
{
    private static readonly Regex _re = new Regex(@"(?<=\D)(?=\d)|(?<=\d)(?=\D)", RegexOptions.Compiled);

    public int Compare(string x, string y)
    {
        x = x.ToLower();
        y = y.ToLower();
        if(string.Compare(x, 0, y, 0, Math.Min(x.Length, y.Length)) == 0)
        {
            if(x.Length == y.Length) return 0;
            return x.Length < y.Length ? -1 : 1;
        }
        var a = _re.Split(x);
        var b = _re.Split(y);
        int i = 0;
        while(true)
        {
            int r = PartCompare(a[i], b[i]);
            if(r != 0) return r;
            ++i;
        }
    }

    private static int PartCompare(string x, string y)
    {
        int a, b;
        if(int.TryParse(x, out a) && int.TryParse(y, out b))
            return a.CompareTo(b);
        return x.CompareTo(y);
    }
}

结果:

1
a2
a3
a4
a10
b4
b5
b400
C1d
c1d2

我喜欢。这很容易理解,不需要Linq。

11

您确实需要小心-我隐约记得阅读过StrCmpLogicalW或类似的东西不是严格可传递的,而且我观察到.NET的sort方法有时会在比较函数违反该规则时陷入无限循环。

如果a <b和b <c,可及比较将始终报告a <c。有一个功能可以进行自然的排序比较,但并不总是满足该条件,但是我不记得它是StrCmpLogicalW还是其他产品。


您对此声明有任何证据吗?谷歌搜索后,我找不到任何迹象表明这是真的。
mhenry1384

1
我已经用StrCmpLogicalW体验了那些无限循环。
THD


Visual Studio反馈项目236900不再存在,但是这里有一个更新的可确认问题的新项目: connect.microsoft.com/VisualStudio/feedback/details/774540/… 它还提供了一种解决方法:CultureInfo具有一个属性CompareInfo,它返回的对象可以为您提供SortKey对象。反过来,这些可以进行比较并保证可传递性。
乔纳森·吉尔伯特

9

这是我的代码,用于对同时包含字母和数字字符的字符串进行排序。

首先,此扩展方法:

public static IEnumerable<string> AlphanumericSort(this IEnumerable<string> me)
{
    return me.OrderBy(x => Regex.Replace(x, @"\d+", m => m.Value.PadLeft(50, '0')));
}

然后,只需在代码中的任何地方使用它,如下所示:

List<string> test = new List<string>() { "The 1st", "The 12th", "The 2nd" };
test = test.AlphanumericSort();

它是如何运作的?用零代替:

  Original  | Regex Replace |      The      |   Returned
    List    | Apply PadLeft |    Sorting    |     List
            |               |               |
 "The 1st"  |  "The 001st"  |  "The 001st"  |  "The 1st"
 "The 12th" |  "The 012th"  |  "The 002nd"  |  "The 2nd"
 "The 2nd"  |  "The 002nd"  |  "The 012th"  |  "The 12th"

使用倍数:

 Alphabetical Sorting | Alphanumeric Sorting
                      |
 "Page 21, Line 42"   | "Page 3, Line 7"
 "Page 21, Line 5"    | "Page 3, Line 32"
 "Page 3, Line 32"    | "Page 21, Line 5"
 "Page 3, Line 7"     | "Page 21, Line 42"

希望这会有所帮助。


6

添加到Greg Beech的答案中(因为我一直在搜索该答案),如果您想在Linq中使用它OrderBy,则可以使用带有的IComparer。例如:

var items = new List<MyItem>();

// fill items

var sorted = items.OrderBy(item => item.Name, new NaturalStringComparer());

2

这是一个相对简单的示例,它不使用P / Invoke并避免在执行过程中进行任何分配。

internal sealed class NumericStringComparer : IComparer<string>
{
    public static NumericStringComparer Instance { get; } = new NumericStringComparer();

    public int Compare(string x, string y)
    {
        // sort nulls to the start
        if (x == null)
            return y == null ? 0 : -1;
        if (y == null)
            return 1;

        var ix = 0;
        var iy = 0;

        while (true)
        {
            // sort shorter strings to the start
            if (ix >= x.Length)
                return iy >= y.Length ? 0 : -1;
            if (iy >= y.Length)
                return 1;

            var cx = x[ix];
            var cy = y[iy];

            int result;
            if (char.IsDigit(cx) && char.IsDigit(cy))
                result = CompareInteger(x, y, ref ix, ref iy);
            else
                result = cx.CompareTo(y[iy]);

            if (result != 0)
                return result;

            ix++;
            iy++;
        }
    }

    private static int CompareInteger(string x, string y, ref int ix, ref int iy)
    {
        var lx = GetNumLength(x, ix);
        var ly = GetNumLength(y, iy);

        // shorter number first (note, doesn't handle leading zeroes)
        if (lx != ly)
            return lx.CompareTo(ly);

        for (var i = 0; i < lx; i++)
        {
            var result = x[ix++].CompareTo(y[iy++]);
            if (result != 0)
                return result;
        }

        return 0;
    }

    private static int GetNumLength(string s, int i)
    {
        var length = 0;
        while (i < s.Length && char.IsDigit(s[i++]))
            length++;
        return length;
    }
}

它不会忽略前导零,所以01在之后2

对应的单元测试:

public class NumericStringComparerTests
{
    [Fact]
    public void OrdersCorrectly()
    {
        AssertEqual("", "");
        AssertEqual(null, null);
        AssertEqual("Hello", "Hello");
        AssertEqual("Hello123", "Hello123");
        AssertEqual("123", "123");
        AssertEqual("123Hello", "123Hello");

        AssertOrdered("", "Hello");
        AssertOrdered(null, "Hello");
        AssertOrdered("Hello", "Hello1");
        AssertOrdered("Hello123", "Hello124");
        AssertOrdered("Hello123", "Hello133");
        AssertOrdered("Hello123", "Hello223");
        AssertOrdered("123", "124");
        AssertOrdered("123", "133");
        AssertOrdered("123", "223");
        AssertOrdered("123", "1234");
        AssertOrdered("123", "2345");
        AssertOrdered("0", "1");
        AssertOrdered("123Hello", "124Hello");
        AssertOrdered("123Hello", "133Hello");
        AssertOrdered("123Hello", "223Hello");
        AssertOrdered("123Hello", "1234Hello");
    }

    private static void AssertEqual(string x, string y)
    {
        Assert.Equal(0, NumericStringComparer.Instance.Compare(x, y));
        Assert.Equal(0, NumericStringComparer.Instance.Compare(y, x));
    }

    private static void AssertOrdered(string x, string y)
    {
        Assert.Equal(-1, NumericStringComparer.Instance.Compare(x, y));
        Assert.Equal( 1, NumericStringComparer.Instance.Compare(y, x));
    }
}

2

实际上,我已经将其实现为的扩展方法,StringComparer因此您可以执行以下操作:

  • StringComparer.CurrentCulture.WithNaturalSort() 要么
  • StringComparer.OrdinalIgnoreCase.WithNaturalSort()

所产生的IComparer<string>可以在所有的地方,如使用OrderByOrderByDescendingThenByThenByDescendingSortedSet<string>,等你仍然可以轻松地调整大小写,文化等

该实现相当琐碎,即使在较大的序列上也应该表现良好。


我也把它作为一个很小的东西发布了 NuGet包,因此您可以这样做:

Install-Package NaturalSort.Extension

该代码包括XML文档注释和测试套件可在NaturalSort.Extension GitHub存储库中找到


整个代码是这样的(如果您仍然不能使用C#7,则只需安装NuGet软件包):

public static class StringComparerNaturalSortExtension
{
    public static IComparer<string> WithNaturalSort(this StringComparer stringComparer) => new NaturalSortComparer(stringComparer);

    private class NaturalSortComparer : IComparer<string>
    {
        public NaturalSortComparer(StringComparer stringComparer)
        {
            _stringComparer = stringComparer;
        }

        private readonly StringComparer _stringComparer;
        private static readonly Regex NumberSequenceRegex = new Regex(@"(\d+)", RegexOptions.Compiled | RegexOptions.CultureInvariant);
        private static string[] Tokenize(string s) => s == null ? new string[] { } : NumberSequenceRegex.Split(s);
        private static ulong ParseNumberOrZero(string s) => ulong.TryParse(s, NumberStyles.None, CultureInfo.InvariantCulture, out var result) ? result : 0;

        public int Compare(string s1, string s2)
        {
            var tokens1 = Tokenize(s1);
            var tokens2 = Tokenize(s2);

            var zipCompare = tokens1.Zip(tokens2, TokenCompare).FirstOrDefault(x => x != 0);
            if (zipCompare != 0)
                return zipCompare;

            var lengthCompare = tokens1.Length.CompareTo(tokens2.Length);
            return lengthCompare;
        }
        
        private int TokenCompare(string token1, string token2)
        {
            var number1 = ParseNumberOrZero(token1);
            var number2 = ParseNumberOrZero(token2);

            var numberCompare = number1.CompareTo(number2);
            if (numberCompare != 0)
                return numberCompare;

            var stringCompare = _stringComparer.Compare(token1, token2);
            return stringCompare;
        }
    }
}

2

这是一种简单的单行无正则表达式LINQ方式(从python借用):

var alphaStrings = new List<string>() { "10","2","3","4","50","11","100","a12","b12" };
var orderedString = alphaStrings.OrderBy(g => new Tuple<int, string>(g.ToCharArray().All(char.IsDigit)? int.Parse(g) : int.MaxValue, g));
// Order Now: ["2","3","4","10","11","50","100","a12","b12"]

删除了Dump()并分配给var,这就像一个魅力!
Arne S

@ArneS:它是用LinQPad编写的;而且我忘记删除了Dump()。感谢您指出。
mshsayem

1

在扩展了之前的几个答案并利用扩展方法的基础上,我想到了以下内容,但没有潜在的多个可枚举枚举的注意事项,也没有涉及使用多个regex对象或不必要地调用regex的性能问题,话说回来,它确实使用了ToList(),它可以抵消较大集合中的好处。

选择器支持通用类型,以允许分配任何委托,选择器对source集合中的元素进行更改,然后使用ToString()将其转换为字符串。

    private static readonly Regex _NaturalOrderExpr = new Regex(@"\d+", RegexOptions.Compiled);

    public static IEnumerable<TSource> OrderByNatural<TSource, TKey>(
        this IEnumerable<TSource> source, Func<TSource, TKey> selector)
    {
        int max = 0;

        var selection = source.Select(
            o =>
            {
                var v = selector(o);
                var s = v != null ? v.ToString() : String.Empty;

                if (!String.IsNullOrWhiteSpace(s))
                {
                    var mc = _NaturalOrderExpr.Matches(s);

                    if (mc.Count > 0)
                    {
                        max = Math.Max(max, mc.Cast<Match>().Max(m => m.Value.Length));
                    }
                }

                return new
                {
                    Key = o,
                    Value = s
                };
            }).ToList();

        return
            selection.OrderBy(
                o =>
                String.IsNullOrWhiteSpace(o.Value) ? o.Value : _NaturalOrderExpr.Replace(o.Value, m => m.Value.PadLeft(max, '0')))
                     .Select(o => o.Key);
    }

    public static IEnumerable<TSource> OrderByDescendingNatural<TSource, TKey>(
        this IEnumerable<TSource> source, Func<TSource, TKey> selector)
    {
        int max = 0;

        var selection = source.Select(
            o =>
            {
                var v = selector(o);
                var s = v != null ? v.ToString() : String.Empty;

                if (!String.IsNullOrWhiteSpace(s))
                {
                    var mc = _NaturalOrderExpr.Matches(s);

                    if (mc.Count > 0)
                    {
                        max = Math.Max(max, mc.Cast<Match>().Max(m => m.Value.Length));
                    }
                }

                return new
                {
                    Key = o,
                    Value = s
                };
            }).ToList();

        return
            selection.OrderByDescending(
                o =>
                String.IsNullOrWhiteSpace(o.Value) ? o.Value : _NaturalOrderExpr.Replace(o.Value, m => m.Value.PadLeft(max, '0')))
                     .Select(o => o.Key);
    }

1

受迈克尔·帕克(Michael Parker)解决方案的启发,以下是IComparer您可以使用任何linq订购方法的实现:

private class NaturalStringComparer : IComparer<string>
{
    public int Compare(string left, string right)
    {
        int max = new[] { left, right }
            .SelectMany(x => Regex.Matches(x, @"\d+").Cast<Match>().Select(y => (int?)y.Value.Length))
            .Max() ?? 0;

        var leftPadded = Regex.Replace(left, @"\d+", m => m.Value.PadLeft(max, '0'));
        var rightPadded = Regex.Replace(right, @"\d+", m => m.Value.PadLeft(max, '0'));

        return string.Compare(leftPadded, rightPadded);
    }
}

0

我们需要一种自然的排序方式来处理具有以下模式的文本:

"Test 1-1-1 something"
"Test 1-2-3 something"
...

由于某种原因,当我第一次查看SO时,我没有找到这篇文章,而是执行了我们自己的文章。与此处介绍的某些解决方案相比,尽管概念相似,但它可能具有更简单易懂的优势。但是,尽管我确实尝试查看了性能瓶颈,但它的实现仍然比默认实现慢得多OrderBy()

这是我实现的扩展方法:

public static class EnumerableExtensions
{
    // set up the regex parser once and for all
    private static readonly Regex Regex = new Regex(@"\d+|\D+", RegexOptions.Compiled | RegexOptions.Singleline);

    // stateless comparer can be built once
    private static readonly AggregateComparer Comparer = new AggregateComparer();

    public static IEnumerable<T> OrderByNatural<T>(this IEnumerable<T> source, Func<T, string> selector)
    {
        // first extract string from object using selector
        // then extract digit and non-digit groups
        Func<T, IEnumerable<IComparable>> splitter =
            s => Regex.Matches(selector(s))
                      .Cast<Match>()
                      .Select(m => Char.IsDigit(m.Value[0]) ? (IComparable) int.Parse(m.Value) : m.Value);
        return source.OrderBy(splitter, Comparer);
    }

    /// <summary>
    /// This comparer will compare two lists of objects against each other
    /// </summary>
    /// <remarks>Objects in each list are compare to their corresponding elements in the other
    /// list until a difference is found.</remarks>
    private class AggregateComparer : IComparer<IEnumerable<IComparable>>
    {
        public int Compare(IEnumerable<IComparable> x, IEnumerable<IComparable> y)
        {
            return
                x.Zip(y, (a, b) => new {a, b})              // walk both lists
                 .Select(pair => pair.a.CompareTo(pair.b))  // compare each object
                 .FirstOrDefault(result => result != 0);    // until a difference is found
        }
    }
}

想法是将原始字符串分为数字和非数字块("\d+|\D+")块。由于这是一项潜在的昂贵任务,因此每个条目只能执行一次。然后,我们使用可比较对象的比较器(对不起,我找不到更合适的说法)。它将每个块与另一个字符串中的相应块进行比较。

我想就如何改进以及主要缺陷有何反馈。请注意,此时的可维护性对我们很重要,并且当前我们不在大型数据集中使用它。


1
当它尝试比较结构上不同的字符串时,此方法会崩溃-例如,将“ a-1”与“ a-2”进行比较可以正常工作,但将“ a”与“ 1”进行比较则不可行,因为“ a” .CompareTo(1)引发异常。
jimrandomh

@jimrandomh,您是正确的。这种方法特定于我们的模式。
埃里克·利普兰迪

0

一个更易于阅读/维护的版本。

public class NaturalStringComparer : IComparer<string>
{
    public static NaturalStringComparer Instance { get; } = new NaturalStringComparer();

    public int Compare(string x, string y) {
        const int LeftIsSmaller = -1;
        const int RightIsSmaller = 1;
        const int Equal = 0;

        var leftString = x;
        var rightString = y;

        var stringComparer = CultureInfo.CurrentCulture.CompareInfo;

        int rightIndex;
        int leftIndex;

        for (leftIndex = 0, rightIndex = 0;
             leftIndex < leftString.Length && rightIndex < rightString.Length;
             leftIndex++, rightIndex++) {
            var leftChar = leftString[leftIndex];
            var rightChar = rightString[leftIndex];

            var leftIsNumber = char.IsNumber(leftChar);
            var rightIsNumber = char.IsNumber(rightChar);

            if (!leftIsNumber && !rightIsNumber) {
                var result = stringComparer.Compare(leftString, leftIndex, 1, rightString, leftIndex, 1);
                if (result != 0) return result;
            } else if (leftIsNumber && !rightIsNumber) {
                return LeftIsSmaller;
            } else if (!leftIsNumber && rightIsNumber) {
                return RightIsSmaller;
            } else {
                var leftNumberLength = NumberLength(leftString, leftIndex, out var leftNumber);
                var rightNumberLength = NumberLength(rightString, rightIndex, out var rightNumber);

                if (leftNumberLength < rightNumberLength) {
                    return LeftIsSmaller;
                } else if (leftNumberLength > rightNumberLength) {
                    return RightIsSmaller;
                } else {
                    if(leftNumber < rightNumber) {
                        return LeftIsSmaller;
                    } else if(leftNumber > rightNumber) {
                        return RightIsSmaller;
                    }
                }
            }
        }

        if (leftString.Length < rightString.Length) {
            return LeftIsSmaller;
        } else if(leftString.Length > rightString.Length) {
            return RightIsSmaller;
        }

        return Equal;
    }

    public int NumberLength(string str, int offset, out int number) {
        if (string.IsNullOrWhiteSpace(str)) throw new ArgumentNullException(nameof(str));
        if (offset >= str.Length) throw new ArgumentOutOfRangeException(nameof(offset), offset, "Offset must be less than the length of the string.");

        var currentOffset = offset;

        var curChar = str[currentOffset];

        if (!char.IsNumber(curChar))
            throw new ArgumentException($"'{curChar}' is not a number.", nameof(offset));

        int length = 1;

        var numberString = string.Empty;

        for (currentOffset = offset + 1;
            currentOffset < str.Length;
            currentOffset++, length++) {

            curChar = str[currentOffset];
            numberString += curChar;

            if (!char.IsNumber(curChar)) {
                number = int.Parse(numberString);

                return length;
            }
        }

        number = int.Parse(numberString);

        return length;
    }
}

-2

让我解释一下我的问题以及我如何解决它。

问题:-根据从目录检索的FileInfo对象中的FileName对文件进行排序。

解决方案:-我从FileInfo中选择了文件名,并修剪了文件名的“ .png”部分。现在,只需执行List.Sort(),它将以自然排序顺序对文件名进行排序。根据我的测试,我发现使用.png会弄乱排序顺序。看看下面的代码

var imageNameList = new DirectoryInfo(@"C:\Temp\Images").GetFiles("*.png").Select(x =>x.Name.Substring(0, x.Name.Length - 4)).ToList();
imageNameList.Sort();

我能知道这个答案为-1的原因吗?
girishkatta9 '18
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