任何人都知道一种在 byte[] 数组中搜索/匹配字节模式然后返回位置的好方法。
例如
byte[] pattern = new byte[] {12,3,5,76,8,0,6,125};
byte[] toBeSearched = new byte[] {23,36,43,76,125,56,34,234,12,3,5,76,8,0,6,125,234,56,211,122,22,4,7,89,76,64,12,3,5,76,8,0,6,125}
Anders R
问问题
109315 次
29 回答
58
我可以建议一些不涉及创建字符串、复制数组或不安全代码的东西：
using System;
using System.Collections.Generic;
static class ByteArrayRocks
{
static readonly int[] Empty = new int[0];
public static int[] Locate (this byte[] self, byte[] candidate)
{
if (IsEmptyLocate(self, candidate))
return Empty;
var list = new List<int>();
for (int i = 0; i < self.Length; i++)
{
if (!IsMatch(self, i, candidate))
continue;
list.Add(i);
}
return list.Count == 0 ? Empty : list.ToArray();
}
static bool IsMatch (byte[] array, int position, byte[] candidate)
{
if (candidate.Length > (array.Length - position))
return false;
for (int i = 0; i < candidate.Length; i++)
if (array[position + i] != candidate[i])
return false;
return true;
}
static bool IsEmptyLocate (byte[] array, byte[] candidate)
{
return array == null
|| candidate == null
|| array.Length == 0
|| candidate.Length == 0
|| candidate.Length > array.Length;
}
static void Main()
{
var data = new byte[] { 23, 36, 43, 76, 125, 56, 34, 234, 12, 3, 5, 76, 8, 0, 6, 125, 234, 56, 211, 122, 22, 4, 7, 89, 76, 64, 12, 3, 5, 76, 8, 0, 6, 125 };
var pattern = new byte[] { 12, 3, 5, 76, 8, 0, 6, 125 };
foreach (var position in data.Locate(pattern))
Console.WriteLine(position);
}
}
编辑（由 IAbstract 提供） -将帖子内容移至此处，因为它不是答案
出于好奇，我创建了一个具有不同答案的小型基准测试。
以下是一百万次迭代的结果：
solution [Locate]: 00:00:00.7714027
solution [FindAll]: 00:00:03.5404399
solution [SearchBytePattern]: 00:00:01.1105190
solution [MatchBytePattern]: 00:00:03.0658212
于 2008-11-12T11:21:36.897 回答
34
使用LINQ 方法。
public static IEnumerable<int> PatternAt(byte[] source, byte[] pattern)
{
for (int i = 0; i < source.Length; i++)
{
if (source.Skip(i).Take(pattern.Length).SequenceEqual(pattern))
{
yield return i;
}
}
}
很简单！
于 2013-02-05T16:28:52.977 回答
20
这是我的建议，更简单，更快捷：
int Search(byte[] src, byte[] pattern)
{
int maxFirstCharSlot = src.Length - pattern.Length + 1;
for (int i = 0; i < maxFirstCharSlot; i++)
{
if (src[i] != pattern[0]) // compare only first byte
continue;
// found a match on first byte, now try to match rest of the pattern
for (int j = pattern.Length - 1; j >= 1; j--)
{
if (src[i + j] != pattern[j]) break;
if (j == 1) return i;
}
}
return -1;
}
这段代码背后的逻辑是这样的：首先它只搜索第一个字节（这是关键的改进），当找到第一个字节时，我尝试匹配模式的其余部分
于 2016-07-28T01:29:45.860 回答
15
最初我发布了一些我使用的旧代码，但对 Jb Evain 的基准测试感到好奇。我发现我的解决方案是愚蠢的慢。看来 bruno conde 的SearchBytePattern是最快的。我不知道为什么，尤其是因为他使用了 Array.Copy 和 Extension 方法。但是在 Jb 的测试中有证据，所以对 bruno 表示敬意。
我进一步简化了这些位，所以希望这将是最清晰和最简单的解决方案。（bruno conde 所做的所有辛勤工作）增强功能包括：
- 缓冲区.块复制
- Array.IndexOf<字节>
- while 循环而不是 for 循环
- 开始索引参数
转换为扩展方法
public static List<int> IndexOfSequence(this byte[] buffer, byte[] pattern, int startIndex) { List<int> positions = new List<int>(); int i = Array.IndexOf<byte>(buffer, pattern[0], startIndex); while (i >= 0 && i <= buffer.Length - pattern.Length) { byte[] segment = new byte[pattern.Length]; Buffer.BlockCopy(buffer, i, segment, 0, pattern.Length); if (segment.SequenceEqual<byte>(pattern)) positions.Add(i); i = Array.IndexOf<byte>(buffer, pattern[0], i + 1); } return positions; }
请注意，while块中的最后一条语句应该i = Array.IndexOf<byte>(buffer, pattern[0], i + 1);代替i = Array.IndexOf<byte>(buffer, pattern[0], i + pattern.Length);. 看看约翰的评论。一个简单的测试可以证明：
byte[] pattern = new byte[] {1, 2};
byte[] toBeSearched = new byte[] { 1, 1, 2, 1, 12 };
随着i = Array.IndexOf<byte>(buffer, pattern[0], i + pattern.Length);，没有返回。i = Array.IndexOf<byte>(buffer, pattern[0], i + 1);返回正确的结果。
于 2008-12-02T00:18:54.617 回答
12
于 2008-11-12T13:21:25.843 回答
10
我的解决方案：
class Program
{
public static void Main()
{
byte[] pattern = new byte[] {12,3,5,76,8,0,6,125};
byte[] toBeSearched = new byte[] { 23, 36, 43, 76, 125, 56, 34, 234, 12, 3, 5, 76, 8, 0, 6, 125, 234, 56, 211, 122, 22, 4, 7, 89, 76, 64, 12, 3, 5, 76, 8, 0, 6, 125};
List<int> positions = SearchBytePattern(pattern, toBeSearched);
foreach (var item in positions)
{
Console.WriteLine("Pattern matched at pos {0}", item);
}
}
static public List<int> SearchBytePattern(byte[] pattern, byte[] bytes)
{
List<int> positions = new List<int>();
int patternLength = pattern.Length;
int totalLength = bytes.Length;
byte firstMatchByte = pattern[0];
for (int i = 0; i < totalLength; i++)
{
if (firstMatchByte == bytes[i] && totalLength - i >= patternLength)
{
byte[] match = new byte[patternLength];
Array.Copy(bytes, i, match, 0, patternLength);
if (match.SequenceEqual<byte>(pattern))
{
positions.Add(i);
i += patternLength - 1;
}
}
}
return positions;
}
}
于 2008-11-12T10:55:03.847 回答
6
如果您使用的是 .NET Core 2.1 或更高版本（或 .NET Standard 2.1 或更高版本平台），您可以使用新类型MemoryExtensions.IndexOf的扩展方法：Span
int matchIndex = toBeSearched.AsSpan().IndexOf(pattern);
要查找所有事件，您可以使用以下内容：
public static IEnumerable<int> IndexesOf(this byte[] haystack, byte[] needle,
int startIndex = 0, bool includeOverlapping = false)
{
int matchIndex = haystack.AsSpan(startIndex).IndexOf(needle);
while (matchIndex >= 0)
{
yield return startIndex + matchIndex;
startIndex += matchIndex + (includeOverlapping ? 1 : needle.Length);
matchIndex = haystack.AsSpan(startIndex).IndexOf(needle);
}
}
不幸的是，.NET Core 2.1 - 3.0 中的实现使用迭代的“在第一个字节上优化单字节搜索然后检查余数”方法，而不是快速字符串搜索算法，但这可能会在未来的版本中改变。（参见dotnet/runtime#60866。）
于 2019-10-11T19:30:44.017 回答
4
我错过了 LINQ 方法/答案:-)
/// <summary>
/// Searches in the haystack array for the given needle using the default equality operator and returns the index at which the needle starts.
/// </summary>
/// <typeparam name="T">Type of the arrays.</typeparam>
/// <param name="haystack">Sequence to operate on.</param>
/// <param name="needle">Sequence to search for.</param>
/// <returns>Index of the needle within the haystack or -1 if the needle isn't contained.</returns>
public static IEnumerable<int> IndexOf<T>(this T[] haystack, T[] needle)
{
if ((needle != null) && (haystack.Length >= needle.Length))
{
for (int l = 0; l < haystack.Length - needle.Length + 1; l++)
{
if (!needle.Where((data, index) => !haystack[l + index].Equals(data)).Any())
{
yield return l;
}
}
}
}
于 2012-08-08T14:41:19.293 回答
3
我上面的 Foubar 答案版本，它避免搜索超出大海捞针，并允许指定起始偏移量。假设针不是空的或比干草堆长。
public static unsafe long IndexOf(this byte[] haystack, byte[] needle, long startOffset = 0)
{
fixed (byte* h = haystack) fixed (byte* n = needle)
{
for (byte* hNext = h + startOffset, hEnd = h + haystack.LongLength + 1 - needle.LongLength, nEnd = n + needle.LongLength; hNext < hEnd; hNext++)
for (byte* hInc = hNext, nInc = n; *nInc == *hInc; hInc++)
if (++nInc == nEnd)
return hNext - h;
return -1;
}
}
于 2015-06-29T04:44:31.237 回答
2
这些是您可以使用的最简单、最快的方法，没有比这些更快的方法了。这是不安全的，但这就是我们使用指针的目的是速度。所以在这里我为您提供我的扩展方法，我使用搜索单个和出现的索引列表。我想说这是这里最干净的代码。
public static unsafe long IndexOf(this byte[] Haystack, byte[] Needle)
{
fixed (byte* H = Haystack) fixed (byte* N = Needle)
{
long i = 0;
for (byte* hNext = H, hEnd = H + Haystack.LongLength; hNext < hEnd; i++, hNext++)
{
bool Found = true;
for (byte* hInc = hNext, nInc = N, nEnd = N + Needle.LongLength; Found && nInc < nEnd; Found = *nInc == *hInc, nInc++, hInc++) ;
if (Found) return i;
}
return -1;
}
}
public static unsafe List<long> IndexesOf(this byte[] Haystack, byte[] Needle)
{
List<long> Indexes = new List<long>();
fixed (byte* H = Haystack) fixed (byte* N = Needle)
{
long i = 0;
for (byte* hNext = H, hEnd = H + Haystack.LongLength; hNext < hEnd; i++, hNext++)
{
bool Found = true;
for (byte* hInc = hNext, nInc = N, nEnd = N + Needle.LongLength; Found && nInc < nEnd; Found = *nInc == *hInc, nInc++, hInc++) ;
if (Found) Indexes.Add(i);
}
return Indexes;
}
}
以 Locate 为基准，速度提高 1.2-1.4 倍
于 2011-03-08T00:55:05.780 回答
2
我参加聚会有点晚了如何使用 Boyer Moore 算法但搜索字节而不是字符串。c#代码如下。
眼码公司
class Program {
static void Main(string[] args) {
byte[] text = new byte[] {12,3,5,76,8,0,6,125,23,36,43,76,125,56,34,234,12,4,5,76,8,0,6,125,234,56,211,122,22,4,7,89,76,64,12,3,5,76,8,0,6,123};
byte[] pattern = new byte[] {12,3,5,76,8,0,6,125};
BoyerMoore tmpSearch = new BoyerMoore(pattern,text);
Console.WriteLine(tmpSearch.Match());
Console.ReadKey();
}
public class BoyerMoore {
private static int ALPHABET_SIZE = 256;
private byte[] text;
private byte[] pattern;
private int[] last;
private int[] match;
private int[] suffix;
public BoyerMoore(byte[] pattern, byte[] text) {
this.text = text;
this.pattern = pattern;
last = new int[ALPHABET_SIZE];
match = new int[pattern.Length];
suffix = new int[pattern.Length];
}
/**
* Searches the pattern in the text.
* returns the position of the first occurrence, if found and -1 otherwise.
*/
public int Match() {
// Preprocessing
ComputeLast();
ComputeMatch();
// Searching
int i = pattern.Length - 1;
int j = pattern.Length - 1;
while (i < text.Length) {
if (pattern[j] == text[i]) {
if (j == 0) {
return i;
}
j--;
i--;
}
else {
i += pattern.Length - j - 1 + Math.Max(j - last[text[i]], match[j]);
j = pattern.Length - 1;
}
}
return -1;
}
/**
* Computes the function last and stores its values in the array last.
* last(Char ch) = the index of the right-most occurrence of the character ch
* in the pattern;
* -1 if ch does not occur in the pattern.
*/
private void ComputeLast() {
for (int k = 0; k < last.Length; k++) {
last[k] = -1;
}
for (int j = pattern.Length-1; j >= 0; j--) {
if (last[pattern[j]] < 0) {
last[pattern[j]] = j;
}
}
}
/**
* Computes the function match and stores its values in the array match.
* match(j) = min{ s | 0 < s <= j && p[j-s]!=p[j]
* && p[j-s+1]..p[m-s-1] is suffix of p[j+1]..p[m-1] },
* if such s exists, else
* min{ s | j+1 <= s <= m
* && p[0]..p[m-s-1] is suffix of p[j+1]..p[m-1] },
* if such s exists,
* m, otherwise,
* where p is the pattern and m is its length.
*/
private void ComputeMatch() {
/* Phase 1 */
for (int j = 0; j < match.Length; j++) {
match[j] = match.Length;
} //O(m)
ComputeSuffix(); //O(m)
/* Phase 2 */
//Uses an auxiliary array, backwards version of the KMP failure function.
//suffix[i] = the smallest j > i s.t. p[j..m-1] is a prefix of p[i..m-1],
//if there is no such j, suffix[i] = m
//Compute the smallest shift s, such that 0 < s <= j and
//p[j-s]!=p[j] and p[j-s+1..m-s-1] is suffix of p[j+1..m-1] or j == m-1},
// if such s exists,
for (int i = 0; i < match.Length - 1; i++) {
int j = suffix[i + 1] - 1; // suffix[i+1] <= suffix[i] + 1
if (suffix[i] > j) { // therefore pattern[i] != pattern[j]
match[j] = j - i;
}
else {// j == suffix[i]
match[j] = Math.Min(j - i + match[i], match[j]);
}
}
/* Phase 3 */
//Uses the suffix array to compute each shift s such that
//p[0..m-s-1] is a suffix of p[j+1..m-1] with j < s < m
//and stores the minimum of this shift and the previously computed one.
if (suffix[0] < pattern.Length) {
for (int j = suffix[0] - 1; j >= 0; j--) {
if (suffix[0] < match[j]) { match[j] = suffix[0]; }
}
{
int j = suffix[0];
for (int k = suffix[j]; k < pattern.Length; k = suffix[k]) {
while (j < k) {
if (match[j] > k) {
match[j] = k;
}
j++;
}
}
}
}
}
/**
* Computes the values of suffix, which is an auxiliary array,
* backwards version of the KMP failure function.
*
* suffix[i] = the smallest j > i s.t. p[j..m-1] is a prefix of p[i..m-1],
* if there is no such j, suffix[i] = m, i.e.
* p[suffix[i]..m-1] is the longest prefix of p[i..m-1], if suffix[i] < m.
*/
private void ComputeSuffix() {
suffix[suffix.Length-1] = suffix.Length;
int j = suffix.Length - 1;
for (int i = suffix.Length - 2; i >= 0; i--) {
while (j < suffix.Length - 1 && !pattern[j].Equals(pattern[i])) {
j = suffix[j + 1] - 1;
}
if (pattern[j] == pattern[i]) {
j--;
}
suffix[i] = j + 1;
}
}
}
}
于 2011-08-06T03:15:37.597 回答
1
为什么要让简单变得困难？这可以使用 for 循环在任何语言中完成。这是 C# 中的一个：
使用系统；
使用 System.Collections.Generic；
命名空间 BinarySearch
{
课堂节目
{
静态无效主要（字符串 [] 参数）
{
字节 [] 模式 = 新字节 [] {12,3,5,76,8,0,6,125};
byte[] toBeSearched = new byte[] {23,36,43,76,125,56,34,234,12,3,5,76,8,0,6,125,234,56,211,
122,22,4,7,89,76, 64,12,3,5,76,8,0,6,125};
List<int> 出现次数 = findOccurences(toBeSearched, pattern);
foreach(int 出现次数) {
Console.WriteLine("从 0 开始的索引找到匹配：" + 出现);
}
}
静态列表<int> findOccurences(byte[] haystack, byte[] needle)
{
List<int> 出现次数 = new List<int>();
for (int i = 0; i < haystack.Length; i++)
{
如果（针[0] == 干草堆[i]）
{
布尔发现=真；
整数 j, k;
对于 (j = 0, k = i; j < needle.Length; j++, k++)
{
if (k >= haystack.Length || needle[j] != haystack[k])
{
找到=假；
休息;
}
}
如果（找到）
{
出现次数.Add(i - 1);
我 = k;
}
}
}
返回事件；
}
}
}
于 2008-11-12T14:31:30.103 回答
1
感谢您抽出宝贵的时间...
这是我在问我的问题之前使用/测试的代码......我问这个问题的原因是我确定我没有使用最佳代码来做到这一点......所以再次感谢花时间！
private static int CountPatternMatches(byte[] pattern, byte[] bytes)
{
int counter = 0;
for (int i = 0; i < bytes.Length; i++)
{
if (bytes[i] == pattern[0] && (i + pattern.Length) < bytes.Length)
{
for (int x = 1; x < pattern.Length; x++)
{
if (pattern[x] != bytes[x+i])
{
break;
}
if (x == pattern.Length -1)
{
counter++;
i = i + pattern.Length;
}
}
}
}
return counter;
}
有人在我的代码中看到任何错误吗？这被认为是一种骇人听闻的方法吗？我已经尝试了你们发布的几乎所有样本，我似乎在比赛结果中得到了一些变化。我一直在使用 ~10Mb 字节数组作为我的 toBeSearched 数组来运行我的测试。
于 2008-11-12T14:44:18.767 回答
1
我使用我的答案和 Alnitak 的答案中的提示创建了一个新功能。
public static List<Int32> LocateSubset(Byte[] superSet, Byte[] subSet)
{
if ((superSet == null) || (subSet == null))
{
throw new ArgumentNullException();
}
if ((superSet.Length < subSet.Length) || (superSet.Length == 0) || (subSet.Length == 0))
{
return new List<Int32>();
}
var result = new List<Int32>();
Int32 currentIndex = 0;
Int32 maxIndex = superSet.Length - subSet.Length;
while (currentIndex < maxIndex)
{
Int32 matchCount = CountMatches(superSet, currentIndex, subSet);
if (matchCount == subSet.Length)
{
result.Add(currentIndex);
}
currentIndex++;
if (matchCount > 0)
{
currentIndex += matchCount - 1;
}
}
return result;
}
private static Int32 CountMatches(Byte[] superSet, int startIndex, Byte[] subSet)
{
Int32 currentOffset = 0;
while (currentOffset < subSet.Length)
{
if (superSet[startIndex + currentOffset] != subSet[currentOffset])
{
break;
}
currentOffset++;
}
return currentOffset;
}
唯一让我不高兴的是
currentIndex++;
if (matchCount > 0)
{
currentIndex += matchCount - 1;
}
部分...我想使用 if else 来避免-1，但这会导致更好的分支预测（尽管我不确定它是否会那么重要）..
于 2008-11-12T14:05:54.040 回答
1
这是我的（不是最高效的）解决方案。它依赖于 bytes/latin-1 转换是无损的这一事实，而 bytes/ASCII 或 bytes/UTF8 转换并非如此。
它的优点是它适用于任何字节值（某些其他解决方案无法正确使用字节 0x80-0xff），并且可以扩展以执行更高级的正则表达式匹配。
using System;
using System.Collections.Generic;
using System.Text;
using System.Text.RegularExpressions;
class C {
public static void Main() {
byte[] data = {0, 100, 0, 255, 100, 0, 100, 0, 255};
byte[] pattern = {0, 255};
foreach (int i in FindAll(data, pattern)) {
Console.WriteLine(i);
}
}
public static IEnumerable<int> FindAll(
byte[] haystack,
byte[] needle
) {
// bytes <-> latin-1 conversion is lossless
Encoding latin1 = Encoding.GetEncoding("iso-8859-1");
string sHaystack = latin1.GetString(haystack);
string sNeedle = latin1.GetString(needle);
for (Match m = Regex.Match(sHaystack, Regex.Escape(sNeedle));
m.Success; m = m.NextMatch()) {
yield return m.Index;
}
}
}
于 2008-11-12T11:27:43.717 回答
1
我会使用一个通过转换为字符串来匹配的解决方案......
您应该编写一个实现Knuth-Morris-Pratt 搜索算法的简单函数。这将是您可以用来查找正确索引的最快的简单算法。（您可以使用Boyer-Moore，但它需要更多设置。
优化算法后，您可以尝试寻找其他类型的优化。但是你应该从基础开始。
例如，当前“最快”的是 Jb Evian 的 Locate 解决方案。
如果你看核心
for (int i = 0; i < self.Length; i++) {
if (!IsMatch (self, i, candidate))
continue;
list.Add (i);
}
在子算法匹配后，它将开始在 i + 1 处找到匹配，但您已经知道第一个可能的匹配将是 i + Candidate.Length。所以如果你添加，
i += candidate.Length -2; // -2 instead of -1 because the i++ will add the last index
当您期望在超集中出现很多子集时，它会快很多。（Bruno Conde 已经在他的解决方案中这样做了）
但这只是 KNP 算法的一半，您还应该在 IsMatch 方法中添加一个名为 numberOfValidMatches 的额外参数，这将是一个输出参数。
这将解决以下问题：
int validMatches = 0;
if (!IsMatch (self, i, candidate, out validMatches))
{
i += validMatches - 1; // -1 because the i++ will do the last one
continue;
}
和
static bool IsMatch (byte [] array, int position, byte [] candidate, out int numberOfValidMatches)
{
numberOfValidMatches = 0;
if (candidate.Length > (array.Length - position))
return false;
for (i = 0; i < candidate.Length; i++)
{
if (array [position + i] != candidate [i])
return false;
numberOfValidMatches++;
}
return true;
}
稍微重构一下，您可以使用 numberOfValidMatches 作为循环变量，并使用 while 重写 Locate 循环以避免 -2 和 -1。但我只是想说明如何添加 KMP 算法。
于 2008-11-12T13:14:49.430 回答
1
Jb Evain 的回答是：
for (int i = 0; i < self.Length; i++) {
if (!IsMatch (self, i, candidate))
continue;
list.Add (i);
}
然后 IsMatch 函数首先检查是否candidate超出了正在搜索的数组的长度。
for如果对循环进行编码，这将更有效：
for (int i = 0, n = self.Length - candidate.Length + 1; i < n; ++i) {
if (!IsMatch (self, i, candidate))
continue;
list.Add (i);
}
在这一点上，也可以从一开始就消除测试IsMatch，只要您通过前置条件签订合同，永远不要使用“非法”参数调用它。注意：修复了 2019 年的一个错误。
于 2008-11-12T13:34:51.153 回答
1
速度不是一切。你检查过它们的一致性吗？
我没有测试这里列出的所有代码。我测试了自己的代码（我承认这并不完全一致）和 IndexOfSequence。我发现对于许多测试 IndexOfSequence 比我的代码快很多，但是通过重复测试，我发现它不太一致。特别是在数组末尾查找模式似乎最麻烦，但有时它也会在数组中间错过它们。
我的测试代码不是为了提高效率而设计的，我只是想有一堆随机数据，里面有一些已知的字符串。该测试模式大致类似于 http 表单上传流中的边界标记。这就是我遇到这段代码时所寻找的，所以我想我会用我将要搜索的数据来测试它。看起来模式越长，IndexOfSequence 越经常错过一个值。
private static void TestMethod()
{
Random rnd = new Random(DateTime.Now.Millisecond);
string Pattern = "-------------------------------65498495198498";
byte[] pattern = Encoding.ASCII.GetBytes(Pattern);
byte[] testBytes;
int count = 3;
for (int i = 0; i < 100; i++)
{
StringBuilder TestString = new StringBuilder(2500);
TestString.Append(Pattern);
byte[] buf = new byte[1000];
rnd.NextBytes(buf);
TestString.Append(Encoding.ASCII.GetString(buf));
TestString.Append(Pattern);
rnd.NextBytes(buf);
TestString.Append(Encoding.ASCII.GetString(buf));
TestString.Append(Pattern);
testBytes = Encoding.ASCII.GetBytes(TestString.ToString());
List<int> idx = IndexOfSequence(ref testBytes, pattern, 0);
if (idx.Count != count)
{
Console.Write("change from {0} to {1} on iteration {2}: ", count, idx.Count, i);
foreach (int ix in idx)
{
Console.Write("{0}, ", ix);
}
Console.WriteLine();
count = idx.Count;
}
}
Console.WriteLine("Press ENTER to exit");
Console.ReadLine();
}
（显然，我将 IndexOfSequence 从扩展转换回此测试的常规方法）
这是我的输出的示例运行：
change from 3 to 2 on iteration 1: 0, 2090,
change from 2 to 3 on iteration 2: 0, 1045, 2090,
change from 3 to 2 on iteration 3: 0, 1045,
change from 2 to 3 on iteration 4: 0, 1045, 2090,
change from 3 to 2 on iteration 6: 0, 2090,
change from 2 to 3 on iteration 7: 0, 1045, 2090,
change from 3 to 2 on iteration 11: 0, 2090,
change from 2 to 3 on iteration 12: 0, 1045, 2090,
change from 3 to 2 on iteration 14: 0, 2090,
change from 2 to 3 on iteration 16: 0, 1045, 2090,
change from 3 to 2 on iteration 17: 0, 1045,
change from 2 to 3 on iteration 18: 0, 1045, 2090,
change from 3 to 1 on iteration 20: 0,
change from 1 to 3 on iteration 21: 0, 1045, 2090,
change from 3 to 2 on iteration 22: 0, 2090,
change from 2 to 3 on iteration 23: 0, 1045, 2090,
change from 3 to 2 on iteration 24: 0, 2090,
change from 2 to 3 on iteration 25: 0, 1045, 2090,
change from 3 to 2 on iteration 26: 0, 2090,
change from 2 to 3 on iteration 27: 0, 1045, 2090,
change from 3 to 2 on iteration 43: 0, 1045,
change from 2 to 3 on iteration 44: 0, 1045, 2090,
change from 3 to 2 on iteration 48: 0, 1045,
change from 2 to 3 on iteration 49: 0, 1045, 2090,
change from 3 to 2 on iteration 50: 0, 2090,
change from 2 to 3 on iteration 52: 0, 1045, 2090,
change from 3 to 2 on iteration 54: 0, 1045,
change from 2 to 3 on iteration 57: 0, 1045, 2090,
change from 3 to 2 on iteration 62: 0, 1045,
change from 2 to 3 on iteration 63: 0, 1045, 2090,
change from 3 to 2 on iteration 72: 0, 2090,
change from 2 to 3 on iteration 73: 0, 1045, 2090,
change from 3 to 2 on iteration 75: 0, 2090,
change from 2 to 3 on iteration 76: 0, 1045, 2090,
change from 3 to 2 on iteration 78: 0, 1045,
change from 2 to 3 on iteration 79: 0, 1045, 2090,
change from 3 to 2 on iteration 81: 0, 2090,
change from 2 to 3 on iteration 82: 0, 1045, 2090,
change from 3 to 2 on iteration 85: 0, 2090,
change from 2 to 3 on iteration 86: 0, 1045, 2090,
change from 3 to 2 on iteration 89: 0, 2090,
change from 2 to 3 on iteration 90: 0, 1045, 2090,
change from 3 to 2 on iteration 91: 0, 2090,
change from 2 to 1 on iteration 92: 0,
change from 1 to 3 on iteration 93: 0, 1045, 2090,
change from 3 to 1 on iteration 99: 0,
我并不是要选择 IndexOfSequence，它恰好是我今天开始使用的那个。我在一天结束时注意到它似乎缺少数据中的模式，所以我今晚编写了自己的模式匹配器。虽然它没有那么快。在发布之前，我将对其进行更多调整，看看是否可以使其 100% 一致。
我只是想提醒大家，在你信任生产代码之前，他们应该测试这样的东西，以确保它们给出好的、可重复的结果。
于 2009-08-13T09:51:37.530 回答
1
我尝试了各种解决方案，最终修改了 SearchBytePattern 之一。我在 30k 序列上进行了测试，速度很快 :)
static public int SearchBytePattern(byte[] pattern, byte[] bytes)
{
int matches = 0;
for (int i = 0; i < bytes.Length; i++)
{
if (pattern[0] == bytes[i] && bytes.Length - i >= pattern.Length)
{
bool ismatch = true;
for (int j = 1; j < pattern.Length && ismatch == true; j++)
{
if (bytes[i + j] != pattern[j])
ismatch = false;
}
if (ismatch)
{
matches++;
i += pattern.Length - 1;
}
}
}
return matches;
}
让我知道你的想法。
于 2009-11-27T09:46:03.330 回答
1
这是我想出的解决方案。我包括了我在实施过程中发现的笔记。它可以匹配前向、后向和不同的（in/dec）修正量，例如方向；从大海捞针中的任何偏移量开始。
任何输入都会很棒！
/// <summary>
/// Matches a byte array to another byte array
/// forwards or reverse
/// </summary>
/// <param name="a">byte array</param>
/// <param name="offset">start offset</param>
/// <param name="len">max length</param>
/// <param name="b">byte array</param>
/// <param name="direction">to move each iteration</param>
/// <returns>true if all bytes match, otherwise false</returns>
internal static bool Matches(ref byte[] a, int offset, int len, ref byte[] b, int direction = 1)
{
#region Only Matched from offset Within a and b, could not differ, e.g. if you wanted to mach in reverse for only part of a in some of b that would not work
//if (direction == 0) throw new ArgumentException("direction");
//for (; offset < len; offset += direction) if (a[offset] != b[offset]) return false;
//return true;
#endregion
//Will match if b contains len of a and return a a index of positive value
return IndexOfBytes(ref a, ref offset, len, ref b, len) != -1;
}
///Here is the Implementation code
/// <summary>
/// Swaps two integers without using a temporary variable
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
internal static void Swap(ref int a, ref int b)
{
a ^= b;
b ^= a;
a ^= b;
}
/// <summary>
/// Swaps two bytes without using a temporary variable
/// </summary>
/// <param name="a"></param>
/// <param name="b"></param>
internal static void Swap(ref byte a, ref byte b)
{
a ^= b;
b ^= a;
a ^= b;
}
/// <summary>
/// Can be used to find if a array starts, ends spot Matches or compltely contains a sub byte array
/// Set checkLength to the amount of bytes from the needle you want to match, start at 0 for forward searches start at hayStack.Lenght -1 for reverse matches
/// </summary>
/// <param name="a">Needle</param>
/// <param name="offset">Start in Haystack</param>
/// <param name="len">Length of required match</param>
/// <param name="b">Haystack</param>
/// <param name="direction">Which way to move the iterator</param>
/// <returns>Index if found, otherwise -1</returns>
internal static int IndexOfBytes(ref byte[] needle, ref int offset, int checkLength, ref byte[] haystack, int direction = 1)
{
//If the direction is == 0 we would spin forever making no progress
if (direction == 0) throw new ArgumentException("direction");
//Cache the length of the needle and the haystack, setup the endIndex for a reverse search
int needleLength = needle.Length, haystackLength = haystack.Length, endIndex = 0, workingOffset = offset;
//Allocate a value for the endIndex and workingOffset
//If we are going forward then the bound is the haystackLength
if (direction >= 1) endIndex = haystackLength;
#region [Optomization - Not Required]
//{
//I though this was required for partial matching but it seems it is not needed in this form
//workingOffset = needleLength - checkLength;
//}
#endregion
else Swap(ref workingOffset, ref endIndex);
#region [Optomization - Not Required]
//{
//Otherwise we are going in reverse and the endIndex is the needleLength - checkLength
//I though the length had to be adjusted but it seems it is not needed in this form
//endIndex = needleLength - checkLength;
//}
#endregion
#region [Optomized to above]
//Allocate a value for the endIndex
//endIndex = direction >= 1 ? haystackLength : needleLength - checkLength,
//Determine the workingOffset
//workingOffset = offset > needleLength ? offset : needleLength;
//If we are doing in reverse swap the two
//if (workingOffset > endIndex) Swap(ref workingOffset, ref endIndex);
//Else we are going in forward direction do the offset is adjusted by the length of the check
//else workingOffset -= checkLength;
//Start at the checkIndex (workingOffset) every search attempt
#endregion
//Save the checkIndex (used after the for loop is done with it to determine if the match was checkLength long)
int checkIndex = workingOffset;
#region [For Loop Version]
///Optomized with while (single op)
///for (int checkIndex = workingOffset; checkIndex < endIndex; offset += direction, checkIndex = workingOffset)
///{
///Start at the checkIndex
/// While the checkIndex < checkLength move forward
/// If NOT (the needle at the checkIndex matched the haystack at the offset + checkIndex) BREAK ELSE we have a match continue the search
/// for (; checkIndex < checkLength; ++checkIndex) if (needle[checkIndex] != haystack[offset + checkIndex]) break; else continue;
/// If the match was the length of the check
/// if (checkIndex == checkLength) return offset; //We are done matching
///}
#endregion
//While the checkIndex < endIndex
while (checkIndex < endIndex)
{
for (; checkIndex < checkLength; ++checkIndex) if (needle[checkIndex] != haystack[offset + checkIndex]) break; else continue;
//If the match was the length of the check
if (checkIndex == checkLength) return offset; //We are done matching
//Move the offset by the direction, reset the checkIndex to the workingOffset
offset += direction; checkIndex = workingOffset;
}
//We did not have a match with the given options
return -1;
}
于 2011-07-04T07:10:33.113 回答
1
您可以使用 ORegex：
var oregex = new ORegex<byte>("{0}{1}{2}", x=> x==12, x=> x==3, x=> x==5);
var toSearch = new byte[]{1,1,12,3,5,1,12,3,5,5,5,5};
var found = oregex.Matches(toSearch);
将找到两个匹配项：
i:2;l:3
i:6;l:3
复杂性：在最坏的情况下为 O(n*m)，在现实生活中它是 O(n)，因为内部状态机。在某些情况下，它比 .NET Regex 更快。它结构紧凑、速度快，专为阵列模式匹配而设计。
于 2016-06-27T07:48:09.257 回答
0
只是另一个易于遵循且对于 O(n) 类型的操作非常有效的答案，无需使用不安全的代码或复制部分源数组。
一定要测试。在这个主题上找到的一些建议很容易受到一定情况的影响。
static void Main(string[] args)
{
// 1 1 1 1 1 1 1 1 1 1 2 2 2
// 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
byte[] buffer = new byte[] { 1, 0, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 5, 5, 0, 5, 5, 1, 2 };
byte[] beginPattern = new byte[] { 1, 0, 2 };
byte[] middlePattern = new byte[] { 8, 9, 10 };
byte[] endPattern = new byte[] { 9, 10, 11 };
byte[] wholePattern = new byte[] { 1, 0, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 };
byte[] noMatchPattern = new byte[] { 7, 7, 7 };
int beginIndex = ByteArrayPatternIndex(buffer, beginPattern);
int middleIndex = ByteArrayPatternIndex(buffer, middlePattern);
int endIndex = ByteArrayPatternIndex(buffer, endPattern);
int wholeIndex = ByteArrayPatternIndex(buffer, wholePattern);
int noMatchIndex = ByteArrayPatternIndex(buffer, noMatchPattern);
}
/// <summary>
/// Returns the index of the first occurrence of a byte array within another byte array
/// </summary>
/// <param name="buffer">The byte array to be searched</param>
/// <param name="pattern">The byte array that contains the pattern to be found</param>
/// <returns>If buffer contains pattern then the index of the first occurrence of pattern within buffer; otherwise, -1</returns>
public static int ByteArrayPatternIndex(byte[] buffer, byte[] pattern)
{
if (buffer != null && pattern != null && pattern.Length <= buffer.Length)
{
int resumeIndex;
for (int i = 0; i <= buffer.Length - pattern.Length; i++)
{
if (buffer[i] == pattern[0]) // Current byte equals first byte of pattern
{
resumeIndex = 0;
for (int x = 1; x < pattern.Length; x++)
{
if (buffer[i + x] == pattern[x])
{
if (x == pattern.Length - 1) // Matched the entire pattern
return i;
else if (resumeIndex == 0 && buffer[i + x] == pattern[0]) // The current byte equals the first byte of the pattern so start here on the next outer loop iteration
resumeIndex = i + x;
}
else
{
if (resumeIndex > 0)
i = resumeIndex - 1; // The outer loop iterator will increment so subtract one
else if (x > 1)
i += (x - 1); // Advance the outer loop variable since we already checked these bytes
break;
}
}
}
}
}
return -1;
}
/// <summary>
/// Returns the indexes of each occurrence of a byte array within another byte array
/// </summary>
/// <param name="buffer">The byte array to be searched</param>
/// <param name="pattern">The byte array that contains the pattern to be found</param>
/// <returns>If buffer contains pattern then the indexes of the occurrences of pattern within buffer; otherwise, null</returns>
/// <remarks>A single byte in the buffer array can only be part of one match. For example, if searching for 1,2,1 in 1,2,1,2,1 only zero would be returned.</remarks>
public static int[] ByteArrayPatternIndex(byte[] buffer, byte[] pattern)
{
if (buffer != null && pattern != null && pattern.Length <= buffer.Length)
{
List<int> indexes = new List<int>();
int resumeIndex;
for (int i = 0; i <= buffer.Length - pattern.Length; i++)
{
if (buffer[i] == pattern[0]) // Current byte equals first byte of pattern
{
resumeIndex = 0;
for (int x = 1; x < pattern.Length; x++)
{
if (buffer[i + x] == pattern[x])
{
if (x == pattern.Length - 1) // Matched the entire pattern
indexes.Add(i);
else if (resumeIndex == 0 && buffer[i + x] == pattern[0]) // The current byte equals the first byte of the pattern so start here on the next outer loop iteration
resumeIndex = i + x;
}
else
{
if (resumeIndex > 0)
i = resumeIndex - 1; // The outer loop iterator will increment so subtract one
else if (x > 1)
i += (x - 1); // Advance the outer loop variable since we already checked these bytes
break;
}
}
}
}
if (indexes.Count > 0)
return indexes.ToArray();
}
return null;
}
于 2013-12-13T00:29:25.600 回答
0
这是我仅使用基本数据类型编写的简单代码：（它返回第一次出现的索引）
private static int findMatch(byte[] data, byte[] pattern) {
if(pattern.length > data.length){
return -1;
}
for(int i = 0; i<data.length ;){
int j;
for(j=0;j<pattern.length;j++){
if(pattern[j]!=data[i])
break;
i++;
}
if(j==pattern.length){
System.out.println("Pattern found at : "+(i - pattern.length ));
return i - pattern.length ;
}
if(j!=0)continue;
i++;
}
return -1;
}
于 2012-04-13T11:47:57.470 回答
0
我试图理解 Sanchez 的建议并进行更快的搜索。以下代码的性能几乎相同。但代码更容易理解。
public int Search3(byte[] src, byte[] pattern)
{
int index = -1;
for (int i = 0; i < src.Length; i++)
{
if (src[i] != pattern[0])
{
continue;
}
else
{
bool isContinoue = true;
for (int j = 1; j < pattern.Length; j++)
{
if (src[++i] != pattern[j])
{
isContinoue = true;
break;
}
if(j == pattern.Length - 1)
{
isContinoue = false;
}
}
if ( ! isContinoue)
{
index = i-( pattern.Length-1) ;
break;
}
}
}
return index;
}
于 2017-01-01T08:19:49.430 回答
0
这是我自己对这个话题的看法。我使用指针来确保它在更大的数组上更快。此函数将返回序列的第一次出现（这是我自己的情况所需要的）。
我相信您可以稍微修改它以返回所有出现的列表。
我所做的相当简单。我循环遍历源数组（干草堆），直到找到模式的第一个字节（针）。当找到第一个字节时，我会继续单独检查下一个字节是否与模式的下一个字节匹配。如果没有，我会继续正常搜索，从我之前所在的索引（大海捞针）开始，然后再尝试匹配针。
所以这里是代码：
public unsafe int IndexOfPattern(byte[] src, byte[] pattern)
{
fixed(byte *srcPtr = &src[0])
fixed (byte* patternPtr = &pattern[0])
{
for (int x = 0; x < src.Length; x++)
{
byte currentValue = *(srcPtr + x);
if (currentValue != *patternPtr) continue;
bool match = false;
for (int y = 0; y < pattern.Length; y++)
{
byte tempValue = *(srcPtr + x + y);
if (tempValue != *(patternPtr + y))
{
match = false;
break;
}
match = true;
}
if (match)
return x;
}
}
return -1;
}
安全代码如下：
public int IndexOfPatternSafe(byte[] src, byte[] pattern)
{
for (int x = 0; x < src.Length; x++)
{
byte currentValue = src[x];
if (currentValue != pattern[0]) continue;
bool match = false;
for (int y = 0; y < pattern.Length; y++)
{
byte tempValue = src[x + y];
if (tempValue != pattern[y])
{
match = false;
break;
}
match = true;
}
if (match)
return x;
}
return -1;
}
于 2018-08-17T09:37:08.447 回答
0
我使用一个简单的通用方法
void Main()
{
Console.WriteLine(new[]{255,1,3,4,8,99,92,9,0,5,128}.Position(new[]{9,0}));
Console.WriteLine("Philipp".ToArray().Position("il".ToArray()));
Console.WriteLine(new[] { "Mo", "Di", "Mi", "Do", "Fr", "Sa", "So","Mo", "Di", "Mi", "Do", "Fr", "Sa", "So"}.Position(new[] { "Fr", "Sa" }, 7));
}
static class Extensions
{
public static int Position<T>(this T[] source, T[] pattern, int start = 0)
{
var matchLenght = 0;
foreach (var indexSource in Enumerable.Range(start, source.Length - pattern.Length))
foreach (var indexPattern in Enumerable.Range(0, pattern.Length))
if (source[indexSource + indexPattern].Equals(pattern[indexPattern]))
if (++matchLenght == pattern.Length)
return indexSource;
return -1;
}
}
输出：
7
2
11
于 2021-12-10T19:47:25.617 回答
0
前几天我遇到了这个问题，试试这个：
public static long FindBinaryPattern(byte[] data, byte[] pattern)
{
using (MemoryStream stream = new MemoryStream(data))
{
return FindBinaryPattern(stream, pattern);
}
}
public static long FindBinaryPattern(string filename, byte[] pattern)
{
using (FileStream stream = new FileStream(filename, FileMode.Open))
{
return FindBinaryPattern(stream, pattern);
}
}
public static long FindBinaryPattern(Stream stream, byte[] pattern)
{
byte[] buffer = new byte[1024 * 1024];
int patternIndex = 0;
int read;
while ((read = stream.Read(buffer, 0, buffer.Length)) > 0)
{
for (int bufferIndex = 0; bufferIndex < read; ++bufferIndex)
{
if (buffer[bufferIndex] == pattern[patternIndex])
{
++patternIndex;
if (patternIndex == pattern.Length)
return stream.Position - (read - bufferIndex) - pattern.Length + 1;
}
else
{
patternIndex = 0;
}
}
}
return -1;
}
它没有做任何聪明的事，保持简单。
于 2019-05-14T07:39:05.377 回答
-2
您可以将字节数组放入String并通过 IndexOf 运行匹配。或者您至少可以在字符串匹配上重用现有算法。
[STAThread]
static void Main(string[] args)
{
byte[] pattern = new byte[] {12,3,5,76,8,0,6,125};
byte[] toBeSearched = new byte[] {23,36,43,76,125,56,34,234,12,3,5,76,8,0,6,125,234,56,211,122,22,4,7,89,76,64,12,3,5,76,8,0,6,125};
string needle, haystack;
unsafe
{
fixed(byte * p = pattern) {
needle = new string((SByte *) p, 0, pattern.Length);
} // fixed
fixed (byte * p2 = toBeSearched)
{
haystack = new string((SByte *) p2, 0, toBeSearched.Length);
} // fixed
int i = haystack.IndexOf(needle, 0);
System.Console.Out.WriteLine(i);
}
}
于 2008-11-12T10:00:24.080 回答
-3
toBeSearched.Except(pattern) 将返回差异 toBeSearched.Intersect(pattern) 将产生一组交点 通常，您应该查看 Linq 扩展中的扩展方法
于 2008-11-12T10:22:58.903 回答