我试图计算在另一个字节序列中字节序列发生的所有时间。但是，如果已经计算了它们，则无法重复使用字节。例如给出字符串
k.k.k.k.k.k.我们假设字节序列为k.k，然后它只会发现3次而不是5次，因为它们会像[k.k].[k.k].[k.k].那样被分解，而不是像[k.[k].[k].[k].[k].k]那样它们在一圈，基本上只是向右移动。
理想情况下，我们的想法是了解压缩字典或运行时编码的外观。所以目标是获得
k.k.k.k.k.k.只有2个部分，因为（k.k.k.）是你可以拥有的最大和最好的符号。
到目前为止这是来源：
using System;
using System.Collections.Generic;
using System.Collections;
using System.Linq;
using System.Text;
using System.IO;
static class Compression
{
static int Main(string[] args)
{
List<byte> bytes = File.ReadAllBytes("ok.txt").ToList();
List<List<int>> list = new List<List<int>>();
// Starting Numbers of bytes - This can be changed manually.
int StartingNumBytes = bytes.Count;
for (int i = StartingNumBytes; i > 0; i--)
{
Console.WriteLine("i: " + i);
for (int ii = 0; ii < bytes.Count - i; ii++)
{
Console.WriteLine("ii: " + i);
// New pattern comes with refresh data.
List<byte> pattern = new List<byte>();
for (int iii = 0; iii < i; iii++)
{
pattern.Add(bytes[ii + iii]);
}
DisplayBinary(bytes, "red");
DisplayBinary(pattern, "green");
int matches = 0;
// foreach (var position in bytes.ToArray().Locate(pattern.ToArray()))
for (int position = 0; position < bytes.Count; position++) {
if (pattern.Count > (bytes.Count - position))
{
continue;
}
for (int iiii = 0; iiii < pattern.Count; iiii++)
{
if (bytes[position + iiii] != pattern[iiii])
{
//Have to use goto because C# doesn't support continue <level>
goto outer;
}
}
// If it made it this far, it has found a match.
matches++;
Console.WriteLine("Matches: " + matches + " Orig Count: " + bytes.Count + " POS: " + position);
if (matches > 1)
{
int numBytesToRemove = pattern.Count;
for (int ra = 0; ra < numBytesToRemove; ra++)
{
// Remove it at the position it was found at, once it
// deletes the first one, the list will shift left and you'll need to be here again.
bytes.RemoveAt(position);
}
DisplayBinary(bytes, "red");
Console.WriteLine(pattern.Count + " Bytes removed.");
// Since you deleted some bytes, set the position less because you will need to redo the pos.
position = position - 1;
}
outer:
continue;
}
List<int> sublist = new List<int>();
sublist.Add(matches);
sublist.Add(pattern.Count);
// Some sort of calculation to determine how good the symbol was
sublist.Add(bytes.Count-((matches * pattern.Count)-matches));
list.Add(sublist);
}
}
Display(list);
Console.Read();
return 0;
}
static void DisplayBinary(List<byte> bytes, string color="white")
{
switch(color){
case "green":
Console.ForegroundColor = ConsoleColor.Green;
break;
case "red":
Console.ForegroundColor = ConsoleColor.Red;
break;
default:
break;
}
for (int i=0; i<bytes.Count; i++)
{
if (i % 8 ==0)
Console.WriteLine();
Console.Write(GetIntBinaryString(bytes[i]) + " ");
}
Console.WriteLine();
Console.ResetColor();
}
static string GetIntBinaryString(int n)
{
char[] b = new char[8];
int pos = 7;
int i = 0;
while (i < 8)
{
if ((n & (1 << i)) != 0)
{
b[pos] = '1';
}
else
{
b[pos] = '0';
}
pos--;
i++;
}
//return new string(b).TrimStart('0');
return new string(b);
}
static void Display(List<List<int>> list)
{
//
// Display everything in the List.
//
Console.WriteLine("Elements:");
foreach (var sublist in list)
{
foreach (var value in sublist)
{
Console.Write("{0,4}", value);
}
Console.WriteLine();
}
//
// Display total count.
//
int count = 0;
foreach (var sublist in list)
{
count += sublist.Count;
}
Console.WriteLine("Count:");
Console.WriteLine(count);
}
static public int SearchBytePattern(byte[] pattern, byte[] bytes)
{
int matches = 0;
// precomputing this shaves some seconds from the loop execution
int maxloop = bytes.Length - pattern.Length;
for (int i = 0; i < maxloop; i++)
{
if (pattern[0] == bytes[i])
{
bool ismatch = true;
for (int j = 1; j < pattern.Length; j++)
{
if (bytes[i + j] != pattern[j])
{
ismatch = false;
break;
}
}
if (ismatch)
{
matches++;
i += pattern.Length - 1;
}
}
}
return matches;
}
}
请参考帖子获取文件的非二进制文件，这里是二进制数据：
011010110010111001101011001011100110101100101110011010110010111001101011001011100110101100101110我希望它比它的开始时小。
答案 0 :(得分：6)
private static int CountOccurences(byte[] target, byte[] pattern)
{
var targetString = BitConverter.ToString(target);
var patternString = BitConverter.ToString(pattern);
return new Regex(patternString).Matches(targetString).Count;
}
答案 1 :(得分：2)
使用此解决方案，您可以访问匹配的各个索引（在枚举时），或者您可以在结果上调用Count()以查看有多少匹配：
public static IEnumerable<int> Find<T>(T[] pattern, T[] sequence, bool overlap)
{
int i = 0;
while (i < sequence.Length - pattern.Length + 1)
{
if (pattern.SequenceEqual(sequence.Skip(i).Take(pattern.Length)))
{
yield return i;
i += overlap ? 1 : pattern.Length;
}
else
{
i++;
}
}
}使用overlap: false来解决您的问题，或overlap: true查看重叠的匹配项（如果您有兴趣的话）。
我还有其他两种方法，它们的API略有不同（以及更好的性能）here，包括一个直接处理字节流的方法。
答案 2 :(得分：1)
快速而肮脏，没有正则表达式。虽然我不确定它是否回答了问题的意图，但它应该相对较快。我想我会对正则表达式进行一些时序测试，以确定相对速度：
private int CountOccurrences(string TestString, string TestPattern)
{
int PatternCount = 0;
int SearchIndex = 0;
if (TestPattern.Length == 0)
throw new ApplicationException("CountOccurrences: Unable to process because TestPattern has zero length.");
if (TestString.Length == 0)
return 0;
do
{
SearchIndex = TestString.IndexOf(TestPattern, SearchIndex);
if (SearchIndex >= 0)
{
++PatternCount;
SearchIndex += TestPattern.Length;
}
}
while ((SearchIndex >= 0) && (SearchIndex < TestString.Length));
return PatternCount;
}
private void btnTest_Click(object sender, EventArgs e)
{
string TestString1 = "k.k.k.k.k.k.k.k.k.k.k.k";
string TestPattern1 = "k.k";
System.Console.WriteLine(CountOccurrences(TestString1, TestPattern1).ToString()); // outputs 6
System.Console.WriteLine(CountOccurrences(TestString1 + ".k", TestPattern1).ToString()); // still 6
System.Console.WriteLine(CountOccurrences(TestString1, TestPattern1 + ".").ToString()); // only 5
}