在较长的序列中查找子序列
本文关键字：查找 | 更新日期: 2023-09-27 18:07:38
我需要在其他大序列中找到一个序列，例如，{1,3,2,3}存在于{1,3,2,3,4,3}和{5,1,3,2,3}中。有没有办法用IEnumerable或其他东西快速地做到这一点?
类似于@dlev的，但这也处理{1,1,1,2}.ContainsSubsequence({1,1,2})
public static bool ContainsSubsequence<T>(this IEnumerable<T> parent, IEnumerable<T> target)
{
var pattern = target.ToArray();
var source = new LinkedList<T>();
foreach (var element in parent)
{
source.AddLast(element);
if(source.Count == pattern.Length)
{
if(source.SequenceEqual(pattern))
return true;
source.RemoveFirst();
}
}
return false;
}
此方法将在父序列中找到可通过Equals()进行比较的任何类型的子序列:
public static bool ContainsSubequence<T>(this IEnumerable<T> parent, IEnumerable<T> target)
{
bool foundOneMatch = false;
using (IEnumerator<T> parentEnum = parent.GetEnumerator())
{
using (IEnumerator<T> targetEnum = target.GetEnumerator())
{
// Get the first target instance; empty sequences are trivially contained
if (!targetEnum.MoveNext())
return true;
while (parentEnum.MoveNext())
{
if (targetEnum.Current.Equals(parentEnum.Current))
{
// Match, so move the target enum forward
foundOneMatch = true;
if (!targetEnum.MoveNext())
{
// We went through the entire target, so we have a match
return true;
}
}
else if (foundOneMatch)
{
return false;
}
}
return false;
}
}
}
你可以这样使用:
bool match = new[] {1, 2, 3}.ContainsSubsequence(new[] {1, 2}); // match == true
match = new[] {1, 2, 3}.ContainsSubsequence(new[] {1, 3}); // match == false
注意，它假设目标序列没有null元素。
更新:感谢大家的支持，但实际上在上面的代码中有一个错误 !如果找到了部分匹配，但没有转换为完全匹配，则该过程将以结束，而不是重置(当应用于{1, 2, 1, 2, 3}.ContainsSubsequence({1, 2, 3})时，这显然是不正确的)。
上面的代码对于更常见的子序列定义(即不需要连续性)来说工作得很好，但是为了处理重置(大多数IEnumerators不支持)，需要先枚举目标序列。这导致了以下代码:
public static bool ContainsSubequence<T>(this IEnumerable<T> parent, IEnumerable<T> target)
{
bool foundOneMatch = false;
var enumeratedTarget = target.ToList();
int enumPos = 0;
using (IEnumerator<T> parentEnum = parent.GetEnumerator())
{
while (parentEnum.MoveNext())
{
if (enumeratedTarget[enumPos].Equals(parentEnum.Current))
{
// Match, so move the target enum forward
foundOneMatch = true;
if (enumPos == enumeratedTarget.Count - 1)
{
// We went through the entire target, so we have a match
return true;
}
enumPos++;
}
else if (foundOneMatch)
{
foundOneMatch = false;
enumPos = 0;
if (enumeratedTarget[enumPos].Equals(parentEnum.Current))
{
foundOneMatch = true;
enumPos++;
}
}
}
return false;
}
}
此代码没有任何错误，但对于大(或无限)序列不能很好地工作。
这是一个很好的研究问题，根据我的研究，有两种算法是最适合这项工作的，这取决于你的数据。
即Knuth-Morris- p ratt算法和Boyer-Moore算法。
在这里，我提交了我的KMP算法的实现，最初在这里审查。
用来处理长度不超过Int64.MaxValue的源序列或父序列。
可以看到，内部实现返回子字符串或目标模式所在的索引序列。您可以通过选择外观来呈现这些结果。
你可以像这样简单地使用
var contains = new[] { 1, 3, 2, 3, 4, 3 }.Contains(new[] { 1, 3, 2, 3 });
下面是一个显示实际代码的工作小提琴。
下面是我的答案的完整注释代码。
namespace Code
{
using System;
using System.Collections.Generic;
using System.Linq;
/// <summary>
/// A generic implementation of the Knuth-Morris-Pratt algorithm that searches,
/// in a memory efficient way, over a given <see cref="IEnumerable{T}"/>.
/// </summary>
public static class KMP
{
/// <summary>
/// Determines whether a sequence contains the search string.
/// </summary>
/// <typeparam name="T">
/// The type of elements of <paramref name="source"/>
/// </typeparam>
/// <param name="source">
/// A sequence of elements
/// </param>
/// <param name="pattern">The search string.</param>
/// <param name="equalityComparer">
/// Determines whether the sequence contains a specified element.
/// If <c>null</c>
/// <see cref="EqualityComparer{T}.Default"/> will be used.
/// </param>
/// <returns>
/// <c>true</c> if the source contains the specified pattern;
/// otherwise, <c>false</c>.
/// </returns>
/// <exception cref="ArgumentNullException">pattern</exception>
public static bool Contains<T>(
this IEnumerable<T> source,
IEnumerable<T> pattern,
IEqualityComparer<T> equalityComparer = null)
{
if (pattern == null)
{
throw new ArgumentNullException(nameof(pattern));
}
equalityComparer = equalityComparer ?? EqualityComparer<T>.Default;
return SearchImplementation(source, pattern, equalityComparer).Any();
}
public static IEnumerable<long> IndicesOf<T>(
this IEnumerable<T> source,
IEnumerable<T> pattern,
IEqualityComparer<T> equalityComparer = null)
{
if (pattern == null)
{
throw new ArgumentNullException(nameof(pattern));
}
equalityComparer = equalityComparer ?? EqualityComparer<T>.Default;
return SearchImplementation(source, pattern, equalityComparer);
}
/// <summary>
/// Identifies indices of a pattern string in a given sequence.
/// </summary>
/// <typeparam name="T">
/// The type of elements of <paramref name="source"/>
/// </typeparam>
/// <param name="source">
/// The sequence to search.
/// </param>
/// <param name="patternString">
/// The string to find in the sequence.
/// </param>
/// <param name="equalityComparer">
/// Determines whether the sequence contains a specified element.
/// </param>
/// <returns>
/// A sequence of indices where the pattern can be found
/// in the source.
/// </returns>
/// <exception cref="ArgumentOutOfRangeException">
/// patternSequence - The pattern must contain 1 or more elements.
/// </exception>
private static IEnumerable<long> SearchImplementation<T>(
IEnumerable<T> source,
IEnumerable<T> patternString,
IEqualityComparer<T> equalityComparer)
{
// Pre-process the pattern
(var slide, var pattern) = GetSlide(patternString, equalityComparer);
var patternLength = pattern.Count;
if (patternLength == 0)
{
throw new ArgumentOutOfRangeException(
nameof(patternString),
"The pattern must contain 1 or more elements.");
}
var buffer = new Dictionary<long, T>(patternLength);
var more = true;
long sourceIndex = 0; // index for source
int patternIndex = 0; // index for pattern
using(var sourceEnumerator = source.GetEnumerator())
while (more)
{
more = FillBuffer(
buffer,
sourceEnumerator,
sourceIndex,
patternLength,
out T t);
if (equalityComparer.Equals(pattern[patternIndex], t))
{
patternIndex++;
sourceIndex++;
more = FillBuffer(
buffer,
sourceEnumerator,
sourceIndex,
patternLength,
out t);
}
if (patternIndex == patternLength)
{
yield return sourceIndex - patternIndex;
patternIndex = slide[patternIndex - 1];
}
else if (more && !equalityComparer.Equals(pattern[patternIndex], t))
{
if (patternIndex != 0)
{
patternIndex = slide[patternIndex - 1];
}
else
{
sourceIndex = sourceIndex + 1;
}
}
}
}
/// <summary>
/// Services the buffer and retrieves the value.
/// </summary>
/// <remarks>
/// The buffer is used so that it is not necessary to hold the
/// entire source in memory.
/// </remarks>
/// <typeparam name="T">
/// The type of elements of <paramref name="source"/>.
/// </typeparam>
/// <param name="buffer">The buffer.</param>
/// <param name="source">The source enumerator.</param>
/// <param name="sourceIndex">The element index to retrieve.</param>
/// <param name="patternLength">Length of the search string.</param>
/// <param name="value">The element value retrieved from the source.</param>
/// <returns>
/// <c>true</c> if there is potentially more data to process;
/// otherwise <c>false</c>.
/// </returns>
private static bool FillBuffer<T>(
IDictionary<long, T> buffer,
IEnumerator<T> source,
long sourceIndex,
int patternLength,
out T value)
{
bool more = true;
if (!buffer.TryGetValue(sourceIndex, out value))
{
more = source.MoveNext();
if (more)
{
value = source.Current;
buffer.Remove(sourceIndex - patternLength);
buffer.Add(sourceIndex, value);
}
}
return more;
}
/// <summary>
/// Gets the offset array which acts as a slide rule for the KMP algorithm.
/// </summary>
/// <typeparam name="T">
/// The type of elements of <paramref name="source"/>.
/// </typeparam>
/// <param name="pattern">The search string.</param>
/// <param name="equalityComparer">
/// Determines whether the sequence contains a specified element.
/// If <c>null</c>
/// <see cref="EqualityComparer{T}.Default"/> will be used.
/// </param>
/// <returns>A tuple of the offsets and the enumerated pattern.</returns>
private static (IReadOnlyList<int> Slide, IReadOnlyList<T> Pattern) GetSlide<T>(
IEnumerable<T> pattern,
IEqualityComparer<T> equalityComparer)
{
var patternList = pattern.ToList();
var slide = new int[patternList.Count];
int length = 0;
int patternIndex = 1;
while (patternIndex < patternList.Count)
{
if (equalityComparer.Equals(
patternList[patternIndex],
patternList[length]))
{
length++;
slide[patternIndex] = length;
patternIndex++;
}
else
{
if (length != 0)
{
length = slide[length - 1];
}
else
{
slide[patternIndex] = length;
patternIndex++;
}
}
}
return (slide, patternList);
}
}
}
您可以尝试这样的操作来开始。将此列表转换为字符串后，可以使用子字符串
查找序列:if (String.Join(",", numericList.ConvertAll<string>(x => x.ToString()).ToArray())
{
//get sequence
}
此函数使用LINQ检查列表parent是否包含列表target:
public static bool ContainsSequence<T>(this List<T> parent, List<T> target)
{
for (int fromElement = parent.IndexOf(target.First());
(fromElement != -1) && (fromElement <= parent.Count - target.Count);
fromElement = parent.FindIndex(fromElement + 1, p => p.Equals(target.First())))
{
var comparedSequence = parent.Skip(fromElement).Take(target.Count);
if (comparedSequence.SequenceEqual(target)) return true;
}
return false;
}
如果您正在处理简单的可序列化类型，您可以很容易地做到这一点，如果您将数组转换为字符串:
public static bool ContainsList<T>(this List<T> containingList, List<T> containedList)
{
string strContaining = "," + string.Join(",", containingList) + ",";
string strContained = "," + string.Join(",", containedList) + ",";
return strContaining.Contains(strContained);
}
if (bigList.ContainsList(smallList))
{
...
}
这对我有用
var a1 = new List<int> { 1, 2, 3, 4, 5 };
var a2 = new List<int> { 2, 3, 4 };
int index = -1;
bool res = a2.All(
x => index != -1 ? (++index == a1.IndexOf(x)) : ((index = a1.IndexOf(x)) != -1)
);
这里有一个日常工作的示例，到目前为止还没有问题。我只是迭代一个相当长的序列，寻找匹配的给定子序列。
static bool SubSequenceExist(byte[] longSeq, byte[] subseq)
{
int seq = longSeq.Length;
int seqToFindLen = subseq.Length;
int seqMatchCount = 0;
for (int i = 0; i < seq; i++)
{
if (longSeq[i] == subseq[seqMatchCount])
seqMatchCount++;
else
seqMatchCount = 0;
if (seqMatchCount == seqToFindLen) return true;
}
return false;
}