Question

我有2个字符串

string a = "foo bar";
string b = "bar foo";

，我想检测一下从a到b的变化。 要从a变为b，我必须更改哪些字符？

我认为每个字符都必须进行迭代，并检测它是否被添加，删除或保持相等。这是我意想不到的结果

'f' Remove
'o' Remove
'o' Remove
' ' Remove
'b' Equal
'a' Equal
'r' Equal
' ' Add
'f' Add
'o' Add
'o' Add

结果的类和枚举：

public enum Operation { Add,Equal,Remove };
public class Difference
{
    public Operation op { get; set; }
    public char c { get; set; }
}

这是我的解决方案，但是“删除”案例对我来说尚不清楚

public static List<Difference> CalculateDifferences(string left, string right)
{
    int count = 0;
    List<Difference> result = new List<Difference>();
    foreach (char ch in left)
    {
        int index = right.IndexOf(ch, count);
        if (index == count)
        {
            count++;
            result.Add(new Difference() { c = ch, op = Operation.Equal });
        }
        else if (index > count)
        {
            string add = right.Substring(count, index - count);
            result.AddRange(add.Select(x => new Difference() { c = x, op = Operation.Add }));
            count += add.Length;
        }
        else
        {
            //Remove?
        }
    }
    return result;
}

已删除字符的代码外观如何？

更新-添加了更多示例

示例1：

string a = "foobar";
string b = "fooar";

预期结果：

'f' Equal
'o' Equal
'o' Equal
'b' Remove
'a' Equal
'r' Equal

示例2：

string a = "asdfghjk";
string b = "wsedrftr";

预期结果：

'a' Remove
'w' Add
's' Equal
'e' Add
'd' Equal
'r' Add
'f' Equal
'g' Remove
'h' Remove
'j' Remove
'k' Remove
't' Add
'r' Add

更新：

这是德米特里（Dmitry）和英根（Ingen）的答案之间的一个比较：https://dotnetfiddle.net/MJQDAO

Answer 1

您正在寻找（最小）编辑距离 / （最小）编辑顺序。您可以在此处找到该过程的理论：

https://web.stanford.edu/class/cs124/lec/med.pdf

让我们实现（最简单的）Levenstein距离/序列算法（有关详细信息，请参见https://en.wikipedia.org/wiki/Levenshtein_distance）。让我们从 helper 类开始（我对您的实现做了一些改动）：

  public enum EditOperationKind : byte {
    None,    // Nothing to do
    Add,     // Add new character
    Edit,    // Edit character into character (including char into itself)
    Remove,  // Delete existing character
  };

  public struct EditOperation {
    public EditOperation(char valueFrom, char valueTo, EditOperationKind operation) {
      ValueFrom = valueFrom;
      ValueTo = valueTo;

      Operation = valueFrom == valueTo ? EditOperationKind.None : operation;
    }

    public char ValueFrom { get; }
    public char ValueTo {get ;}
    public EditOperationKind Operation { get; }

    public override string ToString() {
      switch (Operation) {
        case EditOperationKind.None:
          return $"'{ValueTo}' Equal";
        case EditOperationKind.Add:
          return $"'{ValueTo}' Add";
        case EditOperationKind.Remove:
          return $"'{ValueFrom}' Remove";
        case EditOperationKind.Edit:
          return $"'{ValueFrom}' to '{ValueTo}' Edit";
        default:
          return "???";
      }
    }
  }

据我所提供的示例，我们没有任何 edit 操作，但是 add + remove ;这就是为什么我在editCost = 2，insertCost = 1（在 tie 的情况下：int removeCost = 1与insert + remove的情况下放edit的原因insert + remove）。现在我们准备实现Levenstein算法：

public static EditOperation[] EditSequence(
  string source, string target, 
  int insertCost = 1, int removeCost = 1, int editCost = 2) {

  if (null == source)
    throw new ArgumentNullException("source");
  else if (null == target)
    throw new ArgumentNullException("target");

  // Forward: building score matrix

  // Best operation (among insert, update, delete) to perform 
  EditOperationKind[][] M = Enumerable
    .Range(0, source.Length + 1)
    .Select(line => new EditOperationKind[target.Length + 1])
    .ToArray();

  // Minimum cost so far
  int[][] D = Enumerable
    .Range(0, source.Length + 1)
    .Select(line => new int[target.Length + 1])
    .ToArray();

  // Edge: all removes
  for (int i = 1; i <= source.Length; ++i) {
    M[i][0] = EditOperationKind.Remove;
    D[i][0] = removeCost * i;
  }

  // Edge: all inserts 
  for (int i = 1; i <= target.Length; ++i) {
    M[0][i] = EditOperationKind.Add;
    D[0][i] = insertCost * i;
  }

  // Having fit N - 1, K - 1 characters let's fit N, K
  for (int i = 1; i <= source.Length; ++i)
    for (int j = 1; j <= target.Length; ++j) {
      // here we choose the operation with the least cost
      int insert = D[i][j - 1] + insertCost;
      int delete = D[i - 1][j] + removeCost;
      int edit = D[i - 1][j - 1] + (source[i - 1] == target[j - 1] ? 0 : editCost);

      int min = Math.Min(Math.Min(insert, delete), edit);

      if (min == insert) 
        M[i][j] = EditOperationKind.Add;
      else if (min == delete)
        M[i][j] = EditOperationKind.Remove;
      else if (min == edit)
        M[i][j] = EditOperationKind.Edit;

      D[i][j] = min;
    }

  // Backward: knowing scores (D) and actions (M) let's building edit sequence
  List<EditOperation> result = 
    new List<EditOperation>(source.Length + target.Length);

  for (int x = target.Length, y = source.Length; (x > 0) || (y > 0);) {
    EditOperationKind op = M[y][x];

    if (op == EditOperationKind.Add) {
      x -= 1;
      result.Add(new EditOperation('\0', target[x], op));
    }
    else if (op == EditOperationKind.Remove) {
      y -= 1;
      result.Add(new EditOperation(source[y], '\0', op));
    }
    else if (op == EditOperationKind.Edit) {
      x -= 1;
      y -= 1;
      result.Add(new EditOperation(source[y], target[x], op));
    }
    else // Start of the matching (EditOperationKind.None)
      break;
  }

  result.Reverse();

  return result.ToArray();
}

演示：

var sequence = EditSequence("asdfghjk", "wsedrftr"); 

Console.Write(string.Join(Environment.NewLine, sequence));

结果：

'a' Remove
'w' Add
's' Equal
'e' Add
'd' Equal
'r' Add
'f' Equal
'g' Remove
'h' Remove
'j' Remove
'k' Remove
't' Add
'r' Add

Answer 2

在这里，我将竭尽全力，提供一种并不是最有效的算法，但是很容易推理的算法。

让我们先掩盖一下：

1）订单很重要

string before = "bar foo"
string after = "foo bar"

即使两个字符串中都出现了“ bar”和“ foo”，“ bar”也需要删除并稍后再添加。这还告诉我们，after字符串为我们提供了我们感兴趣的字符的顺序，我们首先需要“ foo”。

2）订单超支

另一种看待它的方式是，某些字符可能永远轮不到。

string before = "abracadabra"
string after = "bar bar"

只有“ bar b a r”的粗体字符，在“ a b r a strong> cadab ra ”。即使我们在两个字符串中都有两个b，也只有第一个 counts 。当我们寻找“ ba rb ar”中的第二个b时，“ abracada br a”中的第二个b已经通过出现“ r”。

3）障碍

壁垒是两个字符串中都存在的字符，它考虑了顺序和计数。这已经表明 set 可能不是最合适的数据结构，因为我们会失去计数。

输入

string before = "pinata" string after = "accidental"

我们得到（伪代码）

var barriers = { 'a', 't', 'a' }

“ pin ata ”

“ a 隐藏 ta l”

让我们遵循执行流程：

'a'是第一个障碍，它也是after的第一个字符，因此可以删除before中第一个'a'之前的所有内容。 “ pin a ta”->“ a ta”

第二个障碍是't'，它不在我们的after字符串中的下一个位置，因此我们可以在两者之间插入所有内容。 “ a t a”->“接受 t a”

第三个障碍'a'已经在下一个位置，因此我们可以不做任何实际工作就移到下一个障碍。

没有更多的障碍，但是我们的字符串长度仍然小于after的字符串长度，因此将进行一些后期处理。 “ accidenta”->“ accidenta l ”

请注意，'i'和'n'不再发挥作用，再次按顺序计数。

实施

我们已经确定了顺序和计数的重要性，想到了Queue。

static public List<Difference> CalculateDifferences(string before, string after) { List<Difference> result = new List<Difference>(); Queue<char> barriers = new Queue<char>(); #region Preprocessing int index = 0; for (int i = 0; i < after.Length; i++) { // Look for the first match starting at index int match = before.IndexOf(after[i], index); if (match != -1) { barriers.Enqueue(after[i]); index = match + 1; } } #endregion #region Queue Processing index = 0; while (barriers.Any()) { char barrier = barriers.Dequeue(); // Get the offset to the barrier in both strings, // ignoring the part that's already been handled int offsetBefore = before.IndexOf(barrier, index) - index; int offsetAfter = after.IndexOf(barrier, index) - index; // Remove prefix from 'before' string if (offsetBefore > 0) { RemoveChars(before.Substring(index, offsetBefore), result); before = before.Substring(offsetBefore); } // Insert prefix from 'after' string if (offsetAfter > 0) { string substring = after.Substring(index, offsetAfter); AddChars(substring, result); before = before.Insert(index, substring); index += substring.Length; } // Jump over the barrier KeepChar(barrier, result); index++; } #endregion #region Post Queue processing if (index < before.Length) { RemoveChars(before.Substring(index), result); } if (index < after.Length) { AddChars(after.Substring(index), result); } #endregion return result; } static private void KeepChar(char barrier, List<Difference> result) { result.Add(new Difference() { c = barrier, op = Operation.Equal }); } static private void AddChars(string substring, List<Difference> result) { result.AddRange(substring.Select(x => new Difference() { c = x, op = Operation.Add })); } static private void RemoveChars(string substring, List<Difference> result) { result.AddRange(substring.Select(x => new Difference() { c = x, op = Operation.Remove })); }

Answer 3

我使用上面的3个示例进行了测试，它正确且完美地返回了预期结果。

        int flag = 0;
        int flag_2 = 0;

        string a = "asdfghjk";
        string b = "wsedrftr";

        char[] array_a = a.ToCharArray();
        char[] array_b = b.ToCharArray();

        for (int i = 0,j = 0, n= 0; i < array_b.Count(); i++)
        {   
            //Execute 1 time until reach first equal character   
            if(i == 0 && a.Contains(array_b[0]))
            {
                while (array_a[n] != array_b[0])
                {
                    Console.WriteLine(String.Concat(array_a[n], " : Remove"));
                    n++;
                }
                Console.WriteLine(String.Concat(array_a[n], " : Equal"));
                n++;
            }
            else if(i == 0 && !a.Contains(array_b[0]))
            {
                Console.WriteLine(String.Concat(array_a[n], " : Remove"));
                n++;
                Console.WriteLine(String.Concat(array_b[0], " : Add"));
            }


            else
            {
                if(n < array_a.Count())
                {
                    if (array_a[n] == array_b[i])
                    {
                        Console.WriteLine(String.Concat(array_a[n], " : Equal"));
                        n++;
                    }
                    else
                    {
                        flag = 0;
                        for (int z = n; z < array_a.Count(); z++)
                        {                              
                            if (array_a[z] == array_b[i])
                            {
                                flag = 1;
                                break;
                            }                                                              
                        }

                        if (flag == 0)
                        {
                            flag_2 = 0;
                            for (int aa = i; aa < array_b.Count(); aa++)
                            {
                                for(int bb = n; bb < array_a.Count(); bb++)
                                {
                                    if (array_b[aa] == array_a[bb])
                                    {
                                        flag_2 = 1;
                                        break;
                                    }
                                }
                            }

                            if(flag_2 == 1)
                            {
                                Console.WriteLine(String.Concat(array_b[i], " : Add"));
                            }
                            else
                            {
                                for (int z = n; z < array_a.Count(); z++)
                                {
                                    Console.WriteLine(String.Concat(array_a[z], " : Remove"));
                                    n++;
                                }
                                 Console.WriteLine(String.Concat(array_b[i], " : Add"));
                            }

                        }
                        else
                        {
                            Console.WriteLine(String.Concat(array_a[n], " : Remove"));
                            i--;
                            n++;
                        }

                    }
                }
                else
                {
                    Console.WriteLine(String.Concat(array_b[i], " : Add"));
                }

            }

        }//end for


        MessageBox.Show("Done");


    //OUTPUT CONSOLE:
    /*
    a : Remove
    w : Add
    s : Equal
    e : Add
    d : Equal
    r : Add
    f : Equal
    g : Remove
    h : Remove
    j : Remove
    k : Remove
    t : Add
    r : Add
    */

Answer 4

这里可能是另一个解决方案，完整的代码并带有注释。但是，您的第一个原始示例的结果却是相反的：

class Program
{
    enum CharState
    {
        Add,
        Equal,
        Remove
    }

    struct CharResult
    {
        public char c;
        public CharState state;
    }

    static void Main(string[] args)
    {
        string a = "asdfghjk";
        string b = "wsedrftr";
        while (true)
        {
            Console.WriteLine("Enter string a (enter to quit) :");
            a = Console.ReadLine();
            if (a == string.Empty)
                break;
            Console.WriteLine("Enter string b :");
            b = Console.ReadLine();

            List<CharResult> result = calculate(a, b);
            DisplayResults(result);
        }
        Console.WriteLine("Press a key to exit");
        Console.ReadLine();
    }

    static List<CharResult> calculate(string a, string b)
    {
        List<CharResult> res = new List<CharResult>();
        int i = 0, j = 0;

        char[] array_a = a.ToCharArray();
        char[] array_b = b.ToCharArray();

        while (i < array_a.Length && j < array_b.Length)
        {
            //For the current char in a, we check for the equal in b
            int index = b.IndexOf(array_a[i], j);
            if (index < 0) //not found, this char should be removed
            {
                res.Add(new CharResult() { c = array_a[i], state = CharState.Remove });
                i++;
            }
            else
            {
                //we add all the chars between B's current index and the index
                while (j < index)
                {
                    res.Add(new CharResult() { c = array_b[j], state = CharState.Add });
                    j++;
                }
                //then we say the current is the same
                res.Add(new CharResult() { c = array_a[i], state = CharState.Equal });
                i++;
                j++;
            }
        }

        while (i < array_a.Length)
        {
            //b is now empty, we remove the remains
            res.Add(new CharResult() { c = array_a[i], state = CharState.Remove });
            i++;
        }
        while (j < array_b.Length)
        {
            //a has been treated, we add the remains
            res.Add(new CharResult() { c = array_b[j], state = CharState.Add });
            j++;
        }

        return res;
    }

    static void DisplayResults(List<CharResult> results)
    {
        foreach (CharResult r in results)
        {
            Console.WriteLine($"'{r.c}' - {r.state}");
        }
    }
}

Answer 5

如果要在两个字符串之间进行精确比较，则必须阅读并理解Levenshtein Distance。通过使用此算法，您可以精确地计算两个字符串之间的相似率，还可以回溯该算法以获得第二个字符串的变化链。该算法也是自然语言处理的重要指标。

还有其他好处，需要时间来学习。

在此链接中，有一个Levenshtein Distance的C＃版本：

https://www.dotnetperls.com/levenshtein

检测两个字符串之间的差异

5 个答案:

实施