是否有一个Boyer-Moore字符串搜索和快速搜索和替换功能以及Delphi 2010 String(UnicodeString)的快速字符串计数?

时间:2010-07-22 16:03:36

标签: algorithm delphi delphi-2010 replace boyer-moore

我需要三个快速大字符串函数:快速搜索,快速搜索和替换,以及字符串中子字符串的快速计数。

我已经在C ++和Python中遇到了Boyer-Moore字符串搜索,但是我用过的唯一一个用于实现快速搜索和替换的Delphi Boyer-Moore算法是由Peter Morris(前身为DroopyEyes软件)提供的FastStrings的一部分,他的网站和电子邮件不再有效。

我已经将FastStrings移植到Delphi 2009/2010的AnsiStrings中,其中一个字节等于一个AnsiChar,但是它们也可以在Delphi 2010中使用String(UnicodeString)不重要的。

使用这个Boyer-Moore算法,应该可以轻松地进行不区分大小写的搜索,以及不区分大小写的搜索和替换,没有任何临时字符串(使用StrUpper等),并且不调用较慢的Pos()比需要在同一文本上重复搜索时的Boyer-Moore搜索。

(编辑:我有一个部分解决方案,写作这个问题的答案,几乎100%完成,它甚至有一个快速的字符串替换功能。我相信它必须有bug,特别是认为,因为它假装具有Unicode功能,必须是由于未实现的Unicode承诺而存在故障。)

(编辑2:有趣和意外的结果;堆栈上的unicode代码点表的大堆栈大小 - 下面的代码中的SkipTable严重阻碍了你可以在这里做的双赢优化的数量unicode string boyer-moore string search。感谢Florent Ouchet指出我应该立即注意到的内容。)

2 个答案:

答案 0 :(得分:11)

这个答案现在已经完成,适用于区分大小写的模式,但不适用于不区分大小写的模式,并且可能还有其他错误,因为它没有经过单元测试,并且可能会进一步优化,例如我重复了本地函数__SameChar而不是使用本来更快的比较函数回调,实际上,允许用户传递所有这些的比较函数对于想要提供一些额外逻辑的Unicode用户来说非常好(等效的Unicode字形集合对于某些语言)。

根据Dorin Dominica的代码,我构建了以下内容。

{ _FindStringBoyer:
  Boyer-Moore search algorith using regular String instead of AnsiSTring, and no ASM.
  Credited to Dorin Duminica.
}
function _FindStringBoyer(const sString, sPattern: string;
  const bCaseSensitive: Boolean = True; const fromPos: Integer = 1): Integer;

    function __SameChar(StringIndex, PatternIndex: Integer): Boolean;
    begin
      if bCaseSensitive then
        Result := (sString[StringIndex] = sPattern[PatternIndex])
      else
        Result := (CompareText(sString[StringIndex], sPattern[PatternIndex]) = 0);
    end; // function __SameChar(StringIndex, PatternIndex: Integer): Boolean;

var
  SkipTable: array [Char] of Integer;
  LengthPattern: Integer;
  LengthString: Integer;
  Index: Integer;
  kIndex: Integer;
  LastMarker: Integer;
  Large: Integer;
  chPattern: Char;
begin
  if fromPos < 1 then
    raise Exception.CreateFmt('Invalid search start position: %d.', [fromPos]);
  LengthPattern := Length(sPattern);
  LengthString := Length(sString);
  for chPattern := Low(Char) to High(Char) do
    SkipTable[chPattern] := LengthPattern;
  for Index := 1 to LengthPattern -1 do
    SkipTable[sPattern[Index]] := LengthPattern - Index;
  Large := LengthPattern + LengthString + 1;
  LastMarker := SkipTable[sPattern[LengthPattern]];
  SkipTable[sPattern[LengthPattern]] := Large;
  Index := fromPos + LengthPattern -1;
  Result := 0;
  while Index <= LengthString do begin
    repeat
      Index := Index + SkipTable[sString[Index]];
    until Index > LengthString;
    if Index <= Large then
      Break
    else
      Index := Index - Large;
    kIndex := 1;
    while (kIndex < LengthPattern) and __SameChar(Index - kIndex, LengthPattern - kIndex) do
      Inc(kIndex);
    if kIndex = LengthPattern then begin
      // Found, return.
      Result := Index - kIndex + 1;
      Index := Index + LengthPattern;
      exit;
    end else begin
      if __SameChar(Index, LengthPattern) then
        Index := Index + LastMarker
      else
        Index := Index + SkipTable[sString[Index]];
    end; // if kIndex = LengthPattern then begin
  end; // while Index <= LengthString do begin
end;

{ Written by Warren, using the above code as a starter, we calculate the SkipTable once, and then count the number of instances of
  a substring inside the main string, at a much faster rate than we
  could have done otherwise.  Another thing that would be great is
  to have a function that returns an array of find-locations,
  which would be way faster to do than repeatedly calling Pos.
}
function _StringCountBoyer(const aSourceString, aFindString : String; Const CaseSensitive : Boolean = TRUE) : Integer;
var
  foundPos:Integer;
  fromPos:Integer;
  Limit:Integer;
  guard:Integer;
  SkipTable: array [Char] of Integer;
  LengthPattern: Integer;
  LengthString: Integer;
  Index: Integer;
  kIndex: Integer;
  LastMarker: Integer;
  Large: Integer;
  chPattern: Char;
    function __SameChar(StringIndex, PatternIndex: Integer): Boolean;
    begin
      if CaseSensitive then
        Result := (aSourceString[StringIndex] = aFindString[PatternIndex])
      else
        Result := (CompareText(aSourceString[StringIndex], aFindString[PatternIndex]) = 0);
    end; // function __SameChar(StringIndex, PatternIndex: Integer): Boolean;

begin
  result := 0;
  foundPos := 1;
  fromPos := 1;
  Limit := Length(aSourceString);
  guard := Length(aFindString);
  Index := 0;
  LengthPattern := Length(aFindString);
  LengthString := Length(aSourceString);
  for chPattern := Low(Char) to High(Char) do
    SkipTable[chPattern] := LengthPattern;
  for Index := 1 to LengthPattern -1 do
    SkipTable[aFindString[Index]] := LengthPattern - Index;
  Large := LengthPattern + LengthString + 1;
  LastMarker := SkipTable[aFindString[LengthPattern]];
  SkipTable[aFindString[LengthPattern]] := Large;
  while (foundPos>=1) and (fromPos < Limit) and (Index<Limit) do begin

    Index := fromPos + LengthPattern -1;
    if Index>Limit then
        break;
    kIndex := 0;
    while Index <= LengthString do begin
      repeat
        Index := Index + SkipTable[aSourceString[Index]];
      until Index > LengthString;
      if Index <= Large then
        Break
      else
        Index := Index - Large;
      kIndex := 1;
      while (kIndex < LengthPattern) and __SameChar(Index - kIndex, LengthPattern - kIndex) do
        Inc(kIndex);
      if kIndex = LengthPattern then begin
        // Found, return.
        //Result := Index - kIndex + 1;
        Index := Index + LengthPattern;
        fromPos := Index;
        Inc(Result);
        break;
      end else begin
        if __SameChar(Index, LengthPattern) then
          Index := Index + LastMarker
        else
          Index := Index + SkipTable[aSourceString[Index]];
      end; // if kIndex = LengthPattern then begin
    end; // while Index <= LengthString do begin

  end;
end; 

这是一个很好的算法,因为:

  • 以这种方式计算字符串Y中子字符串X的实例的速度更快,非常好。
  • 仅仅替换Pos(),_FindStringBoyer()比FastCode项目人员为Delphi贡献的纯asm版本更快,目前用于Pos,如果你需要不区分大小写,你可以想象一下,当我们不必在100兆字节的字符串上调用UpperCase时,性能会提升。 (好吧,你的字符串不会那么大。但是,高效算法仍然是美丽的东西。)

好的,我用Boyer-Moore风格写了一个String Replace:

function _StringReplaceBoyer(const aSourceString, aFindString,aReplaceString : String; Flags: TReplaceFlags) : String;
var
  errors:Integer;
  fromPos:Integer;
  Limit:Integer;
  guard:Integer;
  SkipTable: array [Char] of Integer;
  LengthPattern: Integer;
  LengthString: Integer;
  Index: Integer;
  kIndex: Integer;
  LastMarker: Integer;
  Large: Integer;
  chPattern: Char;
  CaseSensitive:Boolean;
  foundAt:Integer;
  lastFoundAt:Integer;
  copyStartsAt:Integer;
  copyLen:Integer;
    function __SameChar(StringIndex, PatternIndex: Integer): Boolean;
    begin
      if CaseSensitive then
        Result := (aSourceString[StringIndex] = aFindString[PatternIndex])
      else
        Result := (CompareText(aSourceString[StringIndex], aFindString[PatternIndex]) = 0);
    end; // function __SameChar(StringIndex, PatternIndex: Integer): Boolean;

begin
  result := '';
  lastFoundAt := 0;
  fromPos := 1;
  errors := 0;
  CaseSensitive := rfIgnoreCase in Flags;
  Limit := Length(aSourceString);
  guard := Length(aFindString);
  Index := 0;
  LengthPattern := Length(aFindString);
  LengthString := Length(aSourceString);
  for chPattern := Low(Char) to High(Char) do
    SkipTable[chPattern] := LengthPattern;
  for Index := 1 to LengthPattern -1 do
    SkipTable[aFindString[Index]] := LengthPattern - Index;
  Large := LengthPattern + LengthString + 1;
  LastMarker := SkipTable[aFindString[LengthPattern]];
  SkipTable[aFindString[LengthPattern]] := Large;
  while (fromPos>=1) and (fromPos <= Limit) and (Index<=Limit) do begin

    Index := fromPos + LengthPattern -1;
    if Index>Limit then
        break;
    kIndex := 0;
    foundAt := 0;
    while Index <= LengthString do begin
      repeat
        Index := Index + SkipTable[aSourceString[Index]];
      until Index > LengthString;
      if Index <= Large then
        Break
      else
        Index := Index - Large;
      kIndex := 1;
      while (kIndex < LengthPattern) and __SameChar(Index - kIndex, LengthPattern - kIndex) do
        Inc(kIndex);
      if kIndex = LengthPattern then begin


        foundAt := Index - kIndex + 1;
        Index := Index + LengthPattern;
        //fromPos := Index;
        fromPos := (foundAt+LengthPattern);
        if lastFoundAt=0 then begin
                copyStartsAt := 1;
                copyLen := foundAt-copyStartsAt;
        end else begin
                copyStartsAt := lastFoundAt+LengthPattern;
                copyLen := foundAt-copyStartsAt;
        end;

        if (copyLen<=0)or(copyStartsAt<=0) then begin
                Inc(errors);
        end;

        Result := Result + Copy(aSourceString, copyStartsAt, copyLen ) + aReplaceString;
        lastFoundAt := foundAt;
        if not (rfReplaceAll in Flags) then
                 fromPos := 0; // break out of outer while loop too!
        break;
      end else begin
        if __SameChar(Index, LengthPattern) then
          Index := Index + LastMarker
        else
          Index := Index + SkipTable[aSourceString[Index]];
      end; // if kIndex = LengthPattern then begin
    end; // while Index <= LengthString do begin
  end;
  if (lastFoundAt=0) then
  begin
     // nothing was found, just return whole original string
      Result := aSourceString;
  end
  else
  if (lastFoundAt+LengthPattern < Limit) then begin
     // the part that didn't require any replacing, because nothing more was found,
     // or rfReplaceAll flag was not specified, is copied at the
     // end as the final step.
    copyStartsAt := lastFoundAt+LengthPattern;
    copyLen := Limit; { this number can be larger than needed to be, and it is harmless }
    Result := Result + Copy(aSourceString, copyStartsAt, copyLen );
  end;

end;

好的,问题:这个堆栈的足迹:

var
  skiptable : array [Char] of Integer;  // 65536*4 bytes stack usage on Unicode delphi

再见CPU地狱,你好堆栈地狱。如果我使用动态数组,那么我必须在运行时调整它的大小。所以这个东西基本上很快,因为你的计算机上的虚拟内存系统不会在堆栈上以256K闪烁,但这并不总是最佳代码。尽管如此,我的电脑并没有像这样大堆叠的东西眨眼。它不会成为Delphi标准库的默认值,也不会在未来赢得任何fastcode挑战。我认为重复搜索是一种情况,上面的代码应该写成一个类,而skiptable应该是该类中的数据字段。然后你可以构建一次boyer-moore表,并且随着时间的推移,如果字符串是不变的,则重复使用该对象进行快速查找。

答案 1 :(得分:1)

因为我只是在寻找相同的东西: Jedi JCL在jclUnicode.pas中使用Boyer-Moore获得了一个识别unicode的搜索引擎。 我不知道它有多好或有多快。