delphi中如何统计多行文本文件中相同字符串出现的次数

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{ 2002.8.5 Kingron }
{ Source:Source string }
{ Sub:Sub string }
{ Return:Count }
{ Ex:StrSubCount( 'abcdbcd ', 'bc ')=2 }
function StrSubCount(const Source, Sub: string): integer;
var
Buf : string;
i : integer;
Len : integer;
begin
Result := 0;
Buf:=Source;
i := Pos(Sub, Buf);
Len := Length(Sub);
while i <> 0 do
begin
Inc(Result);
Delete(Buf, 1, i + Len -1);
i:=Pos(Sub,Buf);
end;
end; { StrSubCount }

{ 下面这个函数返回SubStr在S中指定位置开始后的位置 }
{ 例如：PosExx( 'ab ', 'abcabcab ',3)返回4 }
function PosExx(const substr: AnsiString; const s: AnsiString; const start: Integer): Integer;
type
StrRec = record
allocSiz: Longint;
refCnt: Longint;
length: Longint;
end;
const
skew = sizeof(StrRec);
asm
{ -> EAX Pointer to substr }
{ EDX Pointer to string }
{ ECX Pointer to start //cs }
{ <-EAX Position of substr in s or 0 }

TEST EAX,EAX
JE @@noWork
TEST EDX,EDX
JE @@stringEmpty
TEST ECX,ECX
JE @@stringEmpty

PUSH EBX
PUSH ESI
PUSH EDI

MOV ESI,EAX { Point ESI to }
MOV EDI,EDX { Point EDI to }

MOV EBX,ECX
MOV ECX,[EDI-skew].StrRec.length
PUSH EDI { remembers position to calculate index }

CMP EBX,ECX
JG @@fail

MOV EDX,[ESI-skew].StrRec.length { EDX = bstr }

DEC EDX { EDX = Length(substr) - }
JS @@fail { < 0 ? return }
MOV AL,[ESI] { AL = first char of }
INC ESI { Point ESI to 2 'nd char of substr }
SUB ECX,EDX { #positions in s to look }
{ = Length(s) - Length(substr) + 1 }
JLE @@fail
DEC EBX
SUB ECX,EBX
JLE @@fail
ADD EDI,EBX

@@loop:
REPNE SCASB
JNE @@fail
MOV EBX,ECX { save outer loop }
PUSH ESI { save outer loop substr pointer }
PUSH EDI { save outer loop s }

MOV ECX,EDX
REPE CMPSB
POP EDI { restore outer loop s pointer }
POP ESI { restore outer loop substr pointer }
JE @@found
MOV ECX,EBX { restore outer loop nter }
JMP @@loop

@@fail:
POP EDX { get rid of saved s nter }
XOR EAX,EAX
JMP @@exit

@@stringEmpty:
XOR EAX,EAX
JMP @@noWork

@@found:
POP EDX { restore pointer to first char of s }
MOV EAX,EDI { EDI points of char after match }
SUB EAX,EDX { the difference is the correct index }
@@exit:
POP EDI
POP ESI
POP EBX
@@noWork:
end;

{ PosEx返回Sub在Source中第Index次出现的位置，若出现的次数不足，返回0，若找不到，返回0 }
{ 例如：PosEx( 'abcdbcd ', 'bcd ',2)返回5，PosEx( 'abcdbcd ', 'adb ')返回0,PoxEx( 'abc ', 'a ',2)返回0 }
function PosEx(const Source, Sub: string; Index: integer): integer;
var
Buf : string;
i, Len, C : integer;
begin
C := 0;
Result := 0;
Buf := Source;
i := Pos(Sub, Source);
Len := Length(Sub);
while i <> 0 do
begin
inc(C);
Inc(Result, i);
Delete(Buf, 1, i + Len - 1);
i := Pos(Sub, Buf);
if C > = Index then Break;
if i > 0 then Inc(Result, Len - 1);
end;
if C < Index then Result := 0;
end;
下面是ZsWang写的代码，效率可能要高一些：
function PosEx1(const Source, Sub: string; Index: Integer = 1): Integer;
var
I, J, K, L: Integer;
T: string;
begin
Result := 0;
T := Source;
K := 0;
L := Length(Sub);
for I := 1 to Index do begin
J := Pos(Sub, T);
if J <= 0 then Exit;
Delete(T, 1, J + L - 1);
Inc(K, J + L - 1);
end;
Dec(K, L - 1);
Result := K;
end; { PosEx1 }
下面是Dirk写的，用的实递归：
function PosN(Substring, Mainstring: string; n: Integer): Integer;
{
Function PosN get recursive - the N th position of "Substring " in
"Mainstring ". Does the Mainstring not contain Substrign the result
is 0. Works with chars and strings.
}
begin
if Pos(substring, mainstring) = 0 then
begin
posn := 0;
Exit;
end
else
begin
if n = 1 then
posn := Pos(substring, mainstring)
else
begin
posn := Pos(substring, mainstring) + posn(substring, Copy(mainstring,
(Pos(substring, mainstring) + 1), Length(mainstring)), n - 1);
end;
end;
end;

function SubStrConut(mStr: string; mSub: string): Integer;
{ 返回子字符串出现的次数 }
begin
Result :=
(Length(mStr) - Length(StringReplace(mStr, mSub, ' ', [rfReplaceAll]))) div
Length(mSub);
end