Go Premium for a chance to win a PS4. Enter to Win

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 149
  • Last Modified:

Split a string: needed efficient algoritme

Having no time to do this myself right now,
I would like to have the following slow code modified so that it becomes
much, much quicker.
Function Reserve can go away, but the procedure SplitFromEnd must give the exact same result as it does know.
The typical length of the string to process in SplitFromEnd is about 20 Kb.
So Reserve is very very inefficient.
Please post your answer ASAP, and in the form of a procedure I can cut and paste
in my program right away and do not give mere hints.

With Regards,
 Donker


Function Reverse(Inputstring: String): String;
var x : integer;
    ToProcess : String;
begin
    ToProcess:='';
    for x:= 1 to length(Inputstring)
    do ToProcess:= copy(Inputstring,x,1) + ToProcess;
    Reverse:=ToProcess;

end;

procedure SplitFromEnd(SplitAt: String; var InputString, PostString: String) ;
var TmpPos: integer;
    Before, AfterIncl, ToProcess: String;
    tmpSplitAt: string;
begin
   ToProcess:=Reverse(Inputstring);
   tmpSplitAt:=Reverse(SplitAt);

   TmpPos := pos(tmpSplitAt, ToProcess);

   if tmppos = 0
   then begin
        MessageDlg('SplitFromEnd: NOT FOUND', mtInformation,[mbOk], 0);
        end;

   AfterIncl:= copy(ToProcess, 1, tmppos + length(SplitAt));
   AfterIncl:=reverse(AfterIncl);
   Before :=  copy(ToProcess, tmppos +length(SplitAt), length(ToProcess));
   Before := reverse(Before);

   InputString:=Before;
   Poststring:= AfterIncl;
end;
0
padonker
Asked:
padonker
1 Solution
 
interCommented:
Hi friend,
This routine does no reversion and it only do one copy operation. May be it full fill you needs.

procedure SplitFromEnd(SplitAt: string; var InStr, PostStr: string);
var
  P : PChar;
  C : PChar;
  CLen : integer;
  Found : boolean;
begin
  // Append #0 for null terminated processing
  C := @SplitAt[1];
  CLen := Length(SplitAt);
  P := @InStr[Length(InStr) - Length(SplitAt)];
  SplitAt := SplitAt + #0;
  // try to find the SplitAt from right
  Found := false;
  while (P <> @InStr[1]) and not Found do  //while not hit to the begining
  begin
    Dec(P);
    Found := StrLComp(P, C, CLen) = 0;
  end;
  if Found then
  begin
    //copy post
    PostStr := StrPas(P);
    //Truncate the InStr
    SetLength(InStr, Longint(P) - Longint(@InStr[1]));
  end else
    MessageDlg('SplitFromEnd: NOT FOUND', mtInformation, [mbOk], 0);
end;

Awaiting comments
Regards,
Igor
0
 
ronit051397Commented:
Maybe this Reverse function will work faster:

Function Reverse(Inputstring: String): String;
var P,S: pChar;
begin
  S:=pChar(Inputstring);
  P:=StrEnd(S)-1;
  Dec(S);
  while P<>S do
  begin
    Result:=Result+P[0];
    Dec(P);
  end;
end;
0
 
padonkerAuthor Commented:
I say: Yes, this is a excellent piece of code.
Thank you Inter.



With regards,
   Donker
0
 
Holger101497Commented:
I guess inter's function will be the fastest... although it does 20000 string compares if the string is 20000 long... I don't know the internal implementation of POS, so I don't know how fast it is...

some hints about your code:
copy(Inputstring,x,1) is InputString[x] ! I don't know how "smart" the compiler is in optimization, but you definitely don't need a function call here! InputString[x] is (almost) directly translated into an address


ronit: Your code will only be marginally faster - if at all.
The problem is this: for x:= 1 to length(Inputstring) do ToProcess:= copy(Inputstring,x,1) + ToProcess; OR Result:=Result+P[0];

They both append the characters one by one and re-allocate the string 20000 times! The "correct" way of reversing it would be something like this:

L:=Length(InputString); //store it to save 20000 function calls in the loop
SetLength(result,L); //allocate the string;
FOR P:=1 TO L DO result[L]:=InputString[L+1-P];

you can use PChars, but I'm not sure if that's faster...

P.S.: Ooops. While I was writing, the answer was accepted. I wonder if my comment still makes it :-))
0
 
interCommented:
Thanks,
If you need a faster routine I can program a FindLast routine which immediatelly finds the right most matching string and we can get rid of the StrLComp, but this may take a while. Regards, Igor
0

Featured Post

Vote for the Most Valuable Expert

It’s time to recognize experts that go above and beyond with helpful solutions and engagement on site. Choose from the top experts in the Hall of Fame or on the right rail of your favorite topic page. Look for the blue “Nominate” button on their profile to vote.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now