Nice and pretty fast filter

[Adrian Verkuil]

He MM Thanks,

You made me study Pointers now, still I will compare the two your's and my tally routine to see wich is actually faster
Wondering if the select case is fast enough?

MM a nice example for me to study thanks for your effort.

[Michael Mattias]

It's not a string searching problem if you think about it...

%CHAR_LOW  = 33?    ' high and low ASC() of searched characters
%CHAR_HIGH = 255?

' TRUE ==> The string 'S DOES CONTAIN at least one character in range %CHAR_LOW to %CHAR_HIGH
'          which occurs more than 'limit' times
' FALSE==> It don't.
FUNCTION DoesStringContainMoreThan (s AS STRING, BYVAL limit as LONG) AS LONG

  LOCAL pChar AS BYTE PTR , nOccur() AS LONG 
  LOCAL lStr AS LONG, I AS LONG

  lStr = LEN(S)   ' number of characters to be inspected 

  pChar = STRPTR (S)   ' point at first character
  REDIM   nOccur (%CHAR_HIGH)   ' array big enough to support %CHAR_HIGH 

  FOR I = 1 TO  lStr
    SELECT CASE AS LONG @pChar
      CASE %CHAR_LOW TO %CHAR_HIGH
         INCR  nOccur (@pChar) 
         IF nOccur(@pChar) > limit THEN   ' it happened, we can stop looking now.
            FUNCTION = %TRUE
            EXIT FUNCTION
         END IF
    END SELECT
    INCR pChar    ' point to next character  of S 
   NEXT

   FUNCTION = 0   ' if we get here it didn't happen

END FUNCTION

Yes, there are many many many different ways to speed it up. BUt just getting rid of the string operations should do lots for you.

MCM

[Adrian Verkuil]

Thanks Chris,

Will do that.

[Chris Holbrook]

Adrian, see recent thread here about instr, where faster methods of string matching are described.

[Adrian Verkuil]

Hi guys needed this,

294721 string 64 long in 30 seconds not bad for a filter like this!

scan for to many repeating chars, put result for every char in array
scan array to see if a char (any char in the range 33 to 255)
is more then 5 times in the string.

Works ok, so I thought lets share it.

WHILE NOT EOF(1)
  LINE INPUT #1, a$
                     
   FOR I = 33 TO 255              
     Z(i) = 0
     Z(i) = TALLY (a$, CHR$(I))  
   NEXT
                    
   ARRAY SCAN Z(33), >5, TO R
                     
    IF  R  THEN .......
                    WEND

But if one off you know a faster way ...