1. Re: Replacing characters (Matt: bug)
Matt,
you wrote:
> It seems to replace the last char when I test it.
You're right, both your original and #2 *do* replace the last char,
normally; BUT, if TWO of the same character to be replaced are together in
the text, like "}}", *then* the last of the two aren't replaced by either of
your routines.
The test file I used (one of the supplied routines) happened to have two
"))" in it, and when I saw that the second ")" wasn't replaced, I seemed to
only see that & didn't realize that all other instances of ")" *had* been
replaced, including those at the end of a line. Not sure if your routines
would replace *doubled* at end of line, but I suspect not.
"Doubled chars" not replaced is still a flaw, just not the one I pointed you
towards. Sorry 'bout that.
> Actually, neither one will handle replacements of different lengths.
The test I ran (& zipped previously) *did* show Andy's replacing one with
two. That same instance of doubled parentheses mentioned above ("))") was
replaced by "ZZZZ" in my test.
> <snip> Also, neither routine will handle:
> R/replace_in_string("abc cba", "ab", "12" )
Hmm, that's true, but they're supposed to replace characters, not strings;
seems we may need at least 2 different routines, one like Henri sent, which
does what it intends, taking every instance of each *character* supplied &
replaces it with a different supplied char; and another which should take
every *string* supplied & replace it with a different *string*, like your
example above should take "ab" & replace every instance with "12".
ReplaceChars & ReplaceStrings?
Dan Moyer
----- Original Message -----
From: "Matthew Lewis" <matthewwalkerlewis at YAHOO.COM>
To: "EUforum" <EUforum at topica.com>
Sent: Tuesday, September 17, 2002 4:23 AM
Subject: RE: Replacing characters (Matt: bug)
>
>
> > From: Dan Moyer [mailto:DANIELMOYER at prodigy.net]
>
> > Matt's now doesn't fail on replace last character, but
> > doesn't actually
> > replace the last character, & is *very* much slower than
> > either of the other
> > two
> > (one test: Henri: 3.9, Andy: 2.75 , Matt: 24.06) ;
>
> It seems to replace the last char when I test it. I believe that the
> slowness is due to the use of subscripting within the call to match().
> There was some discussion about this recently. I typically run fairly
short
> strings through this routine (maybe 2 or 3 sentence lengths) at a time, so
> I've never needed any more speed.
>
> > Henri/Mike "vulcan" routine is faster than Matt's but won't
> > replace one char
> > with 2;
> >
> > Andy's is fastest. How much faster seems to vary. On a
> > "long" sequence to
> > peruse, it seemed twice as fast as Henri/Mike "vulcan", but on many
> > repetitions of replacing in smaller sequence, it seems only
> > 1.4 times faster
> > (see test results above).
>
> Actually, neither one will handle replacements of different lengths. This
> was a definite requirement for what I needed, where, for instance you
might
> want to replace "." with "..". Also, neither routine will handle:
>
> R/replace_in_string("abc cba", "ab", "12" )
>
> Both return "12c c21", which is fine if you've got a cipher (did anyone
use
> this method in the contest? :), but not if you're trying to replace words
> with other words, in which case you end up with garbage. Henri talks
about
> this in the docs for the routine. Of course, if the searched for object
> isn't there, or there aren't many of them, my routine will speed up. So
if
> you're doing it on a large file, you might try calling replace_all each
line
> or every few lines. You'd probably have to test to see where the overhead
> of the call cancelled out the quicker return. And even then, it probably
> would depend upon how often your match came up.
>
> Here is a tweaked version that doesn't use a subscripted match. It seems
to
> be about 15-20% faster:
>
> global function replace_all2( sequence text, object a, object b )
> integer ix, jx, m, buf, lent, lena, lenb, dlen
>
> if atom(a) then
> a = {a}
> end if
>
> if atom(b) then
> b = {b}
> end if
>
> ix = match( a, text )
> if not ix then
> return text
> end if
>
> lena = length(a)
> lenb = length(b)
> dlen = lenb - lena
> lent = length(text)
> jx = lena + 1
>
> while jx > 1 do
> text = text[1..ix-1] & b & text[ix+lena..lent]
> ix += lenb
> jx = 1
> lent += dlen
>
> while ix <= lent and jx <= lena do
> if text[ix] = a[jx] then
> jx += 1
> else
> jx = 1
> end if
> ix += 1
> end while
>
> if ix > lent then
> ix -= lena
> end if
> end while
>
> return text
> end function
>
> Matt Lewis
>
>
>
>
