1. Fastest Way to...
I've got a 3-element sequence...
{ 3, -1, 0 }
and need to turn it into this
{ 1, -1, 0 }
That is, any positive values ( > 0 ) should be turned to one.
What's the fastest way to do this? Right now I'm using
if pos[1] > 0 then
pos[1] = 1
end if
if pos[2] > 0 then
pos[2] = 1
end if
if pos[3] > 0 then
pos[3] = 1
end if
I have a feeling this is fastest. Will it be as fast if put in a loop
for arbitrary length?
2. Re: Fastest Way to...
C. K. Lester wrote:
> I've got a 3-element sequence...
>
> { 3, -1, 0 }
>
> and need to turn it into this
>
> { 1, -1, 0 }
>
> That is, any positive values ( > 0 ) should be turned to one.
Okay, here's my attempt:
junk = pos <= 0
pos = pos * junk
pos += not junk
Yes, I know I could combine that into some big convoluted function, but
you get the gist.
3. Re: Fastest Way to...
----- Original Message -----
From: "C. K. Lester" <euphoric at cklester.com>
To: <EUforum at topica.com>
Subject: Fastest Way to...
>
>
> I've got a 3-element sequence...
>
> { 3, -1, 0 }
>
> and need to turn it into this
>
> { 1, -1, 0 }
>
> That is, any positive values ( > 0 ) should be turned to one.
>
> What's the fastest way to do this? Right now I'm using
>
> if pos[1] > 0 then
> pos[1] = 1
> end if
> if pos[2] > 0 then
> pos[2] = 1
> end if
> if pos[3] > 0 then
> pos[3] = 1
> end if
>
> I have a feeling this is fastest. Will it be as fast if put in a loop
> for arbitrary length?
>
Maybe this...
pos = pos * (pos <= 0) + (pos > 0)
--
Derek
4. Re: Fastest Way to...
pos = { 3, -1, 0 }
mask = (pos > 0)
pos -= (pos * mask)
pos += mask
or
mask = (pos > 0)
pos *= (mask = 0)
pos += mask
Lucius L. Hilley III
----- Original Message -----
From: "C. K. Lester" <euphoric at cklester.com>
To: <EUforum at topica.com>
Sent: Monday, October 27, 2003 09:24 PM
Subject: Fastest Way to...
>
>
> I've got a 3-element sequence...
>
> { 3, -1, 0 }
>
> and need to turn it into this
>
> { 1, -1, 0 }
>
> That is, any positive values ( > 0 ) should be turned to one.
>
> What's the fastest way to do this? Right now I'm using
>
> if pos[1] > 0 then
> pos[1] = 1
> end if
> if pos[2] > 0 then
> pos[2] = 1
> end if
> if pos[3] > 0 then
> pos[3] = 1
> end if
>
> I have a feeling this is fastest. Will it be as fast if put in a loop
> for arbitrary length?
>
>
>
> TOPICA - Start your own email discussion group. FREE!
>
>
5. Re: Fastest Way to...
Derek Parnell wrote:
>>I've got a 3-element sequence...
>>
>> { 3, -1, 0 }
>>
>>and need to turn it into this
>>
>> { 1, -1, 0 }
>>
>>That is, any positive values ( > 0 ) should be turned to one.
>>
>>
>Maybe this...
>
> pos = pos * (pos <= 0) + (pos > 0)
>
>
Oh, sweet! I think my version coulda been that if I had stayed with it. :)
6. Re: Fastest Way to...
On Mon, 27 Oct 2003 21:09:04 -0600, "C. K. Lester"
<euphoric at cklester.com> wrote:
>
>
>C. K. Lester wrote:
>
>> I've got a 3-element sequence...
>>
>> { 3, -1, 0 }
>>
>> and need to turn it into this
>>
>> { 1, -1, 0 }
>>
>> That is, any positive values ( > 0 ) should be turned to one.
>
>
>Okay, here's my attempt:
>
>junk =3D pos <=3D 0
>pos =3D pos * junk
>pos +=3D not junk
>
>Yes, I know I could combine that into some big convoluted function, but=
=20
>you get the gist.
That's about 3=BD times slower.
In your first post, you "unrolled the loop" which is a classic way to
make something as fast as possible.
Pete
7. Re: Fastest Way to...
On Mon, 27 Oct 2003 22:12:41 -0600, "C. K. Lester"
<euphoric at cklester.com> wrote:
>Derek Parnell wrote:
>> pos = pos * (pos <= 0) + (pos > 0)
>
>Oh, sweet! I think my version coulda been that if I had stayed with it. :)
Maybe you had better play with this:
constant ipos={3,-1,0}
sequence pos
sequence mask
integer j
atom t
t=time()
for i=1 to 1000000 do
pos=ipos
--============================
if pos[1] > 0 then -- 2.23
pos[1] = 1
end if
if pos[2] > 0 then
pos[2] = 1
end if
if pos[3] > 0 then
pos[3] = 1
end if
--============================
-- mask=pos>0 -- 7.99
-- pos-=(pos*mask)
-- pos+=mask
--============================
-- mask=pos>0 -- 7.86
-- pos*=(mask=0)
-- pos+=mask
--============================
-- pos=pos*(pos<=0)+(pos>0) -- 7.83
--============================
-- while 1 do -- 5.86
-- j=find(1,pos>1)
-- if not j then exit end if
-- pos[j]=1
-- end while
--============================
-- for k=1 to length(pos) do -- 2.68
-- if pos[k]>0 then pos[k]=1 end if
-- end for
end for
t=time()-t
?t
?pos
Those are my timings on the right for six different ways of doing
this. You may need another zero or two on the for loop 
Pete
8. Re: Fastest Way to...
Pete Lomax wrote:
>>Okay, here's my attempt:
>>
>>junk =3D pos <=3D 0
>>pos =3D pos * junk
>>pos +=3D not junk
>>
>>
>That's about 3 times slower.
>
>In your first post, you "unrolled the loop" which is a classic way to
>make something as fast as possible.
>
>Pete
>
>
hey, Pete, thanks for that info. I later realized that it won't be a
static size sequence, so I have to accommodate all sizes.
I bet looping through the sequence when length() is more than ? would be
faster than the sequence operations... right?
9. Re: Fastest Way to...
Hello, can someone tell me what is wrong with this? I'm using the hackserv
as a basis for a mud. I've tried to change it to accomodate telnet, gmud,
zmud, etc..(basically any client) where it only supported telnet in the
beginning. I KNOW this is too much code..but i've just been adding stuff in
trying to make it work and nothing seems to work. I KNOW that I'm
overlooking something really easy..but I can't figure it out. Can someone
tell me if you know what is wrong with this?
elsif action = FD_READ then -- The client said somethin'
junk = ""
owork = ""
while not match({CRLF},junk) do
owork = WsockReadData(sock,1)
if atom(owork) then
exit
end if
if length(owork)<1 then
exit
end if
if atom(owork) then
exit
end if
if length(owork)<1 then
exit
end if
junk = junk & owork
end while
-- junk now contains the client's responce
-- find this user in our list
iwork = getUserFromSocket(sock)
if length(players)>0 then
for x=1 to length(players) do
if players[x][1]=sock then
charnum=x
end if
end for
end if
players[charnum][6]=players[charnum][6]&junk
if compare({10},players[charnum][6][(length(players[charnum][6]))]) or
compare({13},players[charnum][6][length(players[charnum][6])]) then
if compare({10},players[charnum][6][length(players[charnum][6])]) then
players[charnum][6]=players[charnum][6][1..(length(players[charnum][6])-1)]
end if
if compare({13},players[charnum][6][length(players[charnum][6])]) then
players[charnum][6]=players[charnum][6][1..(length(players[charnum][6])-1)]
end if
if compare({10},players[charnum][6][length(players[charnum][6])]) then
players[charnum][6]=players[charnum][6][1..(length(players[charnum][6])-1)]
end if
junk=players[charnum][6]
warnErr("junk:"&sprint(junk))
players[charnum][6]=""
call_proc(players[charnum][7],{sock,junk})
junk=""
end if
Thank you in advance,
Michelle
----- Original Message -----
From: "C. K. Lester" <euphoric at cklester.com>
To: <EUforum at topica.com>
Sent: Monday, October 27, 2003 11:34 PM
Subject: Re: Fastest Way to...
>
>
> Pete Lomax wrote:
>
> >>Okay, here's my attempt:
> >>
> >>junk =3D pos <=3D 0
> >>pos =3D pos * junk
> >>pos +=3D not junk
> >>
> >>
> >That's about 3 times slower.
> >
> >In your first post, you "unrolled the loop" which is a classic way to
> >make something as fast as possible.
> >
> >Pete
> >
> >
> hey, Pete, thanks for that info. I later realized that it won't be a
> static size sequence, so I have to accommodate all sizes.
>
> I bet looping through the sequence when length() is more than ? would be
> faster than the sequence operations... right?
>
>
>
> TOPICA - Start your own email discussion group. FREE!
>
>
