OpenEuphoria

Re: Fastest Way to...

• Posted by Pete Lomax <petelomax at blueyonder.co.uk> Oct 28, 2003
• 387 views

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