A little slow out of the blocks but maybe useful to someone anyway:
For lists up to a certain size, doing the arithmetic using whole lists
instead of individual elements is faster. On my machine, the first
method is _way_ faster on lists with 500 or fewer items, and roughly 10%
faster on
Hi Guys,
I'm hoping someone can help me out with a quick pointer, I've got a
tight deadline and my minds gone a bit blank.
I have a list of points
[point(112,456), point(12,485), ...] list goes to about 50
These are actually co-ordinates of certain pixels on my stage.
I need to find the
Hi Tim,
Just off the top of my head this might get you started (email lingo, watch
for typos)
on getNearestPoint aListOfPoints, aRefPoint
d = the maxInteger
tWhichPoint = 0
repeat with i in aListOfPoints
a = aRefPoint[1] - i[1]
b = aRefPoint[2] - i [2]
c = a*a + b*b
if
on getNearestPoint aListOfPoints, aRefPoint
...
end
depending on the exact characteristics of your project, you might be
able to speed up the process of finding the minimal distance (as
proposed by rob) by using vectors (with z=0) instead of points, and
their distanceto method. but this
Just had a bet in the pub I can get the previous script shorter:)))
It's using vector maths (not sure how quick it is but it's shorter code)
on getNearest pList,pCenter
pStartVector = vector(pCenter[1],pCenter[2],0)
pCount = pList.count
if pCount 0 then
pNearest = 0
repeat with y
On Tue, 11 Apr 2006 21:41:33 +0200, Valentin Schmidt [EMAIL PROTECTED]
wrote:
but this means you first have to convert your poibnts to vectors, so it
doesn't make sense if all of the points change all the time.
Hey,
I took Valentin's code and threw in the point to vector conversion just
Quoting Pedja [EMAIL PROTECTED]:
Just had a bet in the pub I can get the previous script shorter:)))
It's using vector maths (not sure how quick it is but it's shorter code)
on getNearest pList,pCenter
pStartVector = vector(pCenter[1],pCenter[2],0)
pCount = pList.count
-- if
