| Re: Astronomys` Sixth Neighbour Needs Help [message #36020 is a reply to message #35884] |
Mon, 11 August 2003 07:53   |
google_forums
Messages: 3
Registered: August 2003
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Junior Member |
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A slight twist to the Astronomer's problem...
I am running a similar loop to Astronomer, the exception is that
rather than always needing the distance to the sixth closest point,
the point that I need depends on another variable. What I am using
this for is to spread points out across a map so that they are evenly
distributed across the map while still maintaining the "best" points.
For example, if I have a list of cities with Latitude and Longitude
and population and I want to mostly display the largest cities, but
where there are clusters of large cities (I.E. around Chicago) that
are near eachother, I would only want to display the biggest, while if
there was a smaller city in the middle of nowhere, I would want it to
show. To do this I take a score factor (in this case population) and
from each point, I calculate the distance to the nearest point that
has a higher score... I then either use this as a weight factor along
with the score, or I use it alone depending on how much I want the
score to come through vs. how evenly I want the points spread out. My
code below works fine, but as in Astronomer's case it is very slow --
I typically need to run this code for around 12,000 points but in some
cases for up to 40,000 cases -- the latter, I have yet to have the
patience for... in any case, here's the code:
pro closestpoint, data
; data is a preexisting structure that has fields for score (the
ranking
; field), latitude, longitude, id, and output fields to hold the
closest
; point with a higher field, the distance to that point, and a
"combined
; score" which weights the distance and the score
distance=data.distance
closest=data.closest
tmp=distance
k=0
for i=0L,n_elements(data)-1 do begin
if k eq 0 then print, 'counter...............................',i, '
',data[i].id
for j=0L,n_elements(data)-1 do begin
lon1=data[j].longitude
lat1=data[j].latitude
lon2=data[i].longitude
lat2=data[i].latitude
if ((data[i].longitude eq data[j].longitude) and
(data[i].latitude eq data[j].latitude) $
and (data[i].id ne data[j].id) and (data[i].score le
data[j].score)) then begin
distance[i]=0.01
closest[i]=data[j].id
print, 'Station Colocation Found: ' + data[i].id + ' ' +
data[j].id + ' distance: ' + string(distance[i])
endif else begin
if ((data[i].score le data[j].score) and (data[i].id ne
data[j].id)) then begin
tstdistance=map_2points(lon1,lat1,lon2,lat2, /miles)
;if k eq 0 then print, 'calculated
distance....',i,data[i].score, j,
data[j].score,tstdistance,distance[i]
if ((tstdistance lt distance[i]) or (distance[i] eq 0))
then begin
distance[i]=tstdistance
closest[i]=data[j].id
if (k eq 0) then print, 'distance to ' +
data[j].id + ' = ' + string(tstdistance) +' closest=' + closest[i]
endif
endif
endelse
endfor
data[i].distance=distance[i]
k=k+1
if k eq 500 then k=0
endfor
data.combined=(data.distance * mean(data.score) / mean(data.distance)
* 4)+data.score
data.closest=closest
END
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