On Jul 31, 7:37 am, Jeremy Bailin <astroco...@gmail.com> wrote:
> On Jul 30, 7:54 am, Bennett <juggernau...@gmail.com> wrote:
>
>
>
>> On Jul 29, 11:50 am, Jeremy Bailin <astroco...@gmail.com> wrote:
>
>>> On Jul 29, 2:32 am, Brian Larsen <balar...@gmail.com> wrote:
>
>>>> We do need some more information but this is just screaming for
>>>> histogram. Have a read throughhttp://www.dfanning.com/tips/histogram_tutorial.html
>>>> . Using histogram to see which x's are common you can step through
>>>> the reverse_indices and see which y's are then common. There is
>>>> probably a more graceful way however.
>
>>>> Cheers,
>
>>>> Brian
>
>>>> ------------------------------------------------------------ --------------
>>>> Brian Larsen
>>>> Boston University
>>>> Center for Space Physicshttp://people.bu.edu/balarsen/Home/IDL
>
>>> In particular, if you're dealing with integers that don't span too big
>>> a range, use HIST_2D and find the maximum element. If you've got
>>> floats or a wide range, use UNIQ to turn each into an integer on a
>>> small range first.
>
>>> -Jeremy.
>
>> I think if I were to be working with small datasets....ie not in the
>> millions of points I would use something like this
>
>> coords = [[10,1],[20,32],[5,7],[6,8],[20,32],[2,14],[20,32],[10,10],
>> [3,1],[21,14]]
>
>> counter = intarr(9)
>
>> FOR i = 0, 8 DO BEGIN
>> FOR j = 0, 8 DO BEGIN
>
>> IF array_equal(coords[*,i],coords[*,j]) THEN counter[i]++
>
>> ENDFOR
>> ENDFOR
>
>> ;- Histogram to find the max bins (no need to measure anything below 2
>> ;- because that would just be a single hit and if all of your pairs
>> ;- only occur once then who cares, right?
>> hist = histogram(counter, min=2, reverse_indices=ri)
>> maxHist = max(hist, mxpos)
>> IF maxHist EQ 1 THEN print, 'Each pair occurs no more than once'
>
>> ;- Use the reverse indices given by histogram to find out exactly
>> ;- where in your counter these maxes are occurring
>> array_index = (counter[ri[ri[1]:ri[2]-1]])[0]
>
>> ;- Find where counter is equal to the array index determined by
>> ;- reverse indices
>> max_index = where(counter EQ array_index)
>
>> ;- Voila with your max pair
>> print, coords[*,max_index[0]]
>
>> Which spits out....
>> 20 32
>
>> This could be tweaked to find the top two or three or whatever as
>> well.
>> Hope this helps.
>
> My version of that would be:
>
> min1=min(coords[0,*], max=max1)
> min2=min(coords[1,*], max=max2)
> arraymap = hist_2d(coords[0,*], coords[1,*], min1=min1, max1=max1,
> bin1=1, min2=min2, max2=max2, bin2=1)
> maxval = max(arraymap, maxelement)
> print, array_indices([max1-min1+1,max2-min2+1], maxelement, /dimen)+
> [min1,min2]
>
> ...which avoids loops, and is more obvious to me.
>
> -Jeremy.
No loops is all and good...but if you put a decimal in coords like
this
coords = [[10.0,1.0],[20.0,32.3],[5,7],[6,8],[20.0,32.3],[2,14],
[20.0,32.3],[10,10],[3,1],[21,14]]
your code still spits out (20.0 32.0) where it should spit out (20.0
32.3)
By the way the code I presented up there should have the following
line replaced
array_index = (counter[ri[ri[1]:ri[2]-1]])[0]
with
array_index = (counter[ri[ri[mxpos]:ri[mxpos+1]-1]])[0]
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