On Aug 6, 7:28 am, Jeremy Bailin <astroco...@gmail.com> wrote:
> On Aug 5, 10:14 am, Bennett <juggernau...@gmail.com> wrote:
>
>
>
>> On Aug 1, 7:02 am, Jeremy Bailin <astroco...@gmail.com> wrote:
>
>>> On Jul 31, 10:50 am, Bennett <juggernau...@gmail.com> wrote:
>
>>>> 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]
>
>>> Like I said, if you have floats (or a very large range of integers),
>>> you should map them into integers first using SORT and UNIQ...
>
>>> coordsize = size(coords,/dimen)
>>> coords0_sorted = coords[0,sort(coords[0,*])]
>>> map0 = uniq(coords0_sorted)
>>> nmap = n_elements(map0)
>>> new_coords0 = lonarr(coordsize[1])
>>> for i=0l,nmap-1 do new_coords0[where(coords[0,*] eq
>>> coords0_sorted[map0[i]])]=i
>
>>> ...and the same for coords[1,*]. There's probably a more efficient way
>>> of doing that, but you get the idea.
>
>>> -Jeremy.
>
>> 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]]
>
>> sz = size(coords, /dimensions)
>
>> result = rebin(coords,2,sz[1],sz[1])
>> result2 = rebin(reform(coords,2,1,sz[1]),2,sz[1],sz[1])
>> indices = array_indices(result/result2,where(result/result2 EQ 1))
>
>> hist = histogram(indices[2,*])
>> maxHist = max(hist, mxpos)
>
>> print, coords[*,mxpos]
>
>> No loops...but definitely limited by size...can't really go with more
>> than a 7500 indices
>
> That doesn't work if you have individual elements that come up much
> more often than elements in the most frequent pair... I fooled it with
> this input:
>
> 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],[10.0,32.3],[10.0,8]]
>
> and it comes up with [10.0,32.3].
>
> -Jeremy.
Dang it...well I had to give it a go.
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