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.
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