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*Subject*: Re: Histogram Hot-shots Required*From*: eddie haskell <haskell(at)see.signature.edu>*Date*: Fri, 16 Jul 1999 23:05:56 -0400*Newsgroups*: comp.lang.idl-pvwave*Organization*: Old Dominion Universityaruba*References*: <MPG.11f876f9f15942a98982b@news.frii.com>*Xref*: news.doit.wisc.edu comp.lang.idl-pvwave:15712

David Fanning wrote: > What I can't figure out tonight is how to find out > what bin that pixel is in, given that I know the pixel > value. David, I believe reverse_indices are the way to go. First the short answer: given: A - the array containing the data (your image) V - the data (pixel) value at the chosen location R - the reverse_indices returned from the histogram function (and I don't care what you chose as a bin size, that is incorporated into R) The number of the bin in which the data value resides can be calculated as: bin=(where(R ge ((where(((where(A eq V))[0]) eq R[R[0]:*]))[0]+R[0])))[0] and the number of elements in that bin is then: num=R[bin+1]-R[bin] Now the long answer: reverse_indices gives you the number of elements in each bin and tells you in which bin each element of your original array falls. For example: A=[1,2,3,4,3,2,2] h=histogram(A,bin=2,reverse=R) ;note: binsize does not matter to this function, I chose 2 to make R smaller. now R = [3, 7, 10, 0, 1, 5, 6, 2, 3, 4] What R is telling you is that the subscripts of the items in A in the first bin are the ones found in R[3:7] or A[0,1,5,6] which are the positions of the 1's and 2's in A. The second bin is R[7:10] or A[2,3,4], the 3's and 4. Choose a value that does exist in the array A: e.g., V = 3 First determine a location in A where V exists, for the sake of ease we will choose the first occurance of V w1=(where(A eq V))[0] ;here w1 = 2 Since we know V has to exist in A we don't have to check for where returning -1 Then we find the location of that position in the vector R w2=(where(w1 eq R[R[0]:*]))[0]+R[0] ;here w2 = 7 R[0] happens to be the index of the first location of the portion of R where the separate array subscripts are stored. Adding R[0] to the where() returns the actual value. Again, we know the value exists so no need to check for a -1 The first part of R lists the indices in R of where the subscripts of A are found in each individual bin. Just look for the first value that is greater than or equal to w2. bin=(where(r ge w2))[0] ;here bin = 1 or the value of 3 is found in the 2nd bin. The long line at the top condenses these steps into one confusing line. As before, finding the number of elements in that bin is straightforward. num=R[bin+1]-R[bin] I hope this helps you. As I noted before, odds are either this solution will not work for you or a better solution will soon present itself. Cheers, eddie ----- ---- --- --- ---- --- -- --- --- -- -- - - - - |\ A G Edward Haskell |\ Center for Coastal Physical Oceanography |\ Old Dominion University, Norfolk VA 23529 |\ Voice 757.683.4816 Fax 757.683.5550 |\ e-mail haskell*ccpo.odu.edu ----- ---- --- ---- --- --- --- --- -- -- -- - - - -

**References**:**Histogram Hot-shots Required***From:*David Fanning

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