Re: Histogram Hot-shots Required

[eddie haskell <haskell(at)see.signature.edu>]

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

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