Marshall,

I am having some difficulty getting my head around your comments involving
interval lists. I am still processing your ideas because I don't see how to
put them into a strategy that deals with my diamonds.

On the other hand, one of your comments regarding "Just picking the minimum
won't work,... " has really thrown me for a little loop even in the
strategy Raul suggested using contigmd. So I wonder if I am missing the
point of much of your comments and you are way ahead of my thinking, to
thinking about the lack of a meaningful global minimum and needing the
local minimum in the middle.

For example in one of my lighting situations I got the following results
where the first column is the sampled thresholds and the second column is
the corresponding lengths of nubbed and flattened contigmd results.
Presumably in this example the threshold of about 118 would work well. But
I am still working on better ways to select the sample thresholds and how
to automate the selection of the best result.

On the other hand, I need to better understand your ideas involving
interval lists, so I am going to think about that for now, too.

Thanks very much,

 90  2
 95  7
100 29
102 30
104 31
106 19
108 10
110  4
112  3
114  3
116  2
118  2
120  2
122  3
124  4
126 11
128 23
130 28
132 33
134 43
136 46
138 57
145 23
150  2



On Tue, Jun 27, 2017 at 4:20 AM, Marshall Lochbaum <[email protected]>
wrote:

> A common strategy that may work for you is just to ignore the spacial
> data and perform some sort of clustering (like k-means, with k=2) on the
> intensities. Then classify pixels according to which cluster they fall
> in.
>
> An approach that does use this data, and which I expect would be much
> slower but somewhat more accurate, would be to use the realization that
> a threshold that maximizes the number of contiguous 1's minimizes the
> number of adjacent pixels which are separated by the threshold (that is,
> adjacent 0-1 pairs). To minimize this number, we need to collect all
> pairs of adjacent pixels, then turn consider these pairs as intervals
> and compute for each possible threshold the number of these intervals
> which contain it.
>
> I don't have any particularly clever ways to find pairs of adjacent
> pixels. For the strictest form of adjacency, with no diagonals, the verb
> ((}. ,.&, }:) , (}."1 ,.&, }:"1)) will get them all. Each pair should
> then be sorted, to put them in a standard interval form.
>
> An efficient way to turn a list of intervals into a list of
> number-of-intervals-containing uses an obverse of monad (I.), which I am
> increasingly coming to believe is a major feature missing from J. To
> wit:
>
>    x =. 4 7 2 4 1 7 8
>    (10$0) (#/.~@])`(~.@])`[}"1 x
> 0 1 1 0 2 0 0 2 1 0
>    I. (10$0) (#/.~@])`(~.@])`[}"1 x
> 1 2 4 4 7 7 8
>    (/:~x) -: I. (10$0) (#/.~@])`(~.@])`[}"1 x
> 1
>
> The verb ((#/.~@])`(~.@])`[}) with an appropriate left argument is an
> left inverse to I. ignoring trailing zeros, and a right inverse to I.
> ignoring ordering. It counts, for each number in (i.n) where n is the
> length of the left argument, how many elements on the right are equal to
> it. To get our interval count, we just take this count for the starts of
> intervals, subtract the counts for ends of intervals, and do a running
> sum:
>
>    ]a =. /:~"1 ]10 2?@$10  NB. intervals obtained from image
> 3 4
> 5 9
> 5 7
> 0 6
> 0 5
> 3 6
> 3 6
> 1 7
> 2 9
> 1 4
>    +/\ -/ (10$0) (#/.~@])`(~.@])`[}"1 |:a
> 2 4 5 8 6 7 4 2 2 0
>
> The above example for n=10 has counts increasing then falling, but if
> the values in the image are clustered like yours are, then they should
> dip in the middle. Just picking the minimum won't work, because
> a threshold of 0 or 256 trivially maximizes contiguous 0's or 1's.
> However, it should be possibly to either penalize values at the edges
> (say, by adding a multiple of (*:128-~i.256) or select a local minimum
> near the middle to obtain a threshold.
>
> ​[snip
> ​]
>
>
> Marshall
>
> --
(B=)
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