On Jul 24, 2012 12:32 PM, <giuseppe.amatu...@gmail.com> wrote: > after i need to calculate unique combination in pairs to count the observations > and obtain > (4,3,2) > (4,5,1) > (5,4,1) > (4,4,2)
I don't know about a numpy solution, but this could be achieved by collections.Counter(zip(a, b)).items(). That gives you: ((4,3),2) ((4,5),1) Etc. If you really want triples instead of pair-value pairs, then you would need to flatten the tuples yourself.
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