On Thu, Feb 27, 2014 at 6:11 PM, Alan G Isaac <alan.is...@gmail.com> wrote:
> I have a bincount array `cts`. > I'd like to produce any one array `a` such that `cts==np.bincounts(a)`. > Easy to do in a loop, but does NumPy offer a better (i.e., faster) way? > >>> cts = np.bincount([1,1,2,3,4,4,6]) >>> np.repeat(np.arange(len(cts)), cts) array([1, 1, 2, 3, 4, 4, 6]) Jaime -- (\__/) ( O.o) ( > <) Este es Conejo. Copia a Conejo en tu firma y ayúdale en sus planes de dominación mundial.
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