On Thu, Feb 27, 2014 at 6:11 PM, Alan G Isaac 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.aran
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?
Thanks,
Alan Isaac
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
