27 Feb
2014
27 Feb
'14
7:11 p.m.
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
27 Feb
27 Feb
8:01 p.m.
On Thu, Feb 27, 2014 at 6:11 PM, Alan G Isaac alan.isaac@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
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