[SciPy-User] Ranking a list of numbers
Keith Goodman
kwgoodman@gmail....
Thu Jun 3 09:36:10 CDT 2010
On Thu, Jun 3, 2010 at 7:32 AM, Keith Goodman <kwgoodman@gmail.com> wrote:
> On Thu, Jun 3, 2010 at 7:20 AM, Jose Gomez-Dans <jgomezdans@gmail.com> wrote:
>> Hi,
>> I've done this with loops, but I am sure there is a much nicer way of doing
>> it with a couple of index tricks.
>> Let's say I've got an array of numbers. I want to convert it into an array
>> where each element is the rank of that element in the starting array (by
>> rank I mean its position when sorted in decreasing order)
>> For example, if my original array is
>> [ 0.012, 0.08, 2, 0.5, 0.010, 0.03]
>> my output array ought to look like this (starting at 1, rather than 0)
>> [ 5, 3, 1, 2, 6, 4 ]
>> meaning "the first element of the array is the 5th largest, the second is
>> the 3rd largest, the third is the largest" and so on.
>> The ordering can be done with argsort[::-1] (to get the decreasing order),
>> but to get the final array, I can only think of clumsy ways of doing loops.
>> Any ideas?
>> Thanks!
>> J
>
> If you don't want to split ties nor handle NaNs:
>
>>> (-a).argsort().argsort() + 1
> array([5, 3, 1, 2, 6, 4])
>
> To handle ties you can use:
>
> from scipy.stats import rankdata
>
> To handle ties and NaNs, you can use the labeled array package, la:
>
>>> import la
>>> lar = la.larry([0.012, 0.08, 2, 0.5, 0.010, 0.03])
>>> (-lar).ranking(norm='0,N-1').A + 1
> array([ 5., 3., 1., 2., 6., 4.])
Oh, actually la has a pure array version:
>> a = np.array([0.012, 0.08, 2, 0.5, 0.010, 0.03])
>> from la.afunc import ranking
>> ranking(-a, norm='0,N-1') + 1
array([ 5., 3., 1., 2., 6., 4.])
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