[Numpy-discussion] unique() should return a sorted array

David Huard david.huard at gmail.com
Mon Jul 3 12:00:35 CDT 2006

```Here is a quick benchmark between numpy's unique, unique1d and sasha's
unique:

x = rand(100000)*100
x = x.astype('i')

%timeit unique(x)
10 loops, best of 3: 525 ms per loop

%timeit unique_sasha(x)
100 loops, best of 3: 10.7 ms per loop

timeit unique1d(x)
100 loops, best of 3: 12.6 ms per loop

So I wonder what is the added value of unique?
Could unique1d simply become unique ?

Cheers,

David

P.S.
I modified sasha's version to account for the case where all elements are
identical, which returned an empty array.

def unique_sasha(x):
s = sort(x)
r = empty(s.shape, float)
r[:-1] = s[1:]
r[-1] = NaN
return s[r != s]

2006/7/3, Robert Cimrman <cimrman3 at ntc.zcu.cz>:
>
> Sasha wrote:
> > On 7/2/06, Norbert Nemec <Norbert.Nemec.list at gmx.de> wrote:
> >> ...
> >> Does anybody know about the internals of the python "set"? How is
> >> .keys() implemented? I somehow have really doubts about the efficiency
> >> of this method.
> >>
> > Set implementation (Objects/setobject.c) is a copy and paste job from
> > dictobject with values removed.  As a result it is heavily optimized
> > for the case of string valued keys - a case that is almost irrelevant
> > for numpy.
> >
> > I think something like the following (untested, 1d only) will probably
> > be much faster and sorted:
> >
> > def unique(x):
> >       s = sort(x)
> >       r = empty_like(s)
> >       r[:-1] = s[1:]
> >       r[-1] = s[0]
> >       return s[r != s]
>
> There are 1d array set operations like this already in numpy
> (numpy/lib/arraysetops.py - unique1d, ...)
>
> r.
>
>
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