[Numpy-discussion] unique rows of array

[josef.pktd@gmai...]

On Tue, Aug 18, 2009 at 2:01 AM, Maria Liukis<liukis@usc.edu> wrote:
> Josef,
> Many thanks for the example! It should become an official NumPy recipe :)
> Thanks again,
> Masha
> --------------------
> liukis@usc.edu

Actually, there is also an implementation of unique rows in
scipy.stats._support. It uses loops (and array concatenation in the
loop), but it preserves the order of the rows in the array.

In general, I don't recommend using scipy.stats._support, since many
or most functions are not tested and only some are used in
scipy.stats. These functions wait for a rewrite or removal. When I
thought about a rewrite last year, I didn't know much about structured
arrays and views.

Josef

>>> cc
array([[10,  1,  2],
       [ 3,  4,  5],
       [ 3,  4,  5],
       [ 9, 10, 11]])
>>> scipy.stats._support.unique(cc)
array([[10,  1,  2],
       [ 3,  4,  5],
       [ 9, 10, 11]])

unique columns using transpose :

>>> cct = cc.T.copy()
>>> cct
array([[10,  3,  3,  9],
       [ 1,  4,  4, 10],
       [ 2,  5,  5, 11]])
>>> scipy.stats._support.unique(cct.T).T
array([[10,  3,  9],
       [ 1,  4, 10],
       [ 2,  5, 11]])

Josef



>
> On Aug 17, 2009, at 10:03 PM, josef.pktd@gmail.com wrote:
>
> On Tue, Aug 18, 2009 at 12:59 AM, Maria Liukis<liukis@usc.edu> wrote:
>
> On Aug 17, 2009, at 9:51 PM, Charles R Harris wrote:
>
> On Mon, Aug 17, 2009 at 10:30 PM, Maria Liukis <liukis@usc.edu> wrote:
>
> Hello everybody,
> While re-implementing some Matlab code in Python, I've run into a problem
> of finding a NumPy function analogous to the Matlab's "unique(array,
> 'rows')" to get unique rows of an array. Searching the web, I've found a
> similar discussion from couple of years ago with an example:
>
> Just to be clear, do you mean finding all rows that only occur once in the
> array?
> Yes.
>
> I interpreted your question as removing duplicates. It keeps rows that
> occur more than once.
> That's what my example is intended to do.
> Josef
>
> <snip>
> Chuck
>
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