ed return
> the
indices of the duplicate (or alternatively unique) values in a.
>
>
> Here is the piece of code that you suggested at the time (Re:
> [Numpy-discussion]
Efficient removal of duplicates, posted on Tue, 16 Dec 2008 01:10:00 -0800)
>
>
> -
There was an discussion about this on the c.l.p a while ago. Using a sort
will scale like O(n log n) or worse, whereas using a set (hash table) will
scale like amortized O(n). How to use a Python set to get a unique
collection of objects I'll leave to your imagination.
Sturla Molden
> On Mon, De
Thanks Daran,
that works like a charm!
Hanno
On Tue, Dec 16, 2008, Daran Rife said:
> Whoops! A hasty cut-and-paste from my IDLE session.
> This should read:
>
> import numpy as np
>
> a = [(x0,y0), (x1,y1), ...] # A numpy array, but could be a list
> l = a.tolist()
> l.sort()
> unique = [x
On Mon, Dec 15, 2008 at 18:24, Daran Rife wrote:
> How about a solution inspired by recipe 18.1 in the Python Cookbook,
> 2nd Ed:
>
> import numpy as np
>
> a = [(x0,y0), (x1,y1), ...]
> l = a.tolist()
> l.sort()
> unique = [x for i, x in enumerate(l) if not i or x != b[l-1]]
> a_unique = np.asarr
Whoops! A hasty cut-and-paste from my IDLE session.
This should read:
import numpy as np
a = [(x0,y0), (x1,y1), ...] # A numpy array, but could be a list
l = a.tolist()
l.sort()
unique = [x for i, x in enumerate(l) if not i or x != l[i-1]] # <
a_unique = np.asarray(unique)
Daran
--
On Dec
How about a solution inspired by recipe 18.1 in the Python Cookbook,
2nd Ed:
import numpy as np
a = [(x0,y0), (x1,y1), ...]
l = a.tolist()
l.sort()
unique = [x for i, x in enumerate(l) if not i or x != b[l-1]]
a_unique = np.asarray(unique)
Performance of this approach should be highly scalable.
On Mon, Dec 15, 2008 at 10:27, Hanno Klemm wrote:
>
> Hi,
>
> I the following problem: I have a relatively long array of points
> [(x0,y0), (x1,y1), ...]. Apparently, I have some duplicate entries, which
> prevents the Delaunay triangulation algorithm from completing its task.
>
> Question, is the
Hanno Klemm wrote:
> Hi,
>
> I the following problem: I have a relatively long array of points
> [(x0,y0), (x1,y1), ...]. Apparently, I have some duplicate entries, which
> prevents the Delaunay triangulation algorithm from completing its task.
>
> Question, is there an efficent way, of getting r
Hanno Klemm wrote:
> I the following problem: I have a relatively long array of points
> [(x0,y0), (x1,y1), ...]. Apparently, I have some duplicate entries, which
> prevents the Delaunay triangulation algorithm from completing its task.
>
> Question, is there an efficent way, of getting rid of the
Hi,
I the following problem: I have a relatively long array of points
[(x0,y0), (x1,y1), ...]. Apparently, I have some duplicate entries, which
prevents the Delaunay triangulation algorithm from completing its task.
Question, is there an efficent way, of getting rid of the duplicate
entries?
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