On Mon, Dec 15, 2008 at 8:37 PM, wrote:
> What's the future of the example list, on the example list with docs
> it says Numpy 1.0.4. It hasn't been updated in a while. When I started
> out with numpy, I used it as a main reference, but now, some examples,
> that I wanted to look at, had outdated
On Mon, Dec 15, 2008 at 10:18 PM, Alan G Isaac wrote:
>> On Mon, Dec 15, 2008 at 9:21 PM, Alan G Isaac wrote:
>>> I noticed that unique1d is not documented on the
>>> Numpy Example List http://www.scipy.org/Numpy_Example_List
>>> but is documented on the Numpy Example List with Doc
>>> http://www.
> On Mon, Dec 15, 2008 at 9:21 PM, Alan G Isaac wrote:
>> I noticed that unique1d is not documented on the
>> Numpy Example List http://www.scipy.org/Numpy_Example_List
>> but is documented on the Numpy Example List with Doc
>> http://www.scipy.org/Numpy_Example_List_With_Doc
>> I thought the l
On Mon, Dec 15, 2008 at 9:21 PM, Alan G Isaac wrote:
> On 12/15/2008 7:53 PM Robert Kern apparently wrote:
>> That basic idea is what unique1d() does; however, it uses numpy
>> primitives to keep the heavy lifting in C instead of Python.
>
>
>
> I noticed that unique1d is not documented on the
> N
On 12/15/2008 7:53 PM Robert Kern apparently wrote:
> That basic idea is what unique1d() does; however, it uses numpy
> primitives to keep the heavy lifting in C instead of Python.
I noticed that unique1d is not documented on the
Numpy Example List http://www.scipy.org/Numpy_Example_List
but is
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 12/15/2008 6:01 PM Michael Gilbert apparently wrote:
> According to wikipedia [1], some common Mersenne twister algorithms
> use a linear congruential gradient (LCG) to generate seeds. LCGs have
> been known to produce poor random numbers. Does numpy's Mersenne
> twister do this? And if so, i
On Monday 15 December 2008 18:01:41 Michael Gilbert wrote:
> According to wikipedia [1], some common Mersenne twister algorithms
> use a linear congruential gradient (LCG) to generate seeds. LCGs have
> been known to produce poor random numbers. Does numpy's Mersenne
> twister do this? And if so
On Mon, Dec 15, 2008 at 6:01 PM, Michael Gilbert <
michael.s.gilb...@gmail.com> wrote:
> According to wikipedia [1], some common Mersenne twister algorithms
> use a linear congruential gradient (LCG) to generate seeds. LCGs have
> been known to produce poor random numbers. Does numpy's Mersenne
According to wikipedia [1], some common Mersenne twister algorithms
use a linear congruential gradient (LCG) to generate seeds. LCGs have
been known to produce poor random numbers. Does numpy's Mersenne
twister do this? And if so, is this potentially a problem?
http://en.wikipedia.org/wiki/Line
2008/12/15 Benjamin Haynor :
> I was wondering if I can concatenate 3 arrays, where the result will be a
> view of the original three arrays, instead of a copy of the data. For
> example, suppose I write the following
> import numpy as n
> a = n.array([[1,2],[3,4]])
> b = n.array([[5,6],[7,8]])
>
On Mon, Dec 15, 2008 at 11:39, Benjamin Haynor wrote:
> Hi,
>
> I was wondering if I can concatenate 3 arrays, where the result will be a
> view of the original three arrays, instead of a copy of the data.
No, this is not possible in general with numpy's memory model.
--
Robert Kern
"I have co
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 was wondering if I can concatenate 3 arrays, where the result will be a view
of the original three arrays, instead of a copy of the data. For example,
suppose I write the following
import numpy as n
a = n.array([[1,2],[3,4]])
b = n.array([[5,6],[7,8]])
c = n.array([[9,10],[11,12]])
c = n
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?
All
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