Hi,
Is there going to be a scipy release anytime soon?
I'm using numpy 1.0.3 with scipy 0.5.2 and I get these ugly warnings all
the time:
c:\apps\python25\lib\site-packages\scipy\misc\__init__.py:25:
DeprecationWarning: ScipyTest is now called NumpyTest; please update
your code
test =
David Cournapeau wrote:
Hi,
I was wondering what an empty matrix is, and what it is useful for
(by empty matrix, I mean something created by numpy.matrix([])) ? Using
those crash some functions (see for example scipy ticket #381), and I am
not sure how to fix this bug.
David
Hi,
I have some arrays of various shapes in which I need to set any NaNs to
0. I have been doing the following:
a[numpy.where(numpy.isnan(a)] = 0.
as you can see here:
In [20]: a=numpy.ones(2)
In [21]: a[1]=numpy.log(-1)
In [22]: a
Out[22]: array([ 1., NaN])
In [23]:
Benjamin Root wrote:
On Tue, Jul 13, 2010 at 12:45 PM, Kurt Smith kwmsm...@gmail.com
mailto:kwmsm...@gmail.com wrote:
On Tue, Jul 13, 2010 at 11:54 AM, John Reid
j.r...@mail.cryst.bbk.ac.uk mailto:j.r...@mail.cryst.bbk.ac.uk
wrote:
Hi,
I have some arrays
Hi,
My array of integers is printed like this by default in numpy:
array([[ 4.7500e+02, 9.5000e+02, -1.e+00],
[ 2.6090e+03, 9.5000e+02, -7.0900e+02]])
Can I set an option so that numpy never uses this scientific notation or
raise the threshold at
Hi,
I recently upgraded to numpy 1.5.0 and now I get warnings when I take
the logarithm of 0.
In [5]: np.log(0.)
Warning: divide by zero encountered in log
Out[5]: -inf
I want to evaluate x * log(x) where its value is defined as 0 when x=0
so I have the following function:
def
I imagine I'm using 64 bit numpy as I made a vanilla install from recent
source on a 64 bit box but how can I tell for sure? I have some problems
creating large arrays.
In [29]: a=numpy.empty((1024, 1024, 1024), dtype=int8)
works just fine
In [30]: a=numpy.empty((1024, 1024, 2048),
Sorry for noise, it is my mistake. My assumption that the box is 64 bit
was wrong :(
At least the processors are 64 bit :
Intel® Core™2 Duo Processor T9600
but uname -m reports:
i686
which as far as I understand means it thinks it is a 32 bit processor.
If anyone knows better please let
David Cournapeau wrote:
from platform import machine
print machine()
Should give you something like x86_64 for 64 bits intel/amd architecture,
In [3]: from platform import machine
In [4]: print machine()
i686
Now I'm wondering why the OS isn't 64 bit but that's not for discussion
here I
Charles R Harris wrote:
What really matters is if python is 64 bits. Most 64 bit systems also
run 32 bit binaries.
Are you saying that even if uname -m gives i686, I still might be able
to build a 64 bit python and numpy?
___
Numpy-discussion
Charles R Harris wrote:
What platform are you on? I'm guessing Mac. You can check python on unix
type systems with
$[char...@f9 ~]$ file `which python`
/usr/bin/python: ELF 32-bit LSB executable, Intel 80386, version 1
(SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.9,
Charles R Harris wrote:
On Sat, Mar 28, 2009 at 5:32 AM, John Reid j.r...@mail.cryst.bbk.ac.uk
mailto:j.r...@mail.cryst.bbk.ac.uk wrote:
Charles R Harris wrote:
What really matters is if python is 64 bits. Most 64 bit systems also
run 32 bit binaries
David Cournapeau wrote:
(Continuing the discussion initiated in the neighborhood iterator thread)
- Chuck suggested to drop python 2.6 support from now on. I am
against it without a very strong and detailed rationale, because many OS
still don't have python 2.6 (RHEL, Ubuntu LTS).
I vote
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