Thank Nadav for testing out the matrix. I wonder if you had a chance to
check if the resulting decomposition contained NaN or Inf values?
As far I understood, numpy.linalg.svd uses routines in LAPACK and ATLAS
(if available) to compute the corresponding SVD. I did some
complementary tests on
Thanks!
Jose
On Thu, Aug 11, 2011 at 8:15 PM, Fernando Perez fperez@gmail.comwrote:
On Thu, Aug 11, 2011 at 4:43 PM, Jose Borreguero borregu...@gmail.com
wrote:
a = random.randn(3,3)
b = a.reshape(1,3,3).repeat(50,axis=0)
scipy.linalg.block_diag( *b )
slightly simpler, but
Hi All,
I am working on something that appeared to be a no-brainer issue (at the
beginning), by my complete ignorance in statistics is overwhelming and I got
stuck.
What I am trying to do can be summarized as follows
Let's assume that I have to generate a sample of a 1,000 values for a
On Fri, Aug 12, 2011 at 4:03 AM, Charanpal Dhanjal
dhan...@telecom-paristech.fr wrote:
Thank Nadav for testing out the matrix. I wonder if you had a chance to
check if the resulting decomposition contained NaN or Inf values?
As far I understood, numpy.linalg.svd uses routines in LAPACK and
There'a a boatload of options for nditer. I need a simple explanation, maybe a
few simple examples. Is there anything that might help?
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Hi Andrea--An easy way to get something like this would be
import numpy as np
import scipy.stats as stats
sigma = #some reasonable standard deviation for your application
x = stats.norm.rvs(size=1000, loc=125, scale=sigma)
x = x[x50]
x = x[x200]
That will give a roughly normal distribution to
I tested all the the result 3 matrices with alltrue(infinite(mat)) and got True
answer for all of them.
Nadav
From: numpy-discussion-boun...@scipy.org [numpy-discussion-boun...@scipy.org]
On Behalf Of Warren Weckesser [warren.weckes...@enthought.com]
Sent:
I'll write up some more introductory-style documentation, you're right that
the examples I put in the reference page aren't a nice simple starting
point. Will post back here for feedback when I have a draft for you to
review.
Cheers,
Mark
On Fri, Aug 12, 2011 at 7:30 AM, Neal Becker
===
Announcing Theano 0.4.1
===
This is an important release, with lots of new features, bug
fixes and some deprecation warning. The upgrade is recommended for everybody.
For those using the bleeding edge version in the
mercurial repository, we
