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I'd like to hit control-c and get out of this hung state.
What would it take to support this?
(I'm running ancient numpy and python at work, so if this is already
supported in later versions, my apologies)
- Tom K.
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David Cournapeau wrote:
(I'm running ancient numpy and python at work, so if this is already
supported in later versions, my apologies)
What does ancient mean ? Could you give us the version (numpy.__version__)
1.0.4
Python 2.4.2
IPython 0.9.1
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This one bit me again, and I am trying to understand it better so I can
anticipate when it will happen.
What I want to do is get rid of singleton dimensions, and index into the
last dimension with an array.
In [1]: import numpy as np
In [2]: x=np.zeros((10,1,1,1,14,1024))
In [3]:
davefallest wrote:
...
In [3]: np.arange(1.01, 1.1, 0.01)
Out[3]: array([ 1.01, 1.02, 1.03, 1.04, 1.05, 1.06, 1.07, 1.08,
1.09, 1.1 ])
Why does the ... np.arange command end up including my stop value?
From the help for arange:
For floating point arguments, the
ANNOUNCEMENT
I am pleased to announce a new release of upfirdn - version 0.2.0. This
package provides an efficient polyphase FIR resampler object (SWIG-ed C++)
and some python wrappers.
This release greatly improves installation with distutils relative to the
initial 0.1.0 release. 0.2.0
(also posted on scipy-user)
ANNOUNCEMENT
I am pleased to announce the initial release of upfirdn. This package
provides an efficient polyphase FIR resampler object (SWIG-ed C++) and some
python wrappers.
https://opensource.motorola.com/sf/projects/upfirdn
MOTIVATION
As a long time
it wrong.
I would also vote +1 for an ND version of this (growing only a single
dimension). Keeping 2x for each of n dimensions, while conceivable, would
be 2**n extra memory, and hence probably too costly.
Cheers,
Tom K.
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jseabold wrote:
On Mon, Jun 8, 2009 at 3:33 PM, Robert Kernrobert.k...@gmail.com wrote:
On Mon, Jun 8, 2009 at 14:10, Alan G Isaacais...@american.edu wrote:
Going back to Alan Isaac's example:
1) beta = (X.T*X).I * X.T * Y
2) beta = np.dot(np.dot(la.inv(np.dot(X.T,X)),X.T),Y)
Robert
bruno Piguet wrote:
Can someone point me to a doc on dot product vectorisation ?
As I posted in the past you can try this one liner:
numpy.array(map(numpy.dot, a, b))
that works for matrix multiply if a, b are (n, 3, 3).
This would also work if a is (n, 3, 3) and b is (n, 3, 1)
). But a .I attribute and its behavior
needn't be bundled with whatever proposal we wish to make to the python
community for a new operator of course.
Regards,
Tom K.
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Sent from
the actual time to implement and test the new python syntax would
probably be much shorter - it is not so much a technical issue as one of
socialization. What do we want to do? How can we convince the larger
python community that this is a good idea for them too?
Cheers,
Tom K.
--
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Robert Kern-2 wrote:
On Sun, Jun 7, 2009 at 07:20, Tom K. t...@kraussfamily.org wrote:
Going back to Alan Isaac's example:
1) beta = (X.T*X).I * X.T * Y
...
4) beta = la.lstsq(X, Y)[0]
I really hate that example.
Understood. Maybe propose a different one?
Robert Kern-2 wrote
lives are sane again. :-)
Long live the numpy ndarray! Thanks to the creators for all your hard work
BTW - I love this stuff!
- Tom K.
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Jason Rennie-2 wrote:
By default, it looks like a 1-dim ndarray gets converted to a row vector
by
the matrix constructor. This seems to lead to some odd behavior such as
a[1] yielding the 2nd element as an ndarray and throwing an IndexError as
a
matrix. Is it possible to set a flag to
do it or how hard it would
be.
Regards,
Tom K.
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Sent from the Numpy-discussion mailing list archive at Nabble.com.
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Numpy
h = zeros((1, 4, 100))
h[0,:,arange(14)].shape
(14, 4)
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