I wouldn't know for sure, but could this be related to changes to the
gcc compiler in Fedora 13 (with respect to implicit DSO linking) or
would that only be an issue at build-time?
http://fedoraproject.org/w/index.php?title=UnderstandingDSOLinkChange
I'm not entirely sure I understand
2010/8/4 Søren Gammelmark gammelm...@phys.au.dk:
I wouldn't know for sure, but could this be related to changes to the
gcc compiler in Fedora 13 (with respect to implicit DSO linking) or
would that only be an issue at build-time?
On Tue, Aug 3, 2010 at 4:10 PM, Bruce Southey bsout...@gmail.com wrote:
Hi,
Since I was testing the 1.5 beta, I also tested the alpha release of Python
3.2 on Linux 64-bit (gcc version 4.4.4 20100630 (Red Hat 4.4.4-10) (GCC)).
While my other Python versions passed the tests (once I copied the
Hi,
given A=empty([10,10]), I would like to keep A[:5,:5] as a contiguous memory
segment.
How to do it efficiently?
Thanks.
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On Wed, Aug 4, 2010 at 09:29, Antoine Dechaume boole...@gmail.com wrote:
Hi,
given A=empty([10,10]), I would like to keep A[:5,:5] as a contiguous memory
segment.
How to do it efficiently?
I'm not sure I understand what you want. Your Subject line and the
body of your email conflict with
I forgot to refer to resize, sorry about that.
A[:5,:5] shows the data I want, but it's not contiguous in memory.
A.resize(5,5) is contiguous, but do not contains the data I want.
How to get both efficiently?
On Wed, Aug 4, 2010 at 5:01 PM, Robert Kern robert.k...@gmail.com wrote:
On Wed,
A[:5,:5] shows the data I want, but it's not contiguous in memory.
A.resize(5,5) is contiguous, but do not contains the data I want.
How to get both efficiently?
A[:5,:5].copy()
will give you a new, contiguous array that has the same contents as
A[5:,5:], but in a new chunk of memory. Is
Yes it is, but is there a way to do it in-place?
On Wed, Aug 4, 2010 at 5:20 PM, Zachary Pincus zachary.pin...@yale.eduwrote:
A[:5,:5] shows the data I want, but it's not contiguous in memory.
A.resize(5,5) is contiguous, but do not contains the data I want.
How to get both efficiently?
On Wed, Aug 4, 2010 at 6:25 AM, Pauli Virtanen p...@iki.fi wrote:
Mon, 02 Aug 2010 23:48:52 +0800, Ralf Gommers wrote:
I'm trying to get building to work with Python 3.1 under Wine on OS X.
The first thing you run into is a python distutils problem, which is
fixed by replacing line 379 of
On Wed, Aug 4, 2010 at 7:03 AM, sam tygier samtyg...@yahoo.co.uk wrote:
On 01/08/10 17:38, Ralf Gommers wrote:
I am pleased to announce the availability of the first beta of NumPy
1.5.0. This will be the first NumPy release to include support for
Python 3, as well as for Python 2.7. Please
On Wed, Aug 4, 2010 at 10:31, Antoine Dechaume boole...@gmail.com wrote:
Yes it is, but is there a way to do it in-place?
No.
--
Robert Kern
I have come to believe that the whole world is an enigma, a harmless
enigma that is made terrible by our own mad attempt to interpret it as
though it
Yes it is, but is there a way to do it in-place?
So you want the first 25 elements of the array (in a flat contiguous
view) to contain the 25 elements of A[:5,:5]? This will do that, but
having to do stuff like this (rather than just copying the memory
region) might be indicative that
Oh and PS. Robert's right that there's no general way to do this! What
I have only works because the data existing in the first 25 elements
of A that get clobbered by the copy operation aren't the same data
that are being copied (or where they are, the new copy is identical to
the old
Wed, 04 Aug 2010 23:34:15 +0800, Ralf Gommers wrote:
[clip]
I haven't started using py3k yet so I'm still a bit fuzzy about bytes
vs string. But it's easy to try in the interpreter:
import re
RE_VERSION = re.compile('(\d+\.\d+(\.\d+)*)')
In the Python 3.1 version I have, this line reads
Are they numbered like that? If so you can index into the first array by the
second one.
x[y[:,0], 1] if you can't get them into an indexable format, I think it's
going to be slow no matter how you do it.
On Wed, Aug 4, 2010 at 4:59 PM, phob...@geosyntec.com wrote:
Hey folks,
I've one array,
Is there a concise, Numpythonic way to copy the values of x[:,1] over to
y[:,1] where x[:,0] = y[:,0]? Resulting in, z:
First use
mask = (x[:,0] == y[:,0]) # integers
or
mask = np.abs(x[:,0] - y[:,0]) eps # floats
and then
y[mask,1] = x[mask,1]
Sturla
John,
Thanks for the quick reply. Unfortunately, no, they're not indexed like that.
The first columns are actually floating-point date numbers from
matplotlib.dates.date2num. Looks like this is just going to be painful...
Thanks for the tip though. That'll definitely be useful elsewhere.
-paul
On Wed, Aug 4, 2010 at 6:59 PM, phob...@geosyntec.com wrote:
Hey folks,
I've one array, x, that you could define as follows:
[[1, 2.25],
[2, 2.50],
[3, 2.25],
[4, 0.00],
[8, 0.00],
[9, 2.75]]
Then my second array, y, is:
[[1, 0.00],
[2, 0.00],
[3, 0.00],
[4, 0.00],
[5,
On Wed, Aug 4, 2010 at 8:00 PM, Gökhan Sever gokhanse...@gmail.com wrote:
On Wed, Aug 4, 2010 at 6:59 PM, phob...@geosyntec.com wrote:
Hey folks,
I've one array, x, that you could define as follows:
[[1, 2.25],
[2, 2.50],
[3, 2.25],
[4, 0.00],
[8, 0.00],
[9, 2.75]]
Then my
How exactly are you looping? That sounds absurdly slow.
What you need is a fast dictionary.
On Wed, Aug 4, 2010 at 6:00 PM, Gökhan Sever gokhanse...@gmail.com wrote:
On Wed, Aug 4, 2010 at 6:59 PM, phob...@geosyntec.com wrote:
Hey folks,
I've one array, x, that you could define as
Perhaps try the following:
1) sort x by x[:,0]
2) sort y by y[:,0]
3) loop through both at the same time building an array of indexes A that
tells you the index of y[i,0] in x or just building a new array z with the
value if you don't need them in order
4) if you do need them in order, unsort A
I've deleted the code b/c it was absurdly slow. It was pretty brute-force.
-Looped through each row (r) of y
-check to see where y[r,0] - x[:,0] eps (call that row r_hit)
-set y[r,1] = x[r_hit,1]
There was kind of a short fuse on this, and I was already reading the data from
a text file. So I
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