Hi there,
I have looked back into the way to convert the existing numpy svn
repository into git. It went quite smoothly using svn2git (developed
by the KDE team for their own transition), but there are a few
questions which need to be answered:
- Shall we keep the old svn branches ? I think
Tue, 01 Jun 2010 16:59:47 +0900, David Cournapeau wrote:
I have looked back into the way to convert the existing numpy svn
repository into git. It went quite smoothly using svn2git (developed by
the KDE team for their own transition), but there are a few questions
which need to be answered:
On 06/01/2010 06:03 PM, Pauli Virtanen wrote:
Personally, I don't see problems in leaving them out.
(in maintenance/***)
Why not release/** or releases/**?
Right, release is a better word.
Does having a prefix here imply something to clones?
Not that I am aware of: it is just that / is
Not sure what to call this.
Any suggestion on computing the vector:
sum(u[i*M:i*M+N]) for i in range (len(u)/M)
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On Tue, Jun 1, 2010 at 6:47 AM, Neal Becker ndbeck...@gmail.com wrote:
Not sure what to call this.
Any suggestion on computing the vector:
sum(u[i*M:i*M+N]) for i in range (len(u)/M)
How about a cumsum and then a loop to take the differences of the
desired indices of the cumsum? Might be
On Tue, Jun 1, 2010 at 6:56 AM, Keith Goodman kwgood...@gmail.com wrote:
On Tue, Jun 1, 2010 at 6:47 AM, Neal Becker ndbeck...@gmail.com wrote:
Not sure what to call this.
Any suggestion on computing the vector:
sum(u[i*M:i*M+N]) for i in range (len(u)/M)
How about a cumsum and then a loop
Hello,
I'm not sure if f2py questions are appropriate here, but I have a question.
I had been using f2py without problems, but recently it stopped working.
It's worth mentioning that I'm working remotely on a cluster, and I don't
have root access, so it's possible that the system admins changed
Hi,
I don't think correcting the email addresses in the SVN history is very
useful. Best probably just use some dummy form, maybe
That's what svn2git already does, so that would be less work for me :)
It may not matter much, but I think there is at least one argument for
having real emails:
One can also try to use photometry software like Daophot, it uses
MIDAS by ESO http://www.eso.org/sci/data-processing/software/esomidas//
, which everyone can download.
It seems that Daophot http://www.star.bris.ac.uk/~mbt/daophot/ isn't
for free :-(, I never cared, since it's installed on our
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of latitudes, a list of longitudes, and list of data values.
All lists are the same length.
I want to compute an average of data values for each lat/lon pair. e.g. if
lat[1001] lon[1001] = lat[2001] [lon
On Tue, Jun 1, 2010 at 1:07 PM, Mathew Yeates mat.yea...@gmail.com wrote:
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of latitudes, a list of longitudes, and list of data values.
All lists are the same length.
I want to compute an average of data
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of latitudes, a list of longitudes, and list of data
values. All lists are the same length.
I want to compute an average of data values for each lat/lon pair.
e.g. if lat[1001] lon[1001] = lat[2001]
On Tue, Jun 1, 2010 at 4:49 PM, Zachary Pincus zachary.pin...@yale.edu wrote:
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of latitudes, a list of longitudes, and list of data
values. All lists are the same length.
I want to compute an average of
I guess it's as fast as I'm going to get. I don't really see any other way.
BTW, the lat/lons are integers)
-Mathew
On Tue, Jun 1, 2010 at 1:49 PM, Zachary Pincus zachary.pin...@yale.eduwrote:
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of
I guess it's as fast as I'm going to get. I don't really see any
other way. BTW, the lat/lons are integers)
You could (in c or cython) try a brain-dead hashtable with no
collision detection:
for lat, long, data in dataset:
bin = (lat ^ long) % num_bins
hashtable[bin] =
On Tue, Jun 1, 2010 at 9:57 PM, Zachary Pincus zachary.pin...@yale.edu wrote:
I guess it's as fast as I'm going to get. I don't really see any
other way. BTW, the lat/lons are integers)
You could (in c or cython) try a brain-dead hashtable with no
collision detection:
for lat, long, data in
Subject is a book title from some many years ago, I wonder if it ever
got to Python? I know there were C and Fortran versions.
--
Wayne Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39°
On 2 June 2010 00:33, Wayne Watson sierra_mtnv...@sbcglobal.net wrote:
Subject is a book title from some many years ago, I wonder if it ever
got to Python? I know there were C and Fortran versions.
There is no Numerical Recipes for python. The main reason there isn't
a NR for python is that
On Tue, Jun 1, 2010 at 1:51 PM, Wes McKinney wesmck...@gmail.com wrote:
On Tue, Jun 1, 2010 at 4:49 PM, Zachary Pincus zachary.pin...@yale.edu
wrote:
Hi
Can anyone think of a clever (non-lopping) solution to the following?
A have a list of latitudes, a list of longitudes, and list of data
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