/3.1/
> [2] http://www.python.org/getit/releases/2.7.3/
> [3] http://dl.acm.org/citation.cfm?id=645413.652131
>
> On Sat, May 12, 2012 at 10:17 PM, Jean-Daniel
> wrote:
>>
>> Hello,
>>
>> I have a long list of n date intervals that gets added or suppressed
&
it fast is to have done some preprocessing at
insertion time, so that not all intervals are processed at query time.
On Sat, May 12, 2012 at 2:30 PM, Karl Knechtel wrote:
> On Sat, May 12, 2012 at 8:17 AM, Jean-Daniel
> wrote:
>> I am looking for a fast way to find the intervals
&
Hello,
I have a long list of n date intervals that gets added or suppressed
intervals regularly. I am looking for a fast way to find the intervals
containing a given date, without having to check all intervals (less
than O(n)).
Do you know the best way to do this in Python with the stdlib?
A var
On Sat, Feb 12, 2011 at 7:25 PM, Peter Otten <__pete...@web.de> wrote:
> Jean-Daniel wrote:
>
>> Hello,
>>
>> I am writing a small framework where the user which writes a function
>> can expect some global variable to be set in the function namespace.
>>
Hello,
I am writing a small framework where the user which writes a function
can expect some global variable to be set in the function namespace.
The user has to write a function like this:
"""
# function.py
from framework import, command, run
@command
def myfunc():
print HOST
if __name__==
Hello,
I live in Paris, my roommate and I would gladly host a poor soul
blocked at the airport due to the ash cloud.
See me for details,
Cheers,
PS: disambiguation: talking about real physical cloud here... :)
--
http://mail.python.org/mailman/listinfo/python-list
Maybe the distutils list is more adapted for this question:
The Zope community uses zc.sourcerelease to build rpm
http://www.mail-archive.com/distutils-...@python.org/msg06599.html
Buildout is said to have undocumented features to build packages.
Tarek Ziade is working debian package with 'distr
Building on the answers of the others, a simple one liner, no side
effect, not the fastest I guess:
>>> d={'a': 'bob', 'b': 'stu'}
>>> set( d.keys() ).difference( [ 'a' ] ).pop()
'b'
Note the square brackets for the parameter of difference(). 'The
string 'a' and the list [ 'a' ] are both iterable
