I have installed the 32 bit 2.6 and the 64 bit 3.1 on my machine
running Vista 64 without any issues.
Which of course has nothing to do with your problem
On Thu, Jul 23, 2009 at 7:06 PM, DwBear75dwbea...@gmail.com wrote:
I just downloaded and attempted to install python 2.6.2. The
On Tue, Jun 23, 2009 at 4:45 PM, Cameron
Pulsfordcameron.pulsf...@gmail.com wrote:
Hey all, I have a dictionary that looks like this (small example version)
{(1, 2): 0} named a
so I can do a[1,2] which returns 0. What I also have is a list of
coordinates into a 2 dimensional array that might
On Wed, Jun 17, 2009 at 1:52 PM, Mark Dickinsondicki...@gmail.com wrote:
Maybe James is thinking of the standard theorem
that says that if a sequence of continuous functions
on an interval converges uniformly then its limit
is continuous?
Jaime was simply plain wrong... The example that
On Wed, Jun 17, 2009 at 4:50 AM, Lawrence
D'Oliveirol...@geek-central.gen.new_zealand wrote:
In message 7x63ew3uo9@ruckus.brouhaha.com, wrote:
Lawrence D'Oliveiro l...@geek-central.gen.new_zealand writes:
I don't think any countable set, even a countably-infinite set, can have
a fractal
On Sun, Jun 14, 2009 at 4:27 PM, Aaron Bradycastiro...@gmail.com wrote:
Before I go and flesh out the entire interfaces for the provided
types, does anyone have a use for it?
A real-world application of persistent data structures can be found here:
http://stevekrenzel.com/persistent-list
On Tue, Jun 9, 2009 at 5:38 PM,
stephen_bredplusbluemakespur...@gmail.com wrote:
I'd like to convert a list of floats to a list of strings constrained
to one .1f format. These don't work. Is there a better way?
[.1f % i for i in l]
or
[(.1f % i) for i in l]
There's a missing %, this does
On Tue, May 26, 2009 at 8:42 PM, Sumitava Mukherjee
sm...@cognobytes.com wrote:
On May 26, 11:39 pm, Sumitava Mukherjee sm...@cognobytes.com wrote:
[Oh, I forgot to mention. I am looking for sampling without replacement.]
That means that, you'll have to code something that updates the sums
of
Hi Dhananjay,
Sort has several optional arguments, the function's signature is as follows:
s.sort([cmp[, key[, reverse]]])
If you store your data as a list of lists, to sort by the third column
you could do something like:
data.sort(None, lambda x : x[2])
For more complex sortings, as the one
On Mon, May 25, 2009 at 10:15 AM, Chris Rebert c...@rebertia.com wrote:
Erm, using a compare function rather than a key function slows down
sorting /significantly/. In fact, the `cmp` parameter to list.sort()
has been *removed* in Python 3.0 because of this.
Makes a lot of sense, as you only
These weekend I've been tearing down to pieces Michele Simionato's
decorator module, that builds signature preserving decorators. At the
heart of it all there is a dynamically generated function, which works
something similar to this...
...
src = def function(a,b,c) :\nreturn
map is creating a new list of 10,000,000 items, not modifying the
values inside list a. That's probably where your time difference comes
from...
Jaime
On Fri, May 15, 2009 at 1:56 PM, Gediminas Kregzde
gediminas.kreg...@gmail.com wrote:
Hello,
I'm Vilnius college II degree student and last
(1) building another throwaway list and
(2) function call overhead for calling doit()
You can avoid (1) by using filter() instead of map()
Are you sure of this? My python returns, when asked for help(filter) :
Help on built-in function filter in module __builtin__:
filter(...)
, May 13, 2009 at 10:49 AM, godshorse chinthak...@gmail.com wrote:
On May 13, 3:19 pm, Jaime Fernandez del Rio jaime.f...@gmail.com
wrote:
Dijkstra's algorithm computes shortest paths between a node and _ALL_
other nodes in the graph. It is usually stopped once computing the
shortest path
Dijkstra's algorithm computes shortest paths between a node and _ALL_
other nodes in the graph. It is usually stopped once computing the
shortest path to the target node is done, but that's simply for
efficiency, not a limitation of the algorithm. So you should be able
to tweak the code you are
This one I think I know... Try with:
for k in sorted(word_count) :
print k,=,word_count[k]
You need to do the sorting before iterating over the keys...
Jaime
On Tue, May 12, 2009 at 1:54 PM, Ronn Ross ronn.r...@gmail.com wrote:
I'm attempting to sort for the results of a dictionary. I
If you simply want to generate an ordered list of months, start with
it in order:
dates = [x_jan,...,x_dec]
and if the desired starting month is
start = 6 # i.e. x_jun
dates = dates[start - 1:] + dates[:start - 1]
If you have to sort the list itself, I would use an intermediate
On Tue, May 12, 2009 at 5:02 PM, MRAB goo...@mrabarnett.plus.com wrote:
John Machin wrote:
MRAB google at mrabarnett.plus.com writes:
Sort the list, passing a function as the 'key' argument. The function
should return an integer for the month, eg 0 for 'jan', 1 for 'feb'. If
you want to
17 matches
Mail list logo
