I don't use it often in Vista, but I haven't had any issues. Matter-of-
fact, some things seem nicer in Vista... for instance it resets IDLE
whenever I rerun a module.
Mchizi_Crazy wrote:
> Please help with issue... I heard of compatimbiltity issues and would
> like clarification.
--
http://mail.p
Thank you for all your responses. I've tried the permutations road
(thank you to all those of you who have suggested it) and it takes %*&
%^ long :-) As expected. I've solved it a different way, which runs
through the 26 spots by just adding one at a time if available. Still
takes a long time, but
I'm wondering how to do this the most elegant way: I found this quiz
in some magazine. I've already solved it on paper, but want to write a
python program to solve it. It comes down to being able to represent
range(1,27) through a number of formulas. How do I write a loop that
will loop through thi
Great, I think that's exactly what I'm after. Thank you!
Simon Brunning wrote:
2008/8/3 ToshiBoy <[EMAIL PROTECTED]>:
Currently, I'm using iMacro, an add-on to Firefox, which runs a macro
and enters all the info. It's great, but I would like to try and write
a prog
I'm a newbie to Python... well a newbie to programming, really. I know
the basics and try to learn by setting myself simple tasks and goals
just to find out if I can work out a way to code the solutions. Works
for me. However, now I've set my eyes on a more ambitious project:
We sell office machin
On Jun 28, 2:48 pm, Mel <[EMAIL PROTECTED]> wrote:
> ToshiBoy wrote:
> > I have two lists A and B that are both defined as range(1,27) I want
> > to find the entries that are valid for A = BxB
> [ ... ]
> > I get, as expected 1,4,9,16,25 printed out being the o
I am a newbie... and the first to admit it... but this has me stuffed:
I have two lists A and B that are both defined as range(1,27) I want
to find the entries that are valid for A = BxB
so here is my code:
A = range(1,27)
B = range(1,27)
for b in B:
if b*b in A:
print b
