On Mar 4, 2008, at 11:04 AM, Kent Johnson wrote:
> Eric Abrahamsen wrote:
>> Itertools.groupby is totally impenetrable to me
>
> Maybe this will help:
> http://personalpages.tds.net/~kent37/blog/arch_m1_2005_12.html#e69
>
> Kent
>
It did! Thanks very much. I think I understand now what's going o
Eric Abrahamsen wrote:
> Itertools.groupby is totally impenetrable to me
Maybe this will help:
http://personalpages.tds.net/~kent37/blog/arch_m1_2005_12.html#e69
Kent
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Well I expected to learn a thing or two, but I didn't expect not to
understand the suggestions at all! :) Thanks to everyone who
responded, and sorry for the typo (it was meant to be object_id
throughout, not content_type).
So far Michael's solution works and is most comprehensible to me – i
On 04/03/2008, Shuai Jiang (Runiteking1) <[EMAIL PROTECTED]> wrote:
> Hello, I'm trying to create an application that retrieves and displays
> (probably in HTML or PDF format) math problems from a database.
> The problem is that I need some sort of mechanism to display mathematical
> equations.
If
Hello, I'm trying to create an application that retrieves and displays
(probably in HTML or PDF format) math problems from a database.
The problem is that I need some sort of mechanism to display mathematical
equations.
I'm trying to not use Latex as it would cause the users to install Latex
(and
Varsha Purohit wrote:
> Yeahh so by doing this i am counting only the difference part since we
> have grayscaled the image and assuming it will count only the pixels
> that evolve as difference
Yes
> if i use sum2 instead of sum i think it
> will give squared sum which is area... and if
Yeahh so by doing this i am counting only the difference part since we have
grayscaled the image and assuming it will count only the pixels that evolve
as difference if i use sum2 instead of sum i think it will give squared
sum which is area... and if i just use count it would count the number
Hello Eric,
Your basic outlook is fine, but you can do it much more efficiently
with a single sort. Here's the way I'd approach the task (untested):
# --
# first compute the latest date for each id group; uses O(n) time
newest = {}
f
Chris Fuller wrote:
> You could have a hierarchical sort
> function:
>
> def hiersort(a,b):
>if a.attr1 != b.attr1:
> return cmp(a.attr1, b.attr1)
>else:
> if a.attr2 != b.attr2:
> return cmp(a.attr2, b.attr2)
> else:
> return cmp(a.attr3, b.att3)
>
>
Almost. And better than my original idea. You could have a hierarchical sort
function:
def hiersort(a,b):
if a.attr1 != b.attr1:
return cmp(a.attr1, b.attr1)
else:
if a.attr2 != b.attr2:
return cmp(a.attr2, b.attr2)
else:
return cmp(a.attr3, b.att3)
l
Andreas Kostyrka wrote:
> l.sort(key=lambda x: (x.content_type, x.submit_date))
>
> Now, you can construct a sorted list "t":
>
> t = []
> for key, item_iterator in itertools.groupby(l, key=lambda x: (x.content_type,
> x.submit_date)):
> sorted_part = sorted(item_iterator, key=lambda x: x.s
Eric Abrahamsen wrote:
> I have a list of objects, each of which has two attributes, object_id
> and submit_date. What I want is to sort them by content_type, then by
> submit_date within content_type, and then sort each content_type block
> according to which block has the newest object by
Well, this assumes that all named attributes do exist. If not, you need
to replace x.attr with getattr(x, "attr", defaultvalue) ;)
l.sort(key=lambda x: (x.content_type, x.submit_date))
Now, you can construct a sorted list "t":
t = []
for key, item_iterator in itertools.groupby(l, key=lambda x:
I have a grisly little sorting problem to which I've hacked together a
solution, but I'm hoping someone here might have a better suggestion.
I have a list of objects, each of which has two attributes, object_id
and submit_date. What I want is to sort them by content_type, then by
submit_date
Well, actually, ssh can also protect private keys with a cryptographic
pass phrase. But this is often not what is wanted as it implies that the
user needs to provide the pass phrase every time it is used. (Well,
that's not the complete truth, man ssh-agent, but that's completely
different thing
And if it's a string constant, in many cases running strings (Unix
program) on the pyc file will reveal it too.
All this basically turns down to the problem, that it's hard to embed an
encryption key in a program, so that it's not possible to extract it.
Notice the the current crop of HDDVD/Blu
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