On Thu, Jul 22, 2010 at 7:31 PM, Shashwat Anand
wrote:
>> Are you sure that'll work ? Is creating a dict from a sequence guaranteed
>> to be executed serially withing the sequence ?
>
> It is nowhere mentioned in Docs IIRC but I have never seen any counter
> example.
Actually it is guaranteed.
>F
> Out of curiosity I tried benchmarking few of these solutions alongwith the
> one I wrote using itertools.groupby.
> My benchmarking did include large data sets and worst case.
> Results are the shortest solution (by Anand) was way faster than others.
> After that it was Steve's and mine and Navin
On Thu, Jul 22, 2010 at 10:37 PM, Shekhar Tiwatne wrote:
> On Thursday 22 July 2010 09:36 PM, Anand Balachandran Pillai wrote:
>
>> On Thu, Jul 22, 2010 at 7:00 PM, steve wrote:
>>
>>
>>
>>> Hi,
>>>
>>>
>>> On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
>>>
>>>
>>>
On Thu, Jul 22,
On Thursday 22 July 2010 09:36 PM, Anand Balachandran Pillai wrote:
On Thu, Jul 22, 2010 at 7:00 PM, steve wrote:
Hi,
On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
Suppose you have the following list:
>>>
On Thu, Jul 22, 2010 at 7:00 PM, steve wrote:
> Hi,
>
>
> On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
>
>> On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
>>
>> Suppose you have the following list:
>>>
>>> >>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>>
Hey everyone,
I could'nt find any info on whether or not hack sessions are going to be
conducted as part of Pycon. Did I miss something(in the case that we
*are *having
them) or is there a reason we could'nt have them?
Cheers,
zm
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On Thu, Jul 22, 2010 at 7:00 PM, steve wrote:
> Hi,
>
>
> On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
>
>> On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
>>
>> Suppose you have the following list:
>>>
>>> >>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>>
On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
> Suppose you have the following list:
>
> >>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>
> My problem is that i wish to obtain the following two dictionaries:
> xdictstart = {'cat':10, 'dog':1}
> xdictend = {'cat':30, 'dog
many thanks for this to naveen and others who responded.
On Thu, 22 Jul 2010 15:54:14 +0530 wrote
>I suggest that in such cases, avoid the temptation to do something
clever. If it was difficult for you to write the code, it will be even
more difficult to read and understand it.
Unless there
Hi,
On 07/22/2010 05:02 PM, Anand Balachandran Pillai wrote:
On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
Suppose you have the following list:
>>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
My problem is that i wish to obtain the following two dictionaries:
x
Hackday this weekend, end of july sounds good.
On Thu, Jul 22, 2010 at 2:21 PM, Noufal Ibrahim wrote:
> Anand Balachandran Pillai writes:
>
> [...]
>
>> Can we have it on 30th or 31st then ? Personally I prefer
>> next week-end since I am a bit busy this week-end.
>>
>> Noufal ?
>
> [...]
>
> 3
*Tools Development Engineer*
*Job Location: Bangalore*
* *
*Job Description:*
As a member of the Engineering Software Tools group at Juniper you will be
part of a global team with primary responsibility for the tools
infrastructure of the JUNOS software development environment. You will
desig
On Thu, Jul 15, 2010 at 9:50 AM, Srinivas Reddy Thatiparthy <
srinivas_thatipar...@akebonosoft.com> wrote:
> I read from Python 2.6.3 docs that,copied from chm
>
> " Note that filter(function, iterable) is equivalent to [item for item in
> iterable if function(item)] if function is not None an
On Thu, Jul 22, 2010 at 3:21 PM, Vikram wrote:
> Suppose you have the following list:
>
> >>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>
> My problem is that i wish to obtain the following two dictionaries:
> xdictstart = {'cat':10, 'dog':1}
> xdictend = {'cat':30, 'dog
> Suppose you have the following list:
>
x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>
> My problem is that i wish to obtain the following two dictionaries:
> xdictstart = {'cat':10, 'dog':1}
> xdictend = {'cat':30, 'dog':5}
>
> Any nice way to do the above? Thanks.
Do
>>> for subLi in x:
... if(xdictstart.has_key(subLi[0])):
... if(xdictstart[subLi[0]]>subLi[1]):
...xdictstart[subLi[0]]=subLi[1]
... else:
... xdictstart[subLi[0]]=subLi[1]
The opposite can be done to obtain the xdictend.This is one way.There might
be some easier way's but i'm not aware
On 07/22/2010 03:21 PM, Vikram wrote:
Suppose you have the following list:
x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
My problem is that i wish to obtain the following two dictionaries:
xdictstart = {'cat':10, 'dog':1}
xdictend = {'cat':30, 'dog':5}
I don't get it,
On Thursday, July 22, 2010 03:21:38 pm Vikram wrote:
> Suppose you have the following list:
> >>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
>
> My problem is that i wish to obtain the following two dictionaries:
> xdictstart = {'cat':10, 'dog':1}
> xdictend = {'cat':30, '
I suggest that in such cases, avoid the temptation to do something
clever. If it was difficult for you to write the code, it will be even
more difficult to read and understand it.
Unless there is a very good reason, write the simplest, most braindead code.
xdictstart={}
xdictend={}
for item in sor
Suppose you have the following list:
>>> x =[['cat',10],['cat',20],['cat',30],['dog',5],['dog',1],['dog',3]]
My problem is that i wish to obtain the following two dictionaries:
xdictstart = {'cat':10, 'dog':1}
xdictend = {'cat':30, 'dog':5}
Any nice way to do the above? Thanks.
---
Those in
On Thu, Jul 22, 2010 at 2:57 PM, mujeesh.v.s wrote:
> Hi,
>
> Where you guys are going to meet.Can I join in this meeting
Everyone is welcome to BangPypers meeting. This month's meeting
date is not finalized yet.
Regards,
Baiju M
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Hi,
Where you guys are going to meet.Can I join in this meeting
Mujeesh.v.s
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Anand Balachandran Pillai writes:
[...]
> Can we have it on 30th or 31st then ? Personally I prefer
> next week-end since I am a bit busy this week-end.
>
> Noufal ?
[...]
31st is fine by me. Anyone else?
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On Wed, Jul 21, 2010 at 3:35 PM, Sudheer Satyanarayana <
sudhee...@sudheer.net> wrote:
>
> [...]
>>
>> Yes. Is Sunday fine. Any detractors? Clashes?
>>
>>
>>
> This Saturday and Sunday, I'm attending the Yahoo! Open Hack event -
> http://openhackindia.eventbrite.com/ .
>
>
> It would have been ni
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