Hello
-------- Original Message --------
From: Kent Johnson <ken...@tds.net>
Apparently from: kent3...@gmail.com
To: davidwil...@safe-mail.net
Cc: tutor@python.org
Subject: Re: [Tutor] dictionaries help
Date: Thu, 23 Jul 2009 21:44:33 -0400
> On Thu, Jul 23, 2009 at 7:09 PM, <davidwil...@safe-mail.net> wrote:
> > Hello again,
> > Here is my full attempt, can you please tell me if it is OK and if there
> > should be any improvements
> >
> >>>> ws_industry = ('it', 'science')
> >>>> code = ws_industry[0]
> >>>> active = []
> >>>> industries = [{'code': 'it', 'name': 'Information Technology'}, {'code':
> >>>> 'science', 'name': 'Science'}, {'code': 'biology', 'name': 'Biology'}]
> >>>> for industry in industries:
> > if industry['code'] in ws_industry:
> > active.append({
> > 'code': industry['code'],
> > 'name': industry['name'],
> > 'isdefault': code == industry['code']})
> >
> >>>># Not active industry
> >>>> not_active = [x for x in industries if x['code'] not in ws_industry ]
> >>>>not_active.sort(lambda x, y: cmp(x['name'], y['name']))
> >>>> return [active, non-active]
> >
> > Line [6] i was not too sure if it could be written better or more efficient.
> >
> > Basically I have a big list of industries and I want to verify for each
> > member whether they are active or not active.
> >
> > The industry list of dictionaries will stay fixed at about 25 items,
> > whereas the ws_industry tuple will change depending. I have to go through
> > about 20,000 items.
>
> If ws_industry contains 25 items, it will probably be faster to search
> it if it is a set rather than a list. Just use
> ws_industry = set(('it', 'science'))
ws_industry contains only unique values
>
> Which part of the problem has 20,000 items?
the 20,000 items are the number of times the programme will have to loop to
extract this information.
so i have 25 industries, i also have 20,000 text files which contain the
ws_industry values therefore i would like to extract from each file the number
of active industries and number of inactive industries.
i suppose in a relational db i will have something like:
[file] 1 - n [ws-industry] n - n [industry]
hope this make sense ;)
>
> Kent
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
