Help cleaning up some code

[odeits]

On Mar 8, 4:48 am, "andrew cooke" <and... at acooke.org> wrote:
> odeits wrote:
> > On Mar 7, 1:07 pm, Scott David Daniels <Scott.Dani... at Acm.Org> wrote:
> >> odeits wrote:
> >> > I am looking to clean up this code... any help is much appreciated.
> >> > Note: It works just fine, I just think it could be done cleaner.
>
> >> > The result is a stack of dictionaries. the query returns up to
> >> > STACK_SIZE ads for a user. The check which i think is very ugly is
> >> > putting another contraint saying that all of the ni have to be the
> >> > same.
>
> >> Well, the obvious way to get your constraint is by changing your SQL,
> >> but if you are going to do it by fetching rows, try:
>
> >>      FIELDS = 'ni adid rundateid rundate city state status'.split()
> >>      ni = UNSET = object() # use None unless None might be the value
> >>      stack = []
> >>      rows = self.con.execute(adquerystring,
> >> (user,STACK_SIZE)).fetchall()
> >>      for row in rows:
> >>          ad = dict()
> >>          for field in FIELDS:
> >>              ad[field] = row[field]
> >>          for field in 'city', 'state':
> >>              if ad[field] is None:
> >>                  ad[field] = 'None'
> >>          if ni != ad['ni']:
> >>              if ni is UNSET:
> >>                  ni = ad['ni']
> >>              else:
> >>                  break
> >>          stack.append(ad)
>
> >> --Scott David Daniels
> >> Scott.Dani... at Acm.Org
>
> > Taking from several suggestions this is what i have come up with for
> > now:
>
> >          for row in  ifilter(lambda r: r['ni'] == rows[0]['ni'],rows):
>
> not sure what version of python you're using, but it would be more natural
> in recent python to write that as:
>
>          for row in (r for r in rows if r['ni'] == rows[0]['ni']):
>
> (the () create a generator for you).
>
> andrew
>
> >             ad = dict()
>
> >             keys = row.keys() # if python 2.6
> >             keys =
> > ['ni','adid','rundateid','rundate','city','state','status'] # if
> > python 2.5
>
> >             for index in row.keys():
> >                 if row[index] is None:
> >                     ad[index] = 'None'
> >                 else:
> >                     ad[index] = row[index]
> >             stack.append(ad)
> >             print row
>
> > the test to see if the ad is valid is placed in the ifilter so that I
> > dont build the dictionary unnecessarily. and the None special case is
> > fairly simple to read now. The None case would even be irrelevant if i
> > could get the damn xmlrpc to allow null. sigh. anyhow. thanks for all
> > of your input, it is definitely better than it was ;)
> > --
> >http://mail.python.org/mailman/listinfo/python-list

This function is very time critical so i went with itertools "for
efficient looping" and i am runing 2.5