simple question on dictionary usage

[r.grimm at science-computing.de]

On Oct 27, 6:42 am, Karthik Gurusamy <kar1... at gmail.com> wrote:
> On Oct 26, 9:29 pm, Frank Stutzman <stutz... at skywagon.kjsl.com> wrote:
>
>
>
> > My apologies in advance, I'm new to python
>
> > Say, I have a dictionary that looks like this:
>
> >record={'BAT': '14.4', 'USD': '24', 'DIF': '45', 'OAT': '16',
> >         'FF': '3.9', 'C3': '343', 'E4': '1157', 'C1': '339',
> >         'E6': '1182', 'RPM': '996', 'C6': '311', 'C5': '300',
> >         'C4': '349', 'CLD': '0', 'E5': '1148', 'C2': '329',
> >         'MAP': '15', 'OIL': '167', 'HP': '19', 'E1': '1137',
> >         'MARK': '', 'E3': '1163', 'TIME': '15:43:54',
> >         'E2': '1169'}
>
> > From this dictionary I would like to create another dictionary calld
> > 'egt') that has all of the keys that start with the letter 'E'.  In
> > otherwords it should look like this:
>
> > egt = {'E6': '1182','E1': '1137','E4': '1157','E5': '1148',
> >        'E2': '1169','E3': '1163'}
>
> > This should be pretty easy, but somehow with all my googling I've
> > not found a hint.
>
> One possible solution (read list-comprehension if you not familiar
> with it):
>
> >>>record={'BAT': '14.4', 'USD': '24', 'DIF': '45', 'OAT': '16',
>
> ...         'FF': '3.9', 'C3': '343', 'E4': '1157', 'C1': '339',
> ...         'E6': '1182', 'RPM': '996', 'C6': '311', 'C5': '300',
> ...         'C4': '349', 'CLD': '0', 'E5': '1148', 'C2': '329',
> ...         'MAP': '15', 'OIL': '167', 'HP': '19', 'E1': '1137',
> ...         'MARK': '', 'E3': '1163', 'TIME': '15:43:54',
> ...         'E2': '1169'}>>> egt =dict([(k,record[k]) for k inrecordif k.startswith('E')])
> >>> egt
>
> {'E5': '1148', 'E4': '1157', 'E6': '1182', 'E1': '1137', 'E3': '1163',
> 'E2': '1169'}
>
> Karthik
>
>
>
> > Thanks in advance
>
> > --
> > Frank Stutzman

Hallo,
a functional and concise way.

egt= dict( filter( lambda item: item[0][0] == "E" ,
record.iteritems() ))

Rainer