On Oct 27, 6:42 am, Karthik Gurusamy <[EMAIL PROTECTED]> wrote: > On Oct 26, 9:29 pm, Frank Stutzman <[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list