From: "Ian Kelly" <ian.g.ke...@gmail.com>
> On 1/2/2011 6:18 AM, Octavian Rasnita wrote: >> Hi, >> >> If I want to create a dictionary from a list, is there a better way than the >> long line below? >> >> l = [1, 2, 3, 4, 5, 6, 7, 'a', 8, 'b'] >> >> d = dict(zip([l[x] for x in range(len(l)) if x %2 == 0], [l[x] for x in >> range(len(l)) if x %2 == 1])) > > d = dict(zip(l[0::2], l[1::2])) > > Or, using the "grouper" function recipe from the itertools documentation: > > d = dict(grouper(2, l)) > > Cheers, > Ian The grouper-way looks nice, but I tried it and it didn't work: from itertools import * ... d = dict(grouper(2, l)) NameError: name 'grouper' is not defined I use Python 2.7. Should it work with this version? Octavian -- http://mail.python.org/mailman/listinfo/python-list
