On Thu, 24 Sep 2009 19:45:37 +0100, Simon Forman <sajmik...@gmail.com>
wrote:
FWIW this problem is too simple (IMHO) for regular expressions.
Simply carve off the first three digits and check against sets of the
prefixes you're interested in:
#any number starting with these prefixes is not long distance
local_prefixes = set(['832', '877', '888', '713', '866', '011', '001',
'281', '800'])
long_distance= {}
for key1, value1 in x.iteritems():
if key1 == 'dest':
if value1[:3] not in local_prefixes:
long_distance[key1] = value1
You need to allow for the potential leading 1, which can be done with
a bit of straightforward string slicing but still puts REs more
sensibly in the running.
--
Rhodri James *-* Wildebeest Herder to the Masses
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