On Thu, Sep 24, 2009 at 6:31 PM, Rhodri James
<rho...@wildebst.demon.co.uk> wrote:
> 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
> --
> http://mail.python.org/mailman/listinfo/python-list
>
Wow, I was looking right at it ('1?') and I missed it. I need more sleep.
This is slightly less win:
for key1, value1 in x.iteritems():
if key1 == 'dest':
n = value1[0] == '1'
if value1[n:n + 3] not in local_prefixes:
long_distance[key1] = value1
(MRAB's comment about the foolishness of iterating through the dict
but testing for key == string_literal still applies, of course.)
Depending on what the OP's doing I might recommend just "normalizing"
the phone numbers before processing them, (i.e. stripping off the '1's
and possible breaking them up into tuples ('nnn', 'nnn', 'nnnn').
~Simon
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