Re: Regex trouble
MRAB
> I am trying to loop over a dictionary of phone numbers and using a
> python regex to determine if they are long distance or local and then
> adding them to their appropriate dictionary, My regex doesn't appear to
> be working though.
>
> My regex's are these
>
> international__iregex=r'^1?(011|001)'
> local__iregex=r'^1?(281|832|713|800)'
> #long distance
> ld_regex=r'^1?(281|832|713|800|866|877|011|001|888)'
>
> long_distance= {}
> My loop:
> for key1,value1 in x.items():
> if key1 == 'dest':
> if re.search(ld_regex,value1):
> long_distance[key1] = value1
> print long_distance
>
Define "not working".
BTW, x.items() will give the key/value pairs, but key1 == 'dest' will be
true for only one of the keys, so you're really only looking at
x['dest'].
Please provide a simple though complete example showing the problem,
stating what you expected and what you actually got, including any
tracebacks if an exception occurred.
Simon Forman
> I am trying to loop over a dictionary of phone numbers and using a python
> regex to determine if they are long distance or local and then adding them
> to their appropriate dictionary, My regex doesn't appear to be working
> though.
"doesn't appear to be working" is a useless way to characterize the
problem you're having.
Always tell us what you expected and what you got instead. Preferably
with the actual traceback if any.
Also, post a /runnable/ code fragment. That way we can try it out for ourselves.
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
HTH,
~Simon
>
> My regex's are these
>
>
>
> international__iregex=r'^1?(011|001)' #Anything starting with these
> prefixes is International
>
> local__iregex=r'^1?(281|832|713|800)' #anything starting with these are
> local
>
> #long distance
>
> ld_regex=r'^1?(281|832|713|800|866|877|011|001|888)' #any number not
> starting with these prefixes is long distance
>
>
>
> long_distance= {}
>
> My loop:
>
> for key1,value1 in x.items():
>
> if key1 == 'dest':
>
> if not re.search(ld_regex,value1):
>
> long_distance[key1] = value1
>
> print long_distance
Out of curiosity, why are you printing the whole dict for every "key1
== 'dest'" even when you're not modifying the output dict?
>
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>
Rhodri James
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
Simon Forman
> 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