Re: [Tutor] using re to build dictionary
Thank you all, this is great.
Kent Johnson wrote:
On Tue, Feb 24, 2009 at 6:48 AM, Norman Khine wrote:
Hello,
From my previous post on create dictionary from csv, i have broken the
problem further and wanted the lists feedback if it could be done better:
s = 'Association of British Travel Agents (ABTA) No. 56542\nAir Travel
Organisation Licence (ATOL)\nAppointed Agents of IATA (IATA)\nIncentive
Travel & Meet. Association (ITMA)'
licences = re.split("\n+", s)
licence_list = [re.split("\((\w+)\)", licence) for licence in licences]
This is awkward. You can match directly on what you want:
In [7]: import re
In [8]: s = 'Association of British Travel Agents (ABTA) No.
56542\nAir Travel Organisation Licence (ATOL)\nAppointed Agents of
IATA (IATA)\nIncentive Travel & Meet. Association (ITMA)'
In [9]: licenses = re.split("\n+", s)
In [10]: licenseRe = re.compile(r'\(([A-Z]+)\)( No. (\d+))?')
In [11]: for license in licenses:
: m = licenseRe.search(license)
: print m.group(1, 3)
('ABTA', '56542')
('ATOL', None)
('IATA', None)
('ITMA', None)
Kent
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Re: [Tutor] using re to build dictionary
On Tue, Feb 24, 2009 at 6:48 AM, Norman Khine wrote:
> Hello,
> From my previous post on create dictionary from csv, i have broken the
> problem further and wanted the lists feedback if it could be done better:
>
s = 'Association of British Travel Agents (ABTA) No. 56542\nAir Travel
Organisation Licence (ATOL)\nAppointed Agents of IATA (IATA)\nIncentive
Travel & Meet. Association (ITMA)'
licences = re.split("\n+", s)
licence_list = [re.split("\((\w+)\)", licence) for licence in licences]
This is awkward. You can match directly on what you want:
In [7]: import re
In [8]: s = 'Association of British Travel Agents (ABTA) No.
56542\nAir Travel Organisation Licence (ATOL)\nAppointed Agents of
IATA (IATA)\nIncentive Travel & Meet. Association (ITMA)'
In [9]: licenses = re.split("\n+", s)
In [10]: licenseRe = re.compile(r'\(([A-Z]+)\)( No. (\d+))?')
In [11]: for license in licenses:
: m = licenseRe.search(license)
: print m.group(1, 3)
('ABTA', '56542')
('ATOL', None)
('IATA', None)
('ITMA', None)
Kent
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Re: [Tutor] using re to build dictionary
Le Tue, 24 Feb 2009 12:48:51 +0100,
Norman Khine s'exprima ainsi:
> Hello,
> From my previous post on create dictionary from csv, i have broken the
> problem further and wanted the lists feedback if it could be done better:
>
> >>> s = 'Association of British Travel Agents (ABTA) No. 56542\nAir
> Travel Organisation Licence (ATOL)\nAppointed Agents of IATA
> (IATA)\nIncentive Travel & Meet. Association (ITMA)'
> >>> licences = re.split("\n+", s)
> >>> licence_list = [re.split("\((\w+)\)", licence) for licence in licences]
> >>> association = []
> >>> for x in licence_list:
> ... for y in x:
> ... if y.isupper():
> ...association.append(y)
> ...
> >>> association
> ['ABTA', 'ATOL', 'IATA', 'ITMA']
>
>
> In my string 's', I have 'No. 56542', how would I extract the '56542'
> and map it against the 'ABTA' so that I can have a dictionary for example:
>
> >>> my_dictionary = {'ABTA': '56542', 'ATOL': '', 'IATA': '', 'ITMA': ''}
> >>>
>
>
> Here is what I have so far:
>
> >>> my_dictionary = {}
>
> >>> for x in licence_list:
> ... for y in x:
> ... if y.isupper():
> ... my_dictionary[y] = y
> ...
> >>> my_dictionary
> {'ABTA': 'ABTA', 'IATA': 'IATA', 'ITMA': 'ITMA', 'ATOL': 'ATOL'}
>
> This is wrong as the values should be the 'decimal' i.e. 56542 that is
> in the licence_list.
>
> here is where I miss the point as in my licence_list, not all items have
> a code, all but one are empty, for my usecase, I still need to create
> the dictionary so that it is in the form:
>
> >>> my_dictionary = {'ABTA': '56542', 'ATOL': '', 'IATA': '', 'ITMA': ''}
>
> Any advise much appreciated.
>
> Norman
I had a similar problem once. The nice solution was -- I think, don't take this
for granted I have no time to verify -- simply using multiple group with
re.findall again. Build a rule like:
r'.+(code-pattern).+(number_pattern).+\n+'
Then the results will be a list of tuples like
[
(code1, n1),
(code2, n2),
...
]
where some numbers will be missing. from this it's straightforward to
instantiate a dict, maybe using a default None value for n/a numbers. Someone
will probably infirm or confirm this method.
denis
--
la vita e estrany
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[Tutor] using re to build dictionary
Hello,
From my previous post on create dictionary from csv, i have broken the
problem further and wanted the lists feedback if it could be done better:
>>> s = 'Association of British Travel Agents (ABTA) No. 56542\nAir
Travel Organisation Licence (ATOL)\nAppointed Agents of IATA
(IATA)\nIncentive Travel & Meet. Association (ITMA)'
>>> licences = re.split("\n+", s)
>>> licence_list = [re.split("\((\w+)\)", licence) for licence in licences]
>>> association = []
>>> for x in licence_list:
... for y in x:
... if y.isupper():
...association.append(y)
...
>>> association
['ABTA', 'ATOL', 'IATA', 'ITMA']
In my string 's', I have 'No. 56542', how would I extract the '56542'
and map it against the 'ABTA' so that I can have a dictionary for example:
>>> my_dictionary = {'ABTA': '56542', 'ATOL': '', 'IATA': '', 'ITMA': ''}
>>>
Here is what I have so far:
>>> my_dictionary = {}
>>> for x in licence_list:
... for y in x:
... if y.isupper():
... my_dictionary[y] = y
...
>>> my_dictionary
{'ABTA': 'ABTA', 'IATA': 'IATA', 'ITMA': 'ITMA', 'ATOL': 'ATOL'}
This is wrong as the values should be the 'decimal' i.e. 56542 that is
in the licence_list.
here is where I miss the point as in my licence_list, not all items have
a code, all but one are empty, for my usecase, I still need to create
the dictionary so that it is in the form:
>>> my_dictionary = {'ABTA': '56542', 'ATOL': '', 'IATA': '', 'ITMA': ''}
Any advise much appreciated.
Norman
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