greg whittier wrote:
On Fri, Mar 13, 2009 at 11:09 AM, ski wrote:
Hello,
Here is what I have so far:
mylist = [{'index': 0, 'title': 'Association of British Travel Agents',
'selected': False, 'edit_row': '?edit_affiliation=0', &#x
Hello,
Here is what I have so far:
>>> mylist = [{'index': 0, 'title': 'Association of British Travel
Agents', 'selected': False, 'edit_row': '?edit_affiliation=0',
'affiliation': 'ABTA', 'affiliation_no': u'G3903'}, {'index': 1,
'title': 'Appointed Agents of IATA', 'selected': False, 'edit_ro
how would you do this for a specific key instead of all the keys?
greg whittier wrote:
On Thu, Mar 12, 2009 at 4:24 PM, ski wrote:
Hello,
I have this issue, which I am unsure on how to solve.
mylist1 = {'a': 'x123', 'b':'12'}
mylist2 = {'a':
Hello,
I have this issue, which I am unsure on how to solve.
>>> mylist1 = {'a': 'x123', 'b':'12'}
>>> mylist2 = {'a': 'x234', 'c': 'a23'}
>>> for k in mylist2:
... if k in mylist1:
... mylist1[k] = [mylist1[k], mylist2[k]]
... else:
... mylist1[k] = mylist2[k]
...
i was looking to list all the items in list1 that are not in list2 based
on the key['id] and key['affiliation'] respectively.
thanks
Kent Johnson wrote:
On Mon, Mar 9, 2009 at 4:20 PM, ski wrote:
what am i missing here, i would like to get a list of the diff between
hello again,
i am trying to get my head around this but without success, here is what
i have so far:
>>> list1 = [{'is_selected': False, 'id': 'AAC', 'title': 'Association
of Airline Cons.'}, {'is_selected': False, 'id': 'AALA', 'title': 'Adv.
Activity Licence. Auth.'}, {'is_selected': False,
hello,
i have created this function that compares the values of a csv file
against a dictionary, you can see the code, here
http://paste.lisp.org/display/76705
the affiliations list has 130 items.
is there a better way to build the 'items' list?
thanks
norman
___
thank you all, great feedback.
Sander Sweers wrote:
2009/3/9 A.T.Hofkamp :
You can do something like
[d for d in d1 if d['is_selected'] == False]
to get your dicts.
If 'is_selected' is not always present, it gets a bit more complicated, I'll
leave that as an exercise for the interested reader
hello,
i have this list which contains a number of dictionaries.
>>>d1 = [{'is_selected': False, 'id': 'AAC', 'title': 'Association of
Airline Cons.'}, {'is_selected': False, 'id': 'AALA', 'title': 'Adv.
Activity Licence. Auth.'}, {'is_selected': False, 'id': 'ABPCO',
'title': 'Association of
Hello,
I have this:
>>> import re
>>> s = "Association of British Travel Agents (ABTA) No. 56542\nAir
Travel Organisation Licence (ATOL)\nAppointed Agents of IATA
(IATA)\nIncentive Travel & Meet. Association (ITMA)"
>>> licenses = re.split("\n+", s)
>>> licenseRe = re.compile(r'\(([A-Z]+)\)( N
10 matches
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