Re: [Tutor] update list of dictionaries based on key

[ski]

greg whittier wrote:
On Fri, Mar 13, 2009 at 11:09 AM, ski <nor...@khine.net> 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', 'affiliation': 'ABTA',
'affiliation_no': u'G3903'}, {'index': 1, 'title': 'Appointed Agents of
IATA', 'selected': False, 'edit_row': '?edit_affiliation=1', 'affiliation':
'IATA', 'affiliation_no': u'none'}, {'index': 0, 'title': 'Association of
Airline Cons.', 'selected': False, 'edit_row': '?edit_affiliation=0',
'affiliation': 'AAC', 'affiliation_no': u'sss'}, {'index': 1, 'title':
'Association of British Travel Agents', 'selected': False, 'edit_row':
'?edit_affiliation=1', 'affiliation': 'ABTA', 'affiliation_no': u'zser'}]
key = 'affiliation_no'
my_dup_keys = []
merged_list = []
mykeys = [item['affiliation'] for item in mylist]
for x in mykeys:
       if mykeys.count(x) >= 2:
               my_dup_keys.append(x)

for item in mylist:
       if item['affiliation'] in set(my_dup_keys):
               merged_list.append(item)


values = [d[key] for d in merged_list if d.has_key(key)]
for dict in merged_list:
       dict[key] = values
       mylist.append(dict)

print mylist
[{'index': 0, 'title': 'Association of British Travel Agents', 'selected':
False, 'edit_row': '?edit_affiliation=0', 'affiliation': 'ABTA',
'affiliation_no': [u'G3903', u'zser']}, {'index': 1, 'title': 'Appointed
Agents of IATA', 'selected': False, 'edit_row': '?edit_affiliation=1',
'affiliation': 'IATA', 'affiliation_no': u'none'}, {'index': 0, 'title':
'Association of Airline Cons.', 'selected': False, 'edit_row':
'?edit_affiliation=0', 'affiliation': 'AAC', 'affiliation_no': u'sss'},
{'index': 1, 'title': 'Association of British Travel Agents', 'selected':
False, 'edit_row': '?edit_affiliation=1', 'affiliation': 'ABTA',
'affiliation_no': [u'G3903', u'zser']}, {'index': 0, 'title': 'Association
of British Travel Agents', 'selected': False, 'edit_row':
'?edit_affiliation=0', 'affiliation': 'ABTA', 'affiliation_no': [u'G3903',
u'zser']}, {'index': 1, 'title': 'Association of British Travel Agents',
'selected': False, 'edit_row': '?edit_affiliation=1', 'affiliation': 'ABTA',
'affiliation_no': [u'G3903', u'zser']}]


This sort of works but I want to return a list that updates 'mylist' not
append to it, so I will get:

[{'index': 0, 'title': 'Association of British Travel Agents', 'selected':
False, 'edit_row': '?edit_affiliation=0', 'affiliation': 'ABTA',
'affiliation_no': [u'G3903', u'zser']}, {'index': 1, 'title': 'Appointed
Agents of IATA', 'selected': False, 'edit_row': '?edit_affiliation=1',
'affiliation': 'IATA', 'affiliation_no': u'none'}, {'index': 0, 'title':
'Association of Airline Cons.', 'selected': False, 'edit_row':
'?edit_affiliation=0', 'affiliation': 'AAC', 'affiliation_no': u'sss'}]


It's a little hard to figure what you're getting at.  This looks like
a table represented by a list of dictionaries that you'd like to group
by "affiliation."

i have objects in my database, called addresses, each address has a 
metadata called affiliation.

also each address is a member of a company.

in my company class, i wanted to get all the addresses affiliations and 
only list the unique values.



affiliation_nos = {}
for row in mylist:
    
affiliation_nos.set_default(row['affiliation'],[]).append(row['affliation_no'])

will give you a list of affiliation_no's for each affiliation and if
you want a list of dicts, you could do

mynewlist = [dict(affiliation=affiliation, affiliation_nos =
affiliation_nos[affiliation]) for affiliation in
affiliation_nos.keys()]

I don't know what to do about all the other fields though.  In your
example input list you have two dictionaries with affiliation =
'ABTA'.  In your output you kept the one with index=0 and threw away
index=1.  (same for 'edit_row')  How do you determine which to keep?

I want to keep it based on the 'affiliation' key.


In essence, from my original 'mylist' I want to keep all the list items 
as they are, but just update the list items that have more than one 
'affiliation' by updating only the 'affiliation_no' to contain both entries.

Perhaps there is a simpler way to do this ;)

I will change my code so that mylist does not contain the 'index' and 
'edit_row' keys. Perhaps then it will be easier to update the dictionary?




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_______________________________________________
Tutor maillist  -  Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor