"Elena Schulz" <elena.schulz@gmx.net> wrote in
news:mailman.799.1069086944.702.python-list@python.org:
[color=blue]
> I've the following code:
>
> myList = []
> for i in range(3) :
> myDict['id']=i
> myList.append(myDict)
>
> myList becomes: [{'id':2}, {'id':2}, {'id':2}] but I want: [{'id':0},
> {'id':1}, {'id':2}]
>
> thanx for any hint how to achieve that[/color]
Well, firstly I would say that you don't have the code you posted, because
if you did you would be getting:
NameError: name 'myDict' is not defined
Anyway, if you want your list to contain three different dictionaries
instead of three references to the same dictionary, then you have to create
a new dictionary each time. This should do for your example:
myList = []
for i in range(3):
myList.append({'id':i})
Remember that Python never, ever, makes a copy of an object unless you
explicitly tell it to make a copy. Usually all it does is shuffle around
references to the same objects.
If you had a load of other values in your dictionary, and wanted to have
different dictionaries with just the id field different, you could do
something like:
myDict = { 'a': 1, 'b': 2 }
myList = []
for i in range(3):
myDict['id'] = i
myList.append(myDict.copy())
--
Duncan Booth
duncan@rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?
