Pablo Englebienne wrote:
Hi, I'm trying to work with a dictionary of dictionaries and I'm having
trouble accessing a specific element of it:
$ python
Python 2.6 (trunk:66714:66715M, Oct 1 2008, 18:36:04)
[GCC 4.0.1 (Apple Computer, Inc. build 5370)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> r = ('a','b','c')
>>> c = (1,2,3)
>>> d = dict.fromkeys(r,dict.fromkeys(c))
>>> d
This does
d0 = dict.fromkeys(c)
d = dict.fromkeys(r, d0)
{'a': {1: None, 2: None, 3: None}, 'c': {1: None, 2: None, 3: None},
'b': {1: None, 2: None, 3: None}}
>>> d['a'][1]=0
>>> d
{'a': {1: 0, 2: None, 3: None}, 'c': {1: 0, 2: None, 3: None}, 'b': {1:
0, 2: None, 3: None}}
>>> import copy
As you can see, attempting to assign a specific member, d['a'][1]
updates all values in d[*][1], not what I intended.
The d0 value is shared between all keys of d. As a result, you get the
behaviour you noticed.
(ie there is only 1 dict-value, it gets printed 3 times when you print 'd')
I thought the solution could be in using copy.deepcopy, but I might not
be using it right:
>>> d2 = dict.fromkeys(r,copy.deepcopy(dict.fromkeys(c)))
Here you do
d0 = dict.fromkeys(c)
d1 = copy.deepcopy(d0)
d2 = dict.fromkeys(r, d1)
d0 and d1 are completely seperate due to the deepcopy(), but the last line
still uses the same d1 value for all its keys, as in your first attempt. for
this reason, the problem does not disappear.
Any suggestions/pointers would be appreciated!
To solve this, you need to construct a fresh value for each key.
You can do this with an iterator and a dictionary constructor:
>>> c = (1,2,3)
>>> r = ('a','b','c')
>>> d3 = dict(( (rv, dict.fromkeys(c)) for rv in r ))
d3 is created by constructing a dict from a list of tuples (key, val), where
the key rv is from your 'r' tuple, and val is constructed anew each time from 'c'.
>>> d3
{'a': {1: None, 2: None, 3: None}, 'c': {1: None, 2: None, 3: None}, 'b': {1:
None, 2: None, 3: None}}
>>> d3['a'][1]=0
>>> d3
{'a': {1: 0, 2: None, 3: None}, 'c': {1: None, 2: None, 3: None}, 'b': {1:
None, 2: None, 3: None}}
Sincerely,
Albert
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