On Nov 11, 11:31 pm, macm <moura.ma...@gmail.com> wrote:
>
> I pass a nested dictionary to a function.
>
> def Dicty( dict[k1][k2] ):
> print k1
> print k2
>
> There is a fast way (trick) to get k1 and k2 as string.
It might be possible to do something using a reverse dictionary and
getting rid of the nested dictionary.
This is a quick and simple 'two-way' dictionary class that works by
maintaining two dictionaries: the original key/value, and the reversed
value/key. It returns a list of keys, allowing for a value to be
assigned against more than
from collections import defaultdict
class TwoWayDict(dict):
def __init__(self, *args, **kwargs):
self._reversed = defaultdict(list)
for key, val in kwargs.iteritems():
self[key] = val
def __setitem__(self, key, value):
super(TwoWayDict, self).__setitem__(key, value)
self._reversed[value].append(key)
def getkeys(self, match):
return self._reversed[match]
>>> original = TwoWayDict(a=100,b='foo',c=int,d='foo')
>>> original.getkeys(100)
['a']
>>> original.getkeys('foo')
['b', 'd']
As for the nested dictionary, you could replace it with a _single_
dictionary that uses a composite key:
>>> original = TwoWayDict(a=100,b=100)
>>> original.getkeys(100)
['a', 'b']
>>> original = TwoWayDict()
>>> original['record1','user1'] = 'martin'
>>> original['record1','user2'] = 'robert'
>>> original['record2','user1'] = 'robert'
>>> original.getkeys('robert')
[('record1', 'user2'), ('record2', 'user1')]
> Whithout loop all dict. Just it!
The TwoWayDict class removes the need to loop across the dict looking
for keys that match a value by replacing it with another dict lookup.
Reducing the nested dict to a single dict with composite keys removes
the need to traverse the outer dict to compare against its children.
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