How find all childrens values of a nested dictionary, fast!
Peter Otten
__peter__ at web.de
Thu Nov 4 13:26:03 EDT 2010
macm wrote:
> How find all childrens values of a nested dictionary, fast!
>
>>>> a = {'a' : {'b' :{'/' :[1,2,3,4], 'ba' :{'/' :[41,42,44]} ,'bc'
>>>> :{'/':[51,52,54], 'bcd' :{'/':[68,69,66]}}},'c' :{'/' :[5,6,7,8]}},
>>>> 'ab' : {'/' :[12,13,14,15]}, 'ac' :{'/' :[21,22,23]}} a['a']
> {'c': {'/': [5, 6, 7, 8]}, 'b': {'ba': {'/': [41, 42, 44]}, '/': [1,
> 2, 3, 4], 'bc': {'bcd': {'/': [68, 69, 66]}, '/': [51, 52, 54]}}}
>>>> a['a']['b']
> {'ba': {'/': [41, 42, 44]}, '/': [1, 2, 3, 4], 'bc': {'bcd': {'/':
> [68, 69, 66]}, '/': [51, 52, 54]}}
>>>> a['a']['b'].values()
> [{'/': [41, 42, 44]}, [1, 2, 3, 4], {'bcd': {'/': [68, 69, 66]}, '/':
> [51, 52, 54]}]
>
> Now I want find all values of key "/"
>
> Desire result is [41, 42, 44, 1, 2, 3, 4, 68, 69, 66, 51, 52, 54]
>
> I am trying map, reduce, lambda.
Hmm, I'm trying none of these and get a different result:
>>> def f(d):
... stack = [d.iteritems()]
... while stack:
... for k, v in stack[-1]:
... if k == "/":
... yield v
... else:
... stack.append(v.iteritems())
... break
... else:
... stack.pop()
...
>>> b = []
>>> for v in f(a):
... b.extend(v)
...
>>> b
[5, 6, 7, 8, 41, 42, 44, 1, 2, 3, 4, 68, 69, 66, 51, 52, 54, 21, 22, 23, 12,
13, 14, 15]
What am I missing?
Peter
More information about the Python-list
mailing list