On Aug 17, 4:25 am, Paul McGuire <[EMAIL PROTECTED]> wrote:
> In responding to another post on defaultdict, I posted an
> implementation of a 2-level hashtable, by creating a factory method
> that returned a defaultdict(dict). The OP of that other thread was
> trying to build a nested tree from a set of n-tuples, in which the
> first (n-1) values in each tuple were the keys for navigating down the
> tree, and the final n'th value was the value for to be assigned to the
> leaf node. My post worked only if n=2, which fortunately was the test
> case that the OP gave. But it annoyed me that this required advance
> knowledge of the number of keys.
>
> I've hacked out this recursivedefaultdict which is a
> defaultdict(defaultdict(defaultdict(...))), arbitrarily deep depending
> on the keys provided in the reference.
>
> Please comment.
>
> -- Paul
>
> from collections import defaultdict
>
> data = [
> ('A','B','Z',1), ('A','C','Y',2), ('A','C','X',3),
> ('B','A','W',4), ('B','B','V',5), ('B','B','U',6),
> ('B','D','T',7),
> ]
>
> class recursivedefaultdict(object):
> def __init__(self):
> self.__dd = defaultdict(recursivedefaultdict)
> def __getattr__(self,attr):
> return self.__dd.__getattribute__(attr)
> def __getitem__(self,*args):
> return self.__dd.__getitem__(*args)
> def __setitem__(self,*args):
> return self.__dd.__setitem__(*args)
>
> table = recursivedefaultdict()
>
> for k1,k2,k3,v in data:
> table[k1][k2][k3] = v
>
> for kk in sorted(table.keys()):
> print "-",kk
> for jj in sorted(table[kk].keys()):
> print " -",jj
> for ii in sorted(table[kk][jj].keys()):
> print " -",ii,table[kk][jj][ii]
>
> prints:
> - A
> - B
> - Z 1
> - C
> - X 3
> - Y 2
> - B
> - A
> - W 4
> - B
> - U 6
> - V 5
> - D
> - T 7
There is something here:
http://groups.google.com/group/comp.lang.python/msg/92502e0278c2a56e
- Paddy.
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