Thanks Thomas!
To reply the subject's question: I don't have to. The following works
perfectly.
def populate_trie(trie, sequence, position=None):
"""
Populate a trie.
Assume that the counter is always at `position` 0 while the `position`
of the dictionary is the last one.
>>> trie = {}
>>> populate_trie(trie, 'taspython')
{'t': {'a': {'s': {'p': {'y': {'t': {'h': {'o': {'n': {}}}}}}}}}}
>>> trie = {}
>>> populate_trie(trie, 'kalo', 1)
{'k': [1, {'a': [1, {'l': [1, {'o': [1, {}]}]}]}]}
>>> trie = {}
>>> populate_trie(trie, 'heh', 2)
{'h': [1, 0, {'e': [1, 0, {'h': [1, 0, {}]}]}]}
>>> trie = {}
>>> trie = populate_trie(trie, 'heh', 1)
>>> populate_trie(trie, 'hah', 1)
{'h': [2, {'a': [1, {'h': [1, {}]}], 'e': [1, {'h': [1, {}]}]}]}
"""
if (position is not None) and (position >= 1):
embedded_obj = [0] * position
embedded_obj.append({})
else:
embedded_obj = {}
d2 = trie
for i, character in enumerate(sequence):
d2 = access_trie(trie, sequence[:i], position)
if character not in d2:
if position is None:
d2[character] = deepcopy(embedded_obj)
else:
d2[character] = d2.get(character, deepcopy(embedded_obj))
d2[character][0] += 1
elif position is not None:
d2[character][0] += 1
return trie
Best regards,
Dimitris Leventeas
--
Dimitris Leventeas
http://students.ceid.upatras.gr/~lebenteas/
--
http://mail.python.org/mailman/listinfo/python-list
