问题
I am new to Python and trying to learn and advance. I am interested in TRIEs and DAWGs and I have been reading a lot about it but I don\'t understand what should the output TRIE or DAWG file look like.
- Should a TRIE be an object of nested dictionaries? Where each letter is divided in to letters and so on?
- Would a look up performed on such a dictionary be fast if there are 100k or 500k entries?
- How to implement word-blocks consisting of more than one word separated with - or space?
- How to link prefix or suffix of a word to another part in the structure? [for DAWG]
I want to understand the best output structure in order to figure out how to create and use one.
I would also appreciate what should be the output of a DAWG along with TRIE.
I do not want to see graphical representations with bubbles linked to each other, I saw them plenty whilst reading.
I would like to know the output object once a set of words are turned into TRIEs or DAWGs.
Thank you.
回答1:
Unwind is essentially correct that there are many different ways to implement a trie; and for a large, scalable trie, nested dictionaries might become cumbersome -- or at least space inefficient. But since you're just getting started, I think that's the easiest approach; you could code up a simple trie in just a few lines. First, a function to construct the trie:
>>> _end = '_end_'
>>>
>>> def make_trie(*words):
... root = dict()
... for word in words:
... current_dict = root
... for letter in word:
... current_dict = current_dict.setdefault(letter, {})
... current_dict[_end] = _end
... return root
...
>>> make_trie('foo', 'bar', 'baz', 'barz')
{'b': {'a': {'r': {'_end_': '_end_', 'z': {'_end_': '_end_'}},
'z': {'_end_': '_end_'}}},
'f': {'o': {'o': {'_end_': '_end_'}}}}
If you're not familiar with setdefault, it simply looks up a key in the dictionary (here, letter or _end). If the key is present, it returns the associated value; if not, it assigns a default value to that key and returns the value ({} or _end). (It's like a version of get that also updates the dictionary.)
Next, a function to test whether the word is in the trie:
>>> def in_trie(trie, word):
... current_dict = trie
... for letter in word:
... if letter not in current_dict:
... return False
... current_dict = current_dict[letter]
... return _end in current_dict
...
>>> in_trie(make_trie('foo', 'bar', 'baz', 'barz'), 'baz')
True
>>> in_trie(make_trie('foo', 'bar', 'baz', 'barz'), 'barz')
True
>>> in_trie(make_trie('foo', 'bar', 'baz', 'barz'), 'barzz')
False
>>> in_trie(make_trie('foo', 'bar', 'baz', 'barz'), 'bart')
False
>>> in_trie(make_trie('foo', 'bar', 'baz', 'barz'), 'ba')
False
I'll leave insertion and removal to you as an exercise.
Of course, Unwind's suggestion wouldn't be much harder. There might be a slight speed disadvantage in that finding the correct sub-node would require a linear search. But the search would be limited to the number of possible characters -- 27 if we include _end. Also, there's nothing to be gained by creating a massive list of nodes and accessing them by index as he suggests; you might as well just nest the lists.
Finally, I'll add that creating a directed acyclic word graph (DAWG) would be a bit more complex, because you have to detect situations in which your current word shares a suffix with another word in the structure. In fact, this can get rather complex, depending on how you want to structure the DAWG! You may have to learn some stuff about Levenshtein distance to get it right.
回答2:
Have a look at this:
https://github.com/kmike/marisa-trie
Static memory-efficient Trie structures for Python (2.x and 3.x).
String data in a MARISA-trie may take up to 50x-100x less memory than in a standard Python dict; the raw lookup speed is comparable; trie also provides fast advanced methods like prefix search.
Based on marisa-trie C++ library.
Here's a blog post from a company using marisa trie successfully:
https://www.repustate.com/blog/sharing-large-data-structure-across-processes-python/
At Repustate, much of our data models we use in our text analysis can be represented as simple key-value pairs, or dictionaries in Python lingo. In our particular case, our dictionaries are massive, a few hundred MB each, and they need to be accessed constantly. In fact for a given HTTP request, 4 or 5 models might be accessed, each doing 20-30 lookups. So the problem we face is how do we keep things fast for the client as well as light as possible for the server.
...
I found this package, marisa tries, which is a Python wrapper around a C++ implementation of a marisa trie. “Marisa” is an acronym for Matching Algorithm with Recursively Implemented StorAge. What’s great about marisa tries is the storage mechanism really shrinks how much memory you need. The author of the Python plugin claimed 50-100X reduction in size – our experience is similar.
What’s great about the marisa trie package is that the underlying trie structure can be written to disk and then read in via a memory mapped object. With a memory mapped marisa trie, all of our requirements are now met. Our server’s memory usage went down dramatically, by about 40%, and our performance was unchanged from when we used Python’s dictionary implementation.
There are also a couple of pure-python implementations, though unless you're on a restricted platform you'd want to use the C++ backed implementation above for best performance:
- https://github.com/bdimmick/python-trie
- https://pypi.python.org/pypi/PyTrie
回答3:
Here is a list of python packages that implement Trie:
- marisa-trie - a C++ based implementation.
- python-trie - a simple pure python implementation.
- PyTrie - a more advanced pure python implementation.
- pygtrie - a pure python implementation by Google.
- datrie - a double array trie implementation based on libdatrie.
回答4:
There's no "should"; it's up to you. Various implementations will have different performance characteristics, take various amounts of time to implement, understand, and get right. This is typical for software development as a whole, in my opinion.
I would probably first try having a global list of all trie nodes so far created, and representing the child-pointers in each node as a list of indices into the global list. Having a dictionary just to represent the child linking feels too heavy-weight, to me.
回答5:
Modified from senderle's method (above). I found that Python's defaultdict is ideal for creating a trie or a prefix tree.
from collections import defaultdict
class Trie:
"""
Implement a trie with insert, search, and startsWith methods.
"""
def __init__(self):
self.root = defaultdict()
# @param {string} word
# @return {void}
# Inserts a word into the trie.
def insert(self, word):
current = self.root
for letter in word:
current = current.setdefault(letter, {})
current.setdefault("_end")
# @param {string} word
# @return {boolean}
# Returns if the word is in the trie.
def search(self, word):
current = self.root
for letter in word:
if letter not in current:
return False
current = current[letter]
if "_end" in current:
return True
return False
# @param {string} prefix
# @return {boolean}
# Returns if there is any word in the trie
# that starts with the given prefix.
def startsWith(self, prefix):
current = self.root
for letter in prefix:
if letter not in current:
return False
current = current[letter]
return True
# Now test the class
test = Trie()
test.insert('helloworld')
test.insert('ilikeapple')
test.insert('helloz')
print test.search('hello')
print test.startsWith('hello')
print test.search('ilikeapple')
回答6:
If you want a TRIE implemented as a Python class, here is something I wrote after reading about them:
class Trie:
def __init__(self):
self.__final = False
self.__nodes = {}
def __repr__(self):
return 'Trie<len={}, final={}>'.format(len(self), self.__final)
def __getstate__(self):
return self.__final, self.__nodes
def __setstate__(self, state):
self.__final, self.__nodes = state
def __len__(self):
return len(self.__nodes)
def __bool__(self):
return self.__final
def __contains__(self, array):
try:
return self[array]
except KeyError:
return False
def __iter__(self):
yield self
for node in self.__nodes.values():
yield from node
def __getitem__(self, array):
return self.__get(array, False)
def create(self, array):
self.__get(array, True).__final = True
def read(self):
yield from self.__read([])
def update(self, array):
self[array].__final = True
def delete(self, array):
self[array].__final = False
def prune(self):
for key, value in tuple(self.__nodes.items()):
if not value.prune():
del self.__nodes[key]
if not len(self):
self.delete([])
return self
def __get(self, array, create):
if array:
head, *tail = array
if create and head not in self.__nodes:
self.__nodes[head] = Trie()
return self.__nodes[head].__get(tail, create)
return self
def __read(self, name):
if self.__final:
yield name
for key, value in self.__nodes.items():
yield from value.__read(name + [key])
回答7:
This version is using recursion
import pprint
from collections import deque
pp = pprint.PrettyPrinter(indent=4)
inp = raw_input("Enter a sentence to show as trie\n")
words = inp.split(" ")
trie = {}
def trie_recursion(trie_ds, word):
try:
letter = word.popleft()
out = trie_recursion(trie_ds.get(letter, {}), word)
except IndexError:
# End of the word
return {}
# Dont update if letter already present
if not trie_ds.has_key(letter):
trie_ds[letter] = out
return trie_ds
for word in words:
# Go through each word
trie = trie_recursion(trie, deque(word))
pprint.pprint(trie)
Output:
Coool👾 <algos>🚸 python trie.py
Enter a sentence to show as trie
foo bar baz fun
{
'b': {
'a': {
'r': {},
'z': {}
}
},
'f': {
'o': {
'o': {}
},
'u': {
'n': {}
}
}
}
回答8:
from collections import defaultdict
Define Trie:
_trie = lambda: defaultdict(_trie)
Create Trie:
trie = _trie()
for s in ["cat", "bat", "rat", "cam"]:
curr = trie
for c in s:
curr = curr[c]
curr.setdefault("_end")
Lookup:
def word_exist(trie, word):
curr = trie
for w in word:
if w not in curr:
return False
curr = curr[w]
return '_end' in curr
Test:
print(word_exist(trie, 'cam'))
回答9:
class Trie:
head = {}
def add(self,word):
cur = self.head
for ch in word:
if ch not in cur:
cur[ch] = {}
cur = cur[ch]
cur['*'] = True
def search(self,word):
cur = self.head
for ch in word:
if ch not in cur:
return False
cur = cur[ch]
if '*' in cur:
return True
else:
return False
def printf(self):
print (self.head)
dictionary = Trie()
dictionary.add("hi")
#dictionary.add("hello")
#dictionary.add("eye")
#dictionary.add("hey")
print(dictionary.search("hi"))
print(dictionary.search("hello"))
print(dictionary.search("hel"))
print(dictionary.search("he"))
dictionary.printf()
Out
True
False
False
False
{'h': {'i': {'*': True}}}
来源：https://stackoverflow.com/questions/11015320/how-to-create-a-trie-in-python