I'm interested in tries and DAWGs (direct acyclic word graph) and I've been reading a lot about them 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 lookup 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 want to know the output object once a set of words are turned into tries or DAWGs.
115 Answers
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')
FalseI'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.
5Have a look at this:
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:
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:
2Here 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.
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')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.
2from collections import defaultdictDefine 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 currTest:
print(word_exist(trie, 'cam'))Here is full code using a TrieNode class. Also implemented auto_complete method to return the matching words with a prefix.
Since we are using dictionary to store children, there is no need to convert char to integer and vice versa and don't need to allocate array memory in advance.
class TrieNode: def __init__(self): #Dict: Key = letter, Item = TrieNode self.children = {} self.end = False
class Trie: def __init__(self): self.root = TrieNode() def build_trie(self,words): for word in words: self.insert(word) def insert(self,word): node = self.root for char in word: if char not in node.children: node.children[char] = TrieNode() node = node.children[char] node.end = True def search(self, word): node = self.root for char in word: if char in node.children: node = node.children[char] else: return False return node.end def _walk_trie(self, node, word, word_list): if node.children: for char in node.children: word_new = word + char if node.children[char].end: # if node.end: word_list.append( word_new) # word_list.append( word) self._walk_trie(node.children[char], word_new , word_list) def auto_complete(self, partial_word): node = self.root word_list = [ ] #find the node for last char of word for char in partial_word: if char in node.children: node = node.children[char] else: # partial_word not found return return word_list if node.end: word_list.append(partial_word) # word_list will be created in this method for suggestions that start with partial_word self._walk_trie(node, partial_word, word_list) return word_listcreate a Trie
t = Trie()
words = ['hi', 'hieght', 'rat', 'ram', 'rattle', 'hill']
t.build_trie(words)Search for word
words = ['hi', 'hello']
for word in words: print(word, t.search(word))
hi True
hel Falsesearch for words using prefix
partial_word = 'ra'
t.auto_complete(partial_word)
['rat', 'rattle', 'ram']Using defaultdict and reduce function.
Create Trie
from functools import reduce
from collections import defaultdict
T = lambda : defaultdict(T)
trie = T()
reduce(dict.__getitem__,'how',trie)['isEnd'] = TrueTrie :
defaultdict(<function __main__.<lambda>()>, {'h': defaultdict(<function __main__.<lambda>()>, {'o': defaultdict(<function __main__.<lambda>()>, {'w': defaultdict(<function __main__.<lambda>()>, {'isEnd': True})})})})Search In Trie :
curr = trie
for w in 'how': if w in curr: curr = curr[w] else: print("Not Found") break
if curr['isEnd']: print('Found')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])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': {} } }
}This is much like a previous answer but simpler to read:
def make_trie(words): trie = {} for word in words: head = trie for char in word: if char not in head: head[char] = {} head = head[char] head["_end_"] = "_end_" return triePython Class for Trie
Trie Data Structure can be used to store data in O(L) where L is the length of the string so for inserting N strings time complexity would be O(NL) the string can be searched in O(L) only same goes for deletion.
Can be clone from
class Node: def __init__(self): self.children = [None]*26 self.isend = False
class trie: def __init__(self,): self.__root = Node() def __len__(self,): return len(self.search_byprefix('')) def __str__(self): ll = self.search_byprefix('') string = '' for i in ll: string+=i string+='\n' return string def chartoint(self,character): return ord(character)-ord('a') def remove(self,string): ptr = self.__root length = len(string) for idx in range(length): i = self.chartoint(string[idx]) if ptr.children[i] is not None: ptr = ptr.children[i] else: raise ValueError("Keyword doesn't exist in trie") if ptr.isend is not True: raise ValueError("Keyword doesn't exist in trie") ptr.isend = False return def insert(self,string): ptr = self.__root length = len(string) for idx in range(length): i = self.chartoint(string[idx]) if ptr.children[i] is not None: ptr = ptr.children[i] else: ptr.children[i] = Node() ptr = ptr.children[i] ptr.isend = True def search(self,string): ptr = self.__root length = len(string) for idx in range(length): i = self.chartoint(string[idx]) if ptr.children[i] is not None: ptr = ptr.children[i] else: return False if ptr.isend is not True: return False return True def __getall(self,ptr,key,key_list): if ptr is None: key_list.append(key) return if ptr.isend==True: key_list.append(key) for i in range(26): if ptr.children[i] is not None: self.__getall(ptr.children[i],key+chr(ord('a')+i),key_list) def search_byprefix(self,key): ptr = self.__root key_list = [] length = len(key) for idx in range(length): i = self.chartoint(key[idx]) if ptr.children[i] is not None: ptr = ptr.children[i] else: return None self.__getall(ptr,key,key_list) return key_list t = trie()
t.insert("shubham")
t.insert("shubhi")
t.insert("minhaj")
t.insert("parikshit")
t.insert("pari")
t.insert("shubh")
t.insert("minakshi")print(t.search("minhaj"))
print(t.search("shubhk"))
print(t.search_byprefix('m'))
print(len(t))
print(t.remove("minhaj"))
print(t)Code Oputpt
True
False
['minakshi', 'minhaj']
7
minakshi
minhajsir
pari
parikshit
shubh
shubham
shubhi
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}}}With prefix search
Here is @senderle's answer, slightly modified to accept prefix search (and not only whole-word matching):
_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
def in_trie(trie, word): current_dict = trie for letter in word: if _end in current_dict: return True if letter not in current_dict: return False current_dict = current_dict[letter]
t = make_trie(['hello', 'hi', 'foo', 'bar'])
print(in_trie(t, 'hello world'))
# Trueclass TrieNode: def __init__(self): self.keys = {} self.end = False
class Trie: def __init__(self): self.root = TrieNode() def insert(self, word: str, node=None) -> None: if node == None: node = self.root # insertion is a recursive operation # this is base case to exit the recursion if len(word) == 0: node.end = True return # if this key does not exist create a new node elif word[0] not in node.keys: node.keys[word[0]] = TrieNode() self.insert(word[1:], node.keys[word[0]]) # that means key exists else: self.insert(word[1:], node.keys[word[0]]) def search(self, word: str, node=None) -> bool: if node == None: node = self.root # this is positive base case to exit the recursion if len(word) == 0 and node.end == True: return True elif len(word) == 0: return False elif word[0] not in node.keys: return False else: return self.search(word[1:], node.keys[word[0]]) def startsWith(self, prefix: str, node=None) -> bool: if node == None: node = self.root if len(prefix) == 0: return True elif prefix[0] not in node.keys: return False else: return self.startsWith(prefix[1:], node.keys[prefix[0]])
