@SidharthMacherla 让我走上了正轨（NLTK 和标记化），尽管他的解决方案没有解决多词表达的问题，而且这可能是重叠的。
简而言之，我发现的最佳方法是继承 NLTK 的 MWETokenizer 并添加一个使用 util.Trie 计算多词的函数：
import re, regex, timeit
from nltk.tokenize import MWETokenizer
from nltk.util import Trie
class FreqMWETokenizer(MWETokenizer):
"""A tokenizer that processes tokenized text and merges multi-word expressions
into single tokens.
"""
def __init__(self, mwes=None, separator="_"):
super().__init__(mwes, separator)
def freqs(self, text):
"""
:param text: A list containing tokenized text
:type text: list(str)
:return: A frequency dictionary with multi-words merged together as keys
:rtype: dict
:Example:
>>> tokenizer = FreqMWETokenizer([ mw.split() for mw in ['multilayer ceramic', 'multilayer ceramic capacitor', 'ceramic capacitor']], separator=' ')
>>> tokenizer.freqs("Gimme that multilayer ceramic capacitor please!".split())
{'multilayer ceramic': 1, 'multilayer ceramic capacitor': 1, 'ceramic capacitor': 1}
"""
i = 0
n = len(text)
result = {}
while i < n:
if text[i] in self._mwes:
# possible MWE match
j = i
trie = self._mwes
while j < n and text[j] in trie:
if Trie.LEAF in trie:
# success!
mw = self._separator.join(text[i:j])
result[mw]=result.get(mw,0)+1
trie = trie[text[j]]
j = j + 1
else:
if Trie.LEAF in trie:
# success!
mw = self._separator.join(text[i:j])
result[mw]=result.get(mw,0)+1
i += 1
else:
i += 1
return result
>>> tokenizer = FreqMWETokenizer([ mw.split() for mw in ['multilayer ceramic', 'multilayer ceramic capacitor', 'ceramic capacitor']], separator=' ')
>>> tokenizer.freqs("Gimme that multilayer ceramic capacitor please!".split())
{'multilayer ceramic': 1, 'multilayer ceramic capacitor': 1, 'ceramic capacitor': 1}
这是带有速度测量的测试套件：
用 FreqMWETokenizer 计算 10m 个字符中的 10k 个多词术语需要 2 秒，使用 MWETokenizer 需要 4 秒（还提供了完整的标记化，但不计算重叠），使用简单计数方法需要 150 秒，使用 1000 秒大正则表达式。在 100m 字符中尝试 100k 多词术语仍然可以使用标记器而不是计数或正则表达式。
为了测试，请在https://mega.nz/file/PsVVWSzA#5-OHy-L7SO6fzsByiJzeBnAbtJKRVy95YFdjeF_7yxA找到两个大样本文件
def freqtokenizer(thissampledict, thissampletext):
"""
This method uses the above FreqMWETokenizer's function freqs.
It captures overlapping multi-words
counting 1000 terms in 1000000 characters took 0.3222855870008061 seconds. found 0 terms from the list.
counting 10000 terms in 10000000 characters took 2.5309120759993675 seconds. found 21 terms from the list.
counting 100000 terms in 29467534 characters took 10.57763242800138 seconds. found 956 terms from the list.
counting 743274 terms in 29467534 characters took 25.613067482998304 seconds. found 10411 terms from the list.
"""
tokenizer = FreqMWETokenizer([mw.split() for mw in thissampledict], separator=' ')
thissampletext = re.sub(' +',' ', re.sub('[^\s\w\/\-\']+',' ',thissampletext)) # removing punctuation except /-'_
freqs = tokenizer.freqs(thissampletext.split())
return freqs
def nltkmethod(thissampledict, thissampletext):
""" This function first produces a tokenization by means of MWETokenizer.
This takes the biggest matching multi-word, no overlaps.
They could be computed separately on the dictionary.
counting 1000 terms in 1000000 characters took 0.34804968100070255 seconds. found 0 terms from the list.
counting 10000 terms in 10000000 characters took 3.9042628339993826 seconds. found 20 terms from the list.
counting 100000 terms in 29467534 characters took 12.782784996001283 seconds. found 942 terms from the list.
counting 743274 terms in 29467534 characters took 28.684293715999956 seconds. found 9964 terms from the list.
"""
termfreqdic = {}
tokenizer = MWETokenizer([mw.split() for mw in thissampledict], separator=' ')
thissampletext = re.sub(' +',' ', re.sub('[^\s\w\/\-\']+',' ',thissampletext)) # removing punctuation except /-'_
tokens = tokenizer.tokenize(thissampletext.split())
freqdist = FreqDist(tokens)
termsfound = set([t for t in freqdist.keys()]) & set(thissampledict)
for t in termsfound:termfreqdic[t]=freqdist[t]
return termfreqdic
def countmethod(thissampledict, thissampletext):
"""
counting 1000 in 1000000 took 0.9351876619912218 seconds.
counting 10000 in 10000000 took 91.92642056700424 seconds.
counting 100000 in 29467534 took 3185.7411157219904 seconds.
"""
termfreqdic = {}
for term in thissampledict:
termfreqdic[term] = thissampletext.count(term)
return termfreqdic
def regexmethod(thissampledict, thissampletext):
"""
counting 1000 terms in 1000000 characters took 2.298602456023218 seconds.
counting 10000 terms in 10000000 characters took 395.46084802100086 seconds.
counting 100000: impossible
"""
termfreqdic = {}
termregex = re.compile(r'\b'+r'\b|\b'.join(thissampledict))
for m in termregex.finditer(thissampletext):
termfreqdic[m.group(0)]=termfreqdic.get(m.group(0),0)+1
return termfreqdic
def timing():
"""
for testing, find the two large sample files at
https://mega.nz/file/PsVVWSzA#5-OHy-L7SO6fzsByiJzeBnAbtJKRVy95YFdjeF_7yxA
"""
sampletext=open("G06K0019000000.txt").read().lower()
sampledict=open("manyterms.lower.txt").read().strip().split('\n')
print(len(sampletext),'characters',len(sampledict),'terms')
for i in range(4):
for f in [freqtokenizer, nltkmethod, countmethod, regexmethod]:
start = timeit.default_timer()
thissampledict = sampledict[:1000*10**i]
thissampletext = sampletext[:1000000*10**i]
termfreqdic = f(thissampledict, thissampletext)
#termfreqdic = countmethod(thissampledict, thissampletext)
#termfreqdic = regexmethod(thissampledict, thissampletext)
#termfreqdic = nltkmethod(thissampledict, thissampletext)
#termfreqdic = freqtokenizer(thissampledict, thissampletext)
print('{f} counting {terms} terms in {characters} characters took {seconds} seconds. found {termfreqdic} terms from the list.'.format(f=f, terms=len(thissampledict), characters=len(thissampletext), seconds=timeit.default_timer()-start, termfreqdic=len({a:v for (a,v) in termfreqdic.items() if v})))
timing()