python - 使用 Python 计算大文本中多词项的频率

Question

我有一本包含近一百万个多词术语（包含空格的术语）的字典。这看起来像

[..., 
'multilayer ceramic', 
'multilayer ceramic capacitor', 
'multilayer optical disk', 
'multilayer perceptron', 
...]

我想计算它们在许多 GB 文本中的频率。

作为一个小例子，考虑在 Wikipedia 页面中计算这四个多词表达式：

payload = {'action': 'query', 'titles': 'Ceramic_capacitor', 'explaintext':1, 'prop':'extracts', 'format': 'json'}
r = requests.get('https://en.wikipedia.org/w/api.php', params=payload)
sampletext = r.json()['query']['pages']['9221221']['extract'].lower()
sampledict = ['multilayer ceramic', 'multilayer ceramic capacitor', 'multilayer optical disk', 'multilayer perceptron']

termfreqdic = {}
for term in sampledict:
    termfreqdic[term] = sampletext.count(term)
print(termfreqdic)

这给出了类似的东西，{'multilayer ceramic': 7, 'multilayer ceramic capacitor': 2, 'multilayer optical disk': 0, 'multilayer perceptron': 0}但如果字典包含一百万个条目，它似乎不是最佳的。

我尝试过非常大的正则表达式：

termlist = [re.escape(w) for w in open('termlistfile.txt').read().strip().split('\n')]
termregex = re.compile(r'\b'+r'\b|\b'.join(termlist), re.I)
termfreqdic = {}
for i,li in enumerate(open(f)):
    for m in termregex.finditer(li):
        termfreqdic[m.group(0)]=termfreqdic.get(m.group(0),0)+1
open('counted.tsv','w').write('\n'.join([a+'\t'+v for a,v in termfreqdic.items()]))

这太慢了（最近的 i7 上 1000 行文本需要 6 分钟）。但是如果我使用regex而不是re替换前两行，它会下降到每 1000 行文本大约 12 秒，这对于我的需求来说仍然非常慢：

termlist = open(termlistfile).read().strip().split('\n')
termregex = regex.compile(r"\L<options>", options=termlist)
...

请注意，这并不完全符合我的要求，因为一个术语可能是另一个术语的子术语，如示例“多层陶瓷”和“多层陶瓷电容器”（这也排除了第一次标记的方法，如Find multi-word terms in a Python 中的标记化文本）。

这看起来像是一个常见的序列匹配问题，无论是在文本语料库还是在遗传字符串中，都必须有众所周知的解决方案。也许可以用一些单词来解决（我不介意术语列表的初始编译速度很慢）？唉，我似乎没有在寻找合适的条款。也许有人可以指出我正确的方向？

标签： pythonnlpnltkcorpusword-frequency

Answer 1

@SidharthMacherla 让我走上了正确的轨道（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 NLTKMWETokenizer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，使用大型正则表达式需要 1000 秒。在 100m 个字符中尝试 100k 个多词术语仍然可以使用标记器而不是计数或正则表达式。

如需测试，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()