如何在python中有效地计算多个文档的二元组答案

【问题标题】：How to efficiently count bigrams over multiple documents in python如何在python中有效地计算多个文档的二元组
【发布时间】：2018-06-02 08:59:31
【问题描述】：

我有一组文本文档，想计算所有文本文档的二元组数。

首先，我创建一个列表，其中每个元素又是一个列表，表示一个特定文档中的单词：

print(doc_clean)
# [['This', 'is', 'the', 'first', 'doc'], ['And', 'this', 'is', 'the', 'second'], ..]

然后，我从文档中提取二元组并将它们存储在一个列表中：

bigrams = []
for doc in doc_clean:
    bigrams.extend([(doc[i-1], doc[i]) 
                   for i in range(1, len(doc))])
print(bigrams)
# [('This', 'is'), ('is', 'the'), ..]

现在，我想计算每个唯一二元组的频率：

bigrams_freq = [(b, bigrams.count(b)) 
                for b in set(bigrams)]

一般来说，这种方法是有效的，但它太慢了。 bigrams 列表非常大，总共约 5mio 条目和约 300k 独特的 bigrams。在我的笔记本电脑上，当前的方法花费了太多时间进行分析。

谢谢你帮助我！

【问题讨论】：

标签： python nlp nltk

【解决方案1】：

您可以尝试以下方法：

from collections import Counter
from nltk import word_tokenize 
from nltk.util import ngrams
from nltk.stem import WordNetLemmatizer
from nltk.corpus import stopwords

stop_words = set(stopwords.words('english'))

doc_1 = 'Convolutional Neural Networks are very similar to ordinary Neural Networks from the previous chapter'
doc_2 = 'Convolutional Neural Networks take advantage of the fact that the input consists of images and they constrain the architecture in a more sensible way.'
doc_3 = 'In particular, unlike a regular Neural Network, the layers of a ConvNet have neurons arranged in 3 dimensions: width, height, depth.'
docs = [doc_1, doc_2, doc_3]
docs = (' '.join(filter(None, docs))).lower()

tokens = word_tokenize(docs)
tokens = [t for t in tokens if t not in stop_words]
word_l = WordNetLemmatizer()
tokens = [word_l.lemmatize(t) for t in tokens if t.isalpha()]

bi_grams = list(ngrams(tokens, 2)) 
counter = Counter(bi_grams)
counter.most_common(5)

Out[82]: 
[(('neural', 'network'), 4),
 (('convolutional', 'neural'), 2),
 (('network', 'similar'), 1),
 (('similar', 'ordinary'), 1),
 (('ordinary', 'neural'), 1)]

【讨论】：