Nlp 寻找维基百科上的n-gram数据库

我正在有效地解决与此问题相同的问题：

减去单词表示物理对象的要求。答案和经过编辑的问题似乎表明，一个良好的开端是使用维基百科文本作为语料库建立一个n-gram频率列表。在我开始下载庞大的维基百科转储之前，有人知道这样的列表是否已经存在吗

PS如果上一个问题的原始海报看到了这一点，我很想知道您是如何解决这个问题的，因为您的结果看起来非常好：-）

TB n-garam数据库（最多5个）。

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你可以订购6张DVD，也可以找到托管它的torrent。

你可以找到2008年6月的维基百科n-grams。此外，它也有标题词和标记句。我试图创建自己的n-gram，但在bigrams上内存不足（32Gb）（当前的英文维基百科非常庞大）。提取xml大约需要8小时，Unigram需要5小时，bigrams需要8小时

由于mediawiki和Wikipedia在文本之间有很多垃圾，链接的n-gram也有被清理的好处

以下是我的Python代码：

from nltk.tokenize import sent_tokenize
from nltk.tokenize import word_tokenize
from nltk.tokenize import wordpunct_tokenize
from datetime import datetime
from collections import deque
from collections import defaultdict
from collections import OrderedDict
import operator
import os

# Loop through all the English Wikipedia Article files and store their path and filename in a list. 4 minutes.
dir = r'D:\Downloads\Wikipedia\articles'
l = [os.path.join(root, name) for root, _, files in os.walk(dir) for name in files]

t1 = datetime.now()

# For each article (file) loop through all the words and generate unigrams. 1175MB memory use spotted.
# 12 minutes to first output. 4200000: 4:37:24.586706 was last output.
c = 1
d1s = defaultdict(int)
for file in l:
    try:
        with open(file, encoding="utf8") as f_in:
            content = f_in.read()
    except:
        with open(file, encoding="latin-1") as f_in:
            content = f_in.read()        
    words = wordpunct_tokenize(content)    # word_tokenize handles 'n ʼn and ʼn as a single word. wordpunct_tokenize does not.
    # Take all the words from the sentence and count them.
    for i, word in enumerate(words):    
        d1s[word] = d1s[word] + 1   
    c = c + 1
    if c % 200000 == 0:
        t2 = datetime.now()
        print(str(c) + ': ' + str(t2 - t1))

t2 = datetime.now()
print('After unigram: ' + str(t2 - t1))

t1 = datetime.now()
# Sort the defaultdict in descending order and write the unigrams to a file.
# 0:00:27.740082 was output. 3285Mb memory. 165Mb output file.
d1ss = OrderedDict(sorted(d1s.items(), key=operator.itemgetter(1), reverse=True))
with open("D:\\Downloads\\Wikipedia\\en_ngram1.txt", mode="w", encoding="utf-8") as f_out:
    for k, v in d1ss.items():
        f_out.write(k + '┼' + str(v) + "\n")
t2 = datetime.now()
print('After unigram write: ' + str(t2 - t1))

# Determine the lowest 1gram count we are interested in.
low_count = 20 - 1
d1s = {}
# Get all the 1gram counts as a dict.
for word, count in d1ss.items():
    # Stop adding 1gram counts when we reach the lowest 1gram count.
    if count == low_count:
        break
    # Add the count to the dict.
    d1s[word] = count

t1 = datetime.now()

# For each article (file) loop through all the sentences and generate 2grams. 13GB memory use spotted.
# 17 minutes to first output. 4200000: 4:37:24.586706 was last output.
c = 1
d2s = defaultdict(int)
for file in l:
    try:
        with open(file, encoding="utf8") as f_in:
            content = f_in.read()
    except:
        with open(file, encoding="latin-1") as f_in:
            content = f_in.read()   
    # Extract the sentences in the file content.         
    sentences = deque()
    sentences.extend(sent_tokenize(content))            
    # Get all the words for one sentence.
    for sentence in sentences:        
        words = wordpunct_tokenize(sentence)    # word_tokenize handles 'n ʼn and ʼn as a single word. wordpunct_tokenize does not.
        # Take all the words from the sentence with high 1gram count that are next to each other and count them.
        for i, word in enumerate(words):    
            if word in d1s:
                try:
                    word2 = words[i+1]
                    if word2 in d1s:
                        gram2 = word + ' ' + word2
                        d2s[gram2] = d2s[gram2] + 1
                except:
                    pass
    c = c + 1
    if c % 200000 == 0:
        t2 = datetime.now()
        print(str(c) + ': ' + str(t2 - t1))

t2 = datetime.now()
print('After bigram: ' + str(t2 - t1))

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是的，我考虑过这个数据集——甚至比维基百科的垃圾堆还要大！它不可用于商业用途有人发现它的洪流吗？