我正在有效地尝试解决与这个问题相同的问题：
Finding related words (specifically physical objects) to a specific word
减去单词代表物理对象的要求。答案和编辑过的问题似乎表明，一个好的开始是使用维基百科文本作为语料库来构建 n-gram 频率列表。在我开始下载庞大的维基百科转储之前，有谁知道这样的列表是否已经存在？
PS 如果上一个问题的原始发帖人看到了这一点，我很想知道你是如何解决这个问题的，因为你的结果看起来很棒:-)
【问题讨论】：
Google has a publicly available TB n-garam 数据库（最多 5 个）。
您可以订购 6 张 DVD 或找到托管它的 torrent。
【讨论】：
您可以找到 2008 年 6 月的 Wikipedia n-gram here。此外，它还有词条和标记句子。我尝试创建自己的 n-gram，但在 bigram 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 个小时，unigrams 需要 5 个小时，bigrams 需要 8 个小时。
链接的 n-gram 还具有一些被清理的好处，因为 mediawiki 和 Wikipedia 在文本之间有很多垃圾。
这是我的 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))
【讨论】：