我似乎在从 Python 字符串中剥离标点符号时遇到了一些问题。在这里,我得到了一个文本文件(特别是来自 Project Gutenberg 的一本书)和一个停用词列表。我想返回 10 个最常用单词的字典。不幸的是,我总是在我归还的字典中打嗝。
import sys
import collections
from string import punctuation
import operator
#should return a string without punctuation
def strip_punc(s):
return ''.join(c for c in s if c not in punctuation)
def word_cloud(infile, stopwordsfile):
wordcount = {}
#Reads the stopwords into a list
stopwords = [x.strip() for x in open(stopwordsfile, 'r').readlines()]
#reads data from the text file into a list
lines = []
with open(infile) as f:
lines = f.readlines()
lines = [line.split() for line in lines]
#does the wordcount
for line in lines:
for word in line:
word = strip_punc(word).lower()
if word not in stopwords:
if word not in wordcount:
wordcount[word] = 1
else:
wordcount[word] += 1
#sorts the dictionary, grabs 10 most common words
output = dict(sorted(wordcount.items(),
key=operator.itemgetter(1), reverse=True)[:10])
print(output)
if __name__=='__main__':
try:
word_cloud(sys.argv[1], sys.argv[2])
except Exception as e:
print('An exception has occured:')
print(e)
print('Try running as python3 word_cloud.py <input-text> <stopwords>')
这将打印出来
{'said': 659, 'mr': 606, 'one': 418, '“i': 416, 'lorry': 322, 'upon': 288, 'will': 276, 'defarge': 268, 'man': 264, 'little': 263}
“我不应该在那里。我不明白为什么它没有在我的辅助函数中消除。
提前致谢。
请您参考如下方法:
字符“不是”。
string.punctuation 仅包含以下 ASCII 字符:
In [1]: import string
In [2]: string.punctuation
Out[2]: '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
因此您需要扩充要删除的字符列表。
像下面这样的东西应该可以完成你所需要的:
extended_punc = punctuation + '“' # and any other characters you need to strip
def strip_punc(s):
return ''.join(c for c in s if c not in extended_punc)
或者,您可以使用包 unidecode以 ASCII 编码您的文本,而不必担心创建您可能需要处理的 unicode 字符列表:
from unidecode import unidecode
def strip_punc(s):
s = unidecode(s.decode('utf-8'))
return ''.join(c for c in s if c not in punctuation).encode('utf-8')
