似乎应该有一种比以下更简单的方法:
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)
有?
似乎应该有一种比以下更简单的方法:
import string
s = "string. With. Punctuation?" # Sample string
out = s.translate(string.maketrans("",""), string.punctuation)
有?
当前回答
这里有一个使用RegEx的简单方法
import re
punct = re.compile(r'(\w+)')
sentence = 'This ! is : a # sample $ sentence.' # Text with punctuation
tokenized = [m.group() for m in punct.finditer(sentence)]
sentence = ' '.join(tokenized)
print(sentence)
'This is a sample sentence'
其他回答
正则表达式很简单,如果你知道的话。
import re
s = "string. With. Punctuation?"
s = re.sub(r'[^\w\s]','',s)
下面是Python 3.5的一行代码:
import string
"l*ots! o(f. p@u)n[c}t]u[a'ti\"on#$^?/".translate(str.maketrans({a:None for a in string.punctuation}))
考虑unicode。代码已在python3中检查。
from unicodedata import category
text = 'hi, how are you?'
text_without_punc = ''.join(ch for ch in text if not category(ch).startswith('P'))
使用Python从文本文件中删除停止词
print('====THIS IS HOW TO REMOVE STOP WORS====')
with open('one.txt','r')as myFile:
str1=myFile.read()
stop_words ="not", "is", "it", "By","between","This","By","A","when","And","up","Then","was","by","It","If","can","an","he","This","or","And","a","i","it","am","at","on","in","of","to","is","so","too","my","the","and","but","are","very","here","even","from","them","then","than","this","that","though","be","But","these"
myList=[]
myList.extend(str1.split(" "))
for i in myList:
if i not in stop_words:
print ("____________")
print(i,end='\n')
字符串标点符号漏掉了现实世界中常用的大量标点符号。一个适用于非ASCII标点符号的解决方案怎么样?
import regex
s = u"string. With. Some・Really Weird、Non?ASCII。 「(Punctuation)」?"
remove = regex.compile(ur'[\p{C}|\p{M}|\p{P}|\p{S}|\p{Z}]+', regex.UNICODE)
remove.sub(u" ", s).strip()
我个人认为,这是在Python中删除字符串标点符号的最佳方法,因为:
它删除所有Unicode标点符号它很容易修改,例如,如果您想删除标点符号,可以删除\{s},但保留像$这样的符号。您可以非常具体地了解要保留的内容和要删除的内容,例如,Pd只会删除破折号。此正则表达式还规范了空白。它将制表符、回车符和其他奇怪的字符映射到漂亮的单个空格。
这使用了Unicode字符财产,您可以在Wikipedia上阅读更多有关该属性的信息。