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
对于Python 3 str或Python 2 unicode值,str.translate()只使用字典;在该映射中查找代码点(整数),并删除映射到None的任何内容。
要删除(某些?)标点符号,请使用:
import string
remove_punct_map = dict.fromkeys(map(ord, string.punctuation))
s.translate(remove_punct_map)
dict.fromkeys()类方法使创建映射变得简单,根据键的顺序将所有值设置为None。
要删除所有标点符号,而不仅仅是ASCII标点符号,您的表需要稍微大一点;参见J.F.Sebastian的答案(Python 3版本):
import unicodedata
import sys
remove_punct_map = dict.fromkeys(i for i in range(sys.maxunicode)
if unicodedata.category(chr(i)).startswith('P'))
其他回答
这可能不是最好的解决方案,但我就是这样做的。
import string
f = lambda x: ''.join([i for i in x if i not in string.punctuation])
对于Python 3 str或Python 2 unicode值,str.translate()只使用字典;在该映射中查找代码点(整数),并删除映射到None的任何内容。
要删除(某些?)标点符号,请使用:
import string
remove_punct_map = dict.fromkeys(map(ord, string.punctuation))
s.translate(remove_punct_map)
dict.fromkeys()类方法使创建映射变得简单,根据键的顺序将所有值设置为None。
要删除所有标点符号,而不仅仅是ASCII标点符号,您的表需要稍微大一点;参见J.F.Sebastian的答案(Python 3版本):
import unicodedata
import sys
remove_punct_map = dict.fromkeys(i for i in range(sys.maxunicode)
if unicodedata.category(chr(i)).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')
从效率的角度来看,你不会击败
s.translate(None, string.punctuation)
对于更高版本的Python,请使用以下代码:
s.translate(str.maketrans('', '', string.punctuation))
它使用查找表在C语言中执行原始字符串操作——除了编写自己的C代码之外,没有什么能比这更好的了。
如果速度不令人担忧,另一个选择是:
exclude = set(string.punctuation)
s = ''.join(ch for ch in s if ch not in exclude)
这比用每个字符替换s.replace更快,但不会像正则表达式或字符串转换等非纯python方法那样执行得好,正如您从下面的计时中看到的那样。对于这种类型的问题,在尽可能低的水平上解决是有回报的。
计时代码:
import re, string, timeit
s = "string. With. Punctuation"
exclude = set(string.punctuation)
table = string.maketrans("","")
regex = re.compile('[%s]' % re.escape(string.punctuation))
def test_set(s):
return ''.join(ch for ch in s if ch not in exclude)
def test_re(s): # From Vinko's solution, with fix.
return regex.sub('', s)
def test_trans(s):
return s.translate(table, string.punctuation)
def test_repl(s): # From S.Lott's solution
for c in string.punctuation:
s=s.replace(c,"")
return s
print "sets :",timeit.Timer('f(s)', 'from __main__ import s,test_set as f').timeit(1000000)
print "regex :",timeit.Timer('f(s)', 'from __main__ import s,test_re as f').timeit(1000000)
print "translate :",timeit.Timer('f(s)', 'from __main__ import s,test_trans as f').timeit(1000000)
print "replace :",timeit.Timer('f(s)', 'from __main__ import s,test_repl as f').timeit(1000000)
结果如下:
sets : 19.8566138744
regex : 6.86155414581
translate : 2.12455511093
replace : 28.4436721802
我喜欢使用这样的函数:
def scrub(abc):
while abc[-1] is in list(string.punctuation):
abc=abc[:-1]
while abc[0] is in list(string.punctuation):
abc=abc[1:]
return abc