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
# FIRST METHOD
# Storing all punctuations in a variable
punctuation='!?,.:;"\')(_-'
newstring ='' # Creating empty string
word = raw_input("Enter string: ")
for i in word:
if(i not in punctuation):
newstring += i
print ("The string without punctuation is", newstring)
# SECOND METHOD
word = raw_input("Enter string: ")
punctuation = '!?,.:;"\')(_-'
newstring = word.translate(None, punctuation)
print ("The string without punctuation is",newstring)
# Output for both methods
Enter string: hello! welcome -to_python(programming.language)??,
The string without punctuation is: hello welcome topythonprogramminglanguage
其他回答
# FIRST METHOD
# Storing all punctuations in a variable
punctuation='!?,.:;"\')(_-'
newstring ='' # Creating empty string
word = raw_input("Enter string: ")
for i in word:
if(i not in punctuation):
newstring += i
print ("The string without punctuation is", newstring)
# SECOND METHOD
word = raw_input("Enter string: ")
punctuation = '!?,.:;"\')(_-'
newstring = word.translate(None, punctuation)
print ("The string without punctuation is",newstring)
# Output for both methods
Enter string: hello! welcome -to_python(programming.language)??,
The string without punctuation is: hello welcome topythonprogramminglanguage
您也可以这样做:
import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())
我还没有看到这个答案。只需使用正则表达式;它删除了除单词字符(\w)和数字字符(\d)之外的所有字符,后跟一个空白字符(\s):
import re
s = "string. With. Punctuation?" # Sample string
out = re.sub(ur'[^\w\d\s]+', '', s)
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'))
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