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
with open('one.txt','r')as myFile:
str1=myFile.read()
print(str1)
punctuation = ['(', ')', '?', ':', ';', ',', '.', '!', '/', '"', "'"]
for i in punctuation:
str1 = str1.replace(i," ")
myList=[]
myList.extend(str1.split(" "))
print (str1)
for i in myList:
print(i,end='\n')
print ("____________")
myString.translate(None, string.punctuation)
在不太严格的情况下,单行线可能会有所帮助:
''.join([c for c in s if c.isalnum() or c.isspace()])
您也可以这样做:
import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())
显然,我无法对所选答案进行编辑,所以这里有一个适用于Python3的更新。在进行非平凡转换时,转换方法仍然是最有效的选择。
上面的@Brian为最初的繁重工作做出了贡献。感谢@ddejohn对原始测试的改进建议。
#!/usr/bin/env python3
"""Determination of most efficient way to remove punctuation in Python 3.
Results in Python 3.8.10 on my system using the default arguments:
set : 51.897
regex : 17.901
translate : 2.059
replace : 13.209
"""
import argparse
import re
import string
import timeit
parser = argparse.ArgumentParser()
parser.add_argument("--filename", "-f", default=argparse.__file__)
parser.add_argument("--iterations", "-i", type=int, default=10000)
opts = parser.parse_args()
with open(opts.filename) as fp:
s = fp.read()
exclude = set(string.punctuation)
table = str.maketrans("", "", string.punctuation)
regex = re.compile(f"[{re.escape(string.punctuation)}]")
def test_set(s):
return "".join(ch for ch in s if ch not in exclude)
def test_regex(s): # From Vinko's solution, with fix.
return regex.sub("", s)
def test_translate(s):
return s.translate(table)
def test_replace(s): # From S.Lott's solution
for c in string.punctuation:
s = s.replace(c, "")
return s
opts = dict(globals=globals(), number=opts.iterations)
solutions = "set", "regex", "translate", "replace"
for solution in solutions:
elapsed = timeit.timeit(f"test_{solution}(s)", **opts)
print(f"{solution:<10}: {elapsed:6.3f}")