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
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}")
不一定更简单,但如果你更熟悉re家族的话,就另辟蹊径。
import re, string
s = "string. With. Punctuation?" # Sample string
out = re.sub('[%s]' % re.escape(string.punctuation), '', s)
import re
s = "string. With. Punctuation?" # Sample string
out = re.sub(r'[^a-zA-Z0-9\s]', '', s)
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
您也可以这样做:
import string
' '.join(word.strip(string.punctuation) for word in 'text'.split())