python中如何去除标点符号 |
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程序员必备接口测试调试工具:立即使用 Apipost = Postman + Swagger + Mock + Jmeter Api设计、调试、文档、自动化测试工具 后端、前端、测试,同时在线协作,内容实时同步 Python去掉标点符号的方法如下: 方法一: str.isalnum: S.isalnum() -> bool 返回值:如果string至少有一个字符并且所有字符都是字母或数字则返回True,否则返回False。 实例: >>> string = "Special $#! characters spaces 888323" >>> ''.join(e for e in string if e.isalnum()) 'Specialcharactersspaces888323'登录后复制只能识别字母和数字,杀伤力大,会把中文、空格之类的也干掉 方法二: string.punctuation import re, string s ="string. With. Punctuation?" # Sample string # 写法一: out = s.translate(string.maketrans("",""), string.punctuation) # 写法二: out = s.translate(None, string.punctuation) # 写法三: exclude = set(string.punctuation) out = ''.join(ch for ch in s if ch not in exclude) # 写法四: >>> for c in string.punctuation: s = s.replace(c,"") >>> s 'string With Punctuation' # 写法五: out = re.sub('[%s]' % re.escape(string.punctuation), '', s) ## re.escape:对字符串中所有可能被解释为正则运算符的字符进行转义 # 写法六: # string.punctuation 只包括 ascii 格式; 想要一个包含更广(但是更慢)的方法是使用: unicodedata module : from unicodedata import category s = u'String — with - «Punctuation »...' out = re.sub('[%s]' % re.escape(string.punctuation), '', s) print 'Stripped', out # 输出:u'Stripped String \u2014 with \xabPunctuation \xbb' out = ''.join(ch for ch in s if category(ch)[0] != 'P') print 'Stripped', out # 输出:u'Stripped String with Punctuation ' # For Python 3 str or Python 2 unicode values, str.translate() only takes a dictionary; codepoints (integers) are looked up in that mapping and anything mapped to None is removed. # To remove (some?) punctuation then, use: import string remove_punct_map = dict.fromkeys(map(ord, string.punctuation)) s.translate(remove_punct_map) # Your method doesn't work in Python 3, as the translate method doesn't accept the second argument any more. import unicodedata import sys tbl = dict.fromkeys(i for i in range(sys.maxunicode) if unicodedata.category(chr(i)).startswith('P')) def remove_punctuation(text): return text.translate(tbl)登录后复制方法三: re 例: import re s ="string. With. Punctuation?" s = re.sub(r'[^\w\s]','',s)登录后复制测试: 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): return regex.sub('', s) def test_trans(s): return s.translate(table, string.punctuation) def test_repl(s): 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) out_put: # sets : 19.8566138744 # regex : 6.86155414581 # translate : 2.12455511093 # replace : 28.4436721802登录后复制更多Python相关技术文章,请访问Python教程栏目进行学习! 以上就是python中如何去除标点符号的详细内容,更多请关注php中文网其它相关文章! 声明:本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系[email protected]核实处理。 相关标签:python中去除标点符号程序员必备接口测试调试工具:点击使用 Apipost = Postman + Swagger + Mock + Jmeter Api设计、调试、文档、自动化测试工具 后端、前端、测试,同时在线协作,内容实时同步 手把手教你如何写一个自己的MVC框架(40节精讲/巨细/新人进阶必看) ![]() |
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