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這篇文章將為大家詳細(xì)講解有關(guān)python中去除標(biāo)點(diǎn)符號(hào)的方法,小編覺(jué)得挺實(shí)用的,因此分享給大家做個(gè)參考,希望大家閱讀完這篇文章后可以有所收獲。
Python去掉標(biāo)點(diǎn)符號(hào)的方法如下:
方法一:
str.isalnum:
S.isalnum() -> bool
返回值:如果string至少有一個(gè)字符并且所有字符都是字母或數(shù)字則返回True,否則返回False。
實(shí)例:
>>> string = "Special $#! characters spaces 888323" >>> ''.join(e for e in string if e.isalnum()) 'Specialcharactersspaces888323'
只能識(shí)別字母和數(shù)字,殺傷力大,會(huì)把中文、空格之類的也干掉
方法二:
string.punctuation
import re, string s ="string. With. Punctuation?" # Sample string # 寫法一: out = s.translate(string.maketrans("",""), string.punctuation) # 寫法二: out = s.translate(None, string.punctuation) # 寫法三: exclude = set(string.punctuation) out = ''.join(ch for ch in s if ch not in exclude) # 寫法四: >>> for c in string.punctuation: s = s.replace(c,"") >>> s 'string With Punctuation' # 寫法五: out = re.sub('[%s]' % re.escape(string.punctuation), '', s) ## re.escape:對(duì)字符串中所有可能被解釋為正則運(yùn)算符的字符進(jìn)行轉(zhuǎn)義 # 寫法六: # string.punctuation 只包括 ascii 格式; 想要一個(gè)包含更廣(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 # 輸出:u'Stripped String \u2014 with \xabPunctuation \xbb' out = ''.join(ch for ch in s if category(ch)[0] != 'P') print 'Stripped', out # 輸出: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)
方法三:
re
例:
import re s ="string. With. Punctuation?" s = re.sub(r'[^\w\s]','',s)
測(cè)試:
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
關(guān)于python中去除標(biāo)點(diǎn)符號(hào)的方法就分享到這里了,希望以上內(nèi)容可以對(duì)大家有一定的幫助,可以學(xué)到更多知識(shí)。如果覺(jué)得文章不錯(cuò),可以把它分享出去讓更多的人看到。
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