python 正则表达式过滤字符串两边的字符，类似strip功能练习题

比眉伴天荒 2022-05-25 12:04 120阅读 0赞

import re
    
    
    def customerStrip(steament, string):
        '''
        :param steament: 被过滤的字符串
        :param string: 要过滤的字符
        :return: 返回过滤后的数据
        '''
        string = '\\' + string
        customerRegex = re.compile(r'^(((' + string + ')*)?(\s*))(.*?)((\s*)?((' + string + ')*))$')
        try:
            moCustome = customerRegex.search(steament).groups()
            return moCustome[4]
        except AttributeError:
            return '不存在要过滤的字符！'
    
    
    text = '####3213213###'
    
    print(customerStrip(text, '#'))

**自己练习写的很不完善。**  
经测试，strip 这是这个自定义函数customerStrip的80多倍，很差劲，方法写的不好  
import re

import time
    
    
    def leftRegex(steament, string):
        lRgex = re.compile(r'(^(' + string + ')*)')
        moLeft = lRgex.search(steament).group()
        return moLeft
    
    
    def rightRegex(steament, string):
        rRegex = re.compile(r'((' + string + ')*)$')
        moRight = rRegex.search(steament).group()
        return moRight
    
    
    def customerStrip(steament, string, type='center'):
        '''
        :param steament: 
        :param string: 
        :param type: 
        :return: 
        '''
        if string.isspace():
            string = '\s'
        else:
            string = '\\' + string
        if type == 'right':
            right = rightRegex(steament, string)
            return  steament[:-len(right)]
        elif type == 'left':
            left=leftRegex(steament, string)
            return steament[len(left):]
        elif type == 'center':
            right = rightRegex(steament, string)
            left = leftRegex(steament, string)
            steament =steament[:-len(right)]
            steament=steament[len(left):]
            return steament
    
    
    text = '##321333333333333354343243255442343241341324312414321####'
    
    
    
    
    # customer time
    start = time.time()
    print(customerStrip(text, '#'))
    end = time.time()
    
    time1 = end - start
    # strip time
    start1 = time.time()
    print(text.strip('#'))
    ends = time.time()
    time2 = ends - start1
    print(time1/time2)

**此函数经测试，strip 是他的40-50倍，比上面那个提高了不少，依然很差劲**

import re
    
    import time
    
    
    def customerStrip(steament, string):
        if string.isspace():
            string = '\s'
        else:
            string = '\\' + string
        centerRegex = re.compile(r'(^(' + string + ')*)?(.*?)((' + string + ')*)?$')
        result = centerRegex.search(steament).groups()[2]
        return result

**和上面的差不多速度**