Chapter 7, Automate the boring stuff with Python, practice project: regex version of strip()

问题

I am reading the book "Automate the boring stuff with Python'. In Chapter 7, in the project practice: the regex version of strip(), here is my code (I use Python 3.x):

def stripRegex(x,string):
import re
if x == '':
    spaceLeft = re.compile(r'^\s+')
    stringLeft = spaceLeft.sub('',string)
    spaceRight = re.compile(r'\s+$')
    stringRight = spaceRight.sub('',string)
    stringBoth = spaceRight.sub('',stringLeft)
    print(stringLeft)
    print(stringRight)

else:
    charLeft = re.compile(r'^(%s)+'%x)
    stringLeft = charLeft.sub('',string)
    charRight = re.compile(r'(%s)+$'%x)
    stringBoth = charRight.sub('',stringLeft)
print(stringBoth)

x1 = ''
x2 = 'Spam'
x3 = 'pSam'
string1 = '      Hello world!!!   '
string2 = 'SpamSpamBaconSpamEggsSpamSpam'
stripRegex(x1,string1)
stripRegex(x2,string2)
stripRegex(x3,string2)

And here is the output:

Hello world!!!   
      Hello world!!!
Hello world!!!
BaconSpamEggs
SpamSpamBaconSpamEggsSpamSpam

So, my regex version of strip() nearly work as the original version. In the origninal version, the output always is "BaconSpamEggs" no matter you passed in 'Spam', 'pSam', 'mapS', 'Smpa'... So how to fix this in Regex version???

回答1:

You could check for multiple characters in the regex like this:

charLeft = re.compile(r'^([%s]+)' % 'abc') 
print charLeft.sub('',"aaabcfdsfsabca")
>>> fdsfsabca

Or even better, do it in a single regex:

def strip_custom(x=" ", text):
    return re.search(' *[{s}]*(.*?)[{s}]* *$'.format(s=x), text).group(1)

split_custom('abc', ' aaabtestbcaa ')
>>> test

回答2:

import re

def regexStrip(x,y=''):


if y!='':
    yJoin=r'['+y+']*([^'+y+'].*[^'+y+'])['+y+']*'
    cRegex=re.compile(yJoin,re.DOTALL)
    return cRegex.sub(r'\1',x)
else:
    sRegex=re.compile(r'\s*([^\s].*[^\s])\s*',re.DOTALL)
    return sRegex.sub(r'\1',x)

text='  spmaHellow worldspam'
print(regexStrip(text,'spma'))

回答3:

I switched the arguments, but from my quick testing, this seems to work. I gave it an optional argument which defaults to None.

def stripRegex(s,toStrip=None):
    import re
    if toStrip is None:
        toStrip = '\s'
    return re.sub(r'^[{0}]+|[{0}]+$'.format(toStrip), '', s)

x1 = ''
x2 = 'Spam'
x3 = 'pSam'
string1 = '      Hello world!!!   '
string2 = 'SpamSpamBaconSpamEggsSpamSpam'

print(stripRegex(string1)) # 'Hello world!!!'
print(stripRegex(string1, x1)) # '      Hello world!!!   '
print(stripRegex(string2, x2)) # 'BaconSpamEggs'
print(stripRegex(string2, x3)) # 'BaconSpamEggs'

回答4:

I have written two different codes for the same: 1st way:

import re    
def stripfn(string, c):
        if c != '':
            Regex = re.compile(r'^['+ c +']*|['+ c +']*$')
            strippedString = Regex.sub('', string)
            print(strippedString)
        else:
            blankRegex = re.compile(r'^(\s)*|(\s)*$')
            strippedString = blankRegex.sub('', string)
            print(strippedString)

2nd way:

import re
def stripfn(string, c):
    if c != '':
        startRegex = re.compile(r'^['+c+']*')
        endRegex = re.compile(r'['+c+']*$')
        startstrippedString = startRegex.sub('', string)
        endstrippedString = endRegex.sub('', startstrippedString)
        print(endstrippedString)
    else:
        blankRegex = re.compile(r'^(\s)*|(\s)*$')
        strippedString = blankRegex.sub('', string)
        print(strippedString)

回答5:

This seems to work:

def stripp(text, leftright = None):
    import re
    if leftright == None:
        stripRegex = re.compile(r'^\s*|\s*$')
        text = stripRegex.sub('', text)
        print(text)
    else:
        stripRegex = re.compile(r'^.|.$')
        margins = stripRegex.findall(text)
        while margins[0] in leftright:
            text = text[1:]
            margins = stripRegex.findall(text)
        while margins[-1] in leftright:
            text = text[:-2]
            margins = stripRegex.findall(text)
        print(text) 

mo = '    @@@@@@     '
mow = '@&&@#$texttexttext&&^&&&&%%'
bla = '@&#$^%+'

stripp(mo)
stripp(mow, bla)

回答6:

Here my version:

    #!/usr/bin/env python3

import re

def strippp(txt,arg=''): # assigning a default value to arg prevents the error if no argument is passed when calling strippp()
    if arg =='':
        regex1 = re.compile(r'^(\s+)')
        mo = regex1.sub('', txt)
        regex2 = re.compile(r'(\s+)$')
        mo = regex2.sub('', mo)
        print(mo)
    else:
        regex1 = re.compile(arg)
        mo = regex1.sub('', txt)
        print(mo)

text = '        So, you can create the illusion of smooth motion        '
strippp(text, 'e')
strippp(text)

回答7:

Solution by @rtemperv is missing a case when a string starts/ends w/ whitespace characters but such character is not provided for removal.

I.e

>>> var="     foobar"
>>> var.strip('raf')
'     foob'

Hence regex should be a bit different:

def strip_custom(x=" ", text):
    return re.search('^[{s}]*(.*?)[{s}]*$'.format(s=x), text).group(1)

回答8:

See the code below

from re import *
check = '1'
while(check == '1'):
    string = input('Enter the string: ')
    strToStrip = input('Enter the string to strip: ')
    if strToStrip == '':                              #If the string to strip is empty
        exp = compile(r'^[\s]*')                      #Looks for all kinds of spaces in beginning until anything other than that is found
        string = exp.sub('',string)                   #Replaces that with empty string
        exp = compile(r'[\s]*$')                      #Looks for all kinds of spaces in the end until anything other than that is found
        string = exp.sub('',string)                   #Replaces that with empty string
        print('Your Stripped string is \'', end = '')
        print(string, end = '')
        print('\'')
    else:
        exp = compile(r'^[%s]*'%strToStrip)           #Finds all instances of the characters in strToStrip in the beginning until anything other than that is found
        string = exp.sub('',string)                   #Replaces it with empty string
        exp = compile(r'[%s]*$'%strToStrip)           #Finds all instances of the characters in strToStrip in the end until anything other than that is found
        string = exp.sub('',string)                   #Replaces it with empty string
        print('Your Stripped string is \'', end = '')
        print(string, end = '')
        print('\'')
    print('Do you want to continue (1\\0): ', end = '')
    check = input()

Explanation:

The character class [] is used to check the individual instances of the character in the string.
The ^ is used to check whether the characters in the string to strip are in the beginning or not
The $ is used to check whether the characters in the string to strip are in the end or not
If found they are replaced by empty string with the sub()
* is used to match the maximum of the characters in the string to strip until anything other than that is found.
* matches 0 is no instance if found or matches as many as instances if found.

回答9:

#! python
# Regex Version of Strip()
import re
def RegexStrip(mainString,charsToBeRemoved=None):
    if(charsToBeRemoved!=None):
        regex=re.compile(r'[%s]'%charsToBeRemoved)#Interesting TO NOTE
        return regex.sub('',mainString)
    else:
        regex=re.compile(r'^\s+')
        regex1=re.compile(r'$\s+')
        newString=regex1.sub('',mainString)
        newString=regex.sub('',newString)
        return newString

Str='   hello3123my43name is antony    '
print(RegexStrip(Str))

I think this is a rather comfortable code, I found the carets(^) and the dollar($) really effective.

回答10:

import re
def strips(arg, string):
    beginning = re.compile(r"^[{}]+".format(arg))        
    strip_beginning = beginning.sub("", string)
    ending = re.compile(r"[{}]+$".format(arg))
    strip_ending = ending.sub("", strip_beginning)
    return strip_ending

The function strips will strip whatever "arg" refers to irrespective of the occurrence

回答11:

I believe this regexp could be simpler to understand:

import re

strip_reg =  re.compile("\s*(.*?)\s*$")
strip_rep.search(<mystring>).group(1)

How it works? let take it backwards. We look for zeros one more space in the end of the string "\s*$"

the ".*?" Is a special case where you ask the regexp to look for the minimal number of character to match. (most of the time a regexp will try to get the most) We capture this.

we try to capture zeros or more character before the group we capture.

回答12:

The following is my attempt to apply lessons learned from "Clean Code" by R.C. Martin and "Automate the boring stuff" by Al Sweigart. One of the rules of clean code is to write functions that are small and do one thing.

def removeSpacesAndSecondString(text):
    print(text)
    stripSecondStringRegex = re.compile(r'((\w+)\s(\w+)?)')
    for groups in stripSecondStringRegex.findall(text):
        newText = groups[1]
    print(newText)

def removeSpaces(text):
    print(text)
    stripSpaceRegex = re.compile(r'\s')
    mo = stripSpaceRegex.sub('', text)
    print(mo)

text = '"  hjjkhk  "'

if len(text.split()) > 1:
    removeSpacesAndSecondString(text)
else:
    removeSpaces(text)

来源：https://stackoverflow.com/questions/34954040/chapter-7-automate-the-boring-stuff-with-python-practice-project-regex-versio

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