Python: strip function definition using regex

Question

I am a very beginner of programming and reading the book "Automate the boring stuff with Python'. In Chapter 7, there is a project practice: the regex version of strip(). My code below does not work (I use Python 3.6.1). Could anyone help?

import re

string = input("Enter a string to strip: ")
strip_chars = input("Enter the characters you want to be stripped: ")

def strip_fn(string, strip_chars):
    if strip_chars == '':
        blank_start_end_regex = re.compile(r'^(\s)+|(\s)+$')
        stripped_string = blank_start_end_regex.sub('', string)
        print(stripped_string)
    else:
        strip_chars_start_end_regex = re.compile(r'^(strip_chars)*|(strip_chars)*$')
        stripped_string = strip_chars_start_end_regex.sub('', string)
        print(stripped_string)

r'^(strip_chars)*|(strip_chars)*$' -> r'^[{0}]+|[{0}]+$'.format("".join([re.escape(x) for x in strip_chars])). Also, remove the unnecessary ( and ) in r'^(\s)+|(\s)+$'. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Sep 21, 2017 at 13:02

Rishabh Kalakoti · Accepted Answer · 2017-09-21 13:14:04Z

2

You can also use re.sub to substitute the characters in the start or end. Let us say if the char is 'x'

re.sub(r'^x+', "", string)
re.sub(r'x+$', "", string)

The first line as lstrip and the second as rstrip This just looks simpler.

answered Sep 21, 2017 at 13:14

Rishabh Kalakoti

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Wiktor Stribiżew · Accepted Answer · 2017-09-21 13:11:10Z

When using r'^(strip_chars)*|(strip_chars)*$' string literal, the strip_chars is not interpolated, i.e. it is treated as a part of the string. You need to pass it as a variable to the regex. However, just passing it in the current form would result in a "corrupt" regex because (...) in a regex is a grouping construct, while you want to match a single char from the define set of chars stored in the strip_chars variable.

You could just wrap the string with a pair of [ and ] to create a character class, but if the variable contains, say z-a, it would make the resulting pattern invalid. You also need to escape each char to play it safe.

Replace

r'^(strip_chars)*|(strip_chars)*$'

with

r'^[{0}]+|[{0}]+$'.format("".join([re.escape(x) for x in strip_chars]))

I advise to replace * (zero or more occurrences) with + (one or more occurrences) quantifier because in most cases, when we want to remove something, we need to match at least 1 occurrence of the unnecessary string(s).

Also, you may replace r'^(\s)+|(\s)+$' with r'^\s+|\s+$' since the repeated capturing groups will keep on re-writing group values upon each iteration slightly hampering the regex execution.

Ieshaan Saxena · Accepted Answer · 2019-01-15 12:44:22Z

0

#! 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))

Maybe this could help, it can be further simplified of course.

answered Jan 15, 2019 at 12:44

Ieshaan Saxena

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