Discussion on Regex Substitution Challenge

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303 Discussions

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1 day ago+ 0 comments

import re
1. for i in range(int(input())):
i = input()
i=re.sub(r"(?<= )&&(?= )","and",i)
i=re.sub(r"(?<= )||(?= )","or",i)
print(i)

4 weeks ago+ 0 comments

import re and_pattern = re.compile( r"(?<=[\s])[&&]{2}(?=[\s])" ) or_pattern = re.compile( r"(?<=[\s])[||]{2}(?=[\s])" )

for _ in range(int(input())): print( re.sub(and_pattern,'and', re.sub(or_pattern, 'or', input())) )

Could reduce it down, however it would make it less readable.

Sometimes fewer lines isn't better

3 months ago+ 0 comments

import re
N = int(input())
text = "\n".join(input() for _ in range(N))
print(re.sub(r'(?<=\s)&&(?=\s)', 'and', re.sub(r'(?<=\s)\|\|(?=\s)', 'or', text)))

4 months ago+ 0 comments

import os, sys, re

text = sys.stdin.read()
pattern = r"(?<=\s)(&{2}|[|]{2})(?=\s)"

def modified_match(match_object):

    if match_object.group() == '&&':
        return 'and'
    else:
        return 'or'

text = re.sub(pattern, modified_match, text)
print(os.linesep.join(text.splitlines()[1:])) #removes the line count at line n°1

4 months ago+ 0 comments

import re

n = int(input())
pattern = r'(?<= )(&{2}|\|{2})(?= )'
func = lambda x: 'and' if x.group() == '&&' else 'or'

for _ in range(n):
    print(re.sub(pattern, func, input()))

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