import re

input_text = "hay alrededor de 4 coches el dia 7 del mes 5" #example 1
input_text = "Serian 4 unidades de mermelada para el dia 04 del 8 de este año 2023"  #example 2
input_text = "Hay 10 unidades para el dia 15 del 12 y seran 9 ya para el 7 de noviembre"  #example 3
input_text = "Hay 10 unidades para el 15 del 1º mes del año y seran alrededor de 9 para el 7º dia del mes de noviembre"  #example 4


#for days
standard_number_of_digits_re_1 = r"(?:del dia|de el dia|el dia|del|de el|el)[\s|]*(\b\d{1}\b)"
standard_number_of_digits_re_2 = r"(\b\d{1}\b)[\s|]*º[\s|]*dia"

re_1_and_re_2 = r"(?:"   standard_number_of_digits_re_1   r"|"   standard_number_of_digits_re_2    r")"

#for months
#standard_number_of_digits_re_3 = re_1_and_re_2   r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d{1}\b)"
#standard_number_of_digits_re_4 = re_1_and_re_2   r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d{1}\b)[\s|]*(?:º[\s|]*mes del año|º[\s|]*mes)"

standard_number_of_digits_re_3 = r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d)"
standard_number_of_digits_re_4 = r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d)[\s|]*(?:º[\s|]*mes del año|º[\s|]*mes)"


#replacement with this conditions, and put '0' in front a day number only if it is one number and not two(or more numbers)
#  example: '1' --> '01'  or  '10' --> '10'
input_text = re.sub(standard_number_of_digits_re_3, r"0\1", input_text)
input_text = re.sub(standard_number_of_digits_re_4, r"0\1", input_text)
input_text = re.sub(standard_number_of_digits_re_1, r"0\1", input_text)
input_text = re.sub(standard_number_of_digits_re_2, r"0\1", input_text)

print(repr(input_text)) #output

The problem I'm having is that the regex don't seem to work correctly since I can't do the replacements, but trying in several code editors I notice these 2 things:

• In the Regex Debuger
  • In the SublimeText editor, when I put an r in front of "(?:" the whole line is immediately put as if it were a string of characters although you can see how this string is separated with operators in the middle in charge of the concatenation, however in StackOverflow that line is displayed correctly.

  

  I do not know if these observations are important in the malfunction of these regex, but for some reason that I cannot find, the regex cannot extract the text to be edited by placing (if applicable) a 0 in front.

  Leaving the output should be like this when printing them:

  "hay alrededor de 4 coches el dia 07 del mes 05" #for example 1
  "Serian 4 unidades de mermelada para el dia 04 del 08 de este año 2023"  #for example 2
  "Hay 10 unidades para el dia 15 del 12 y seran 9 ya para el 07 de noviembre"  #for example 3
  "Hay 10 unidades para el 15 del 01º mes del año y seran alrededor de 9 para el 07º dia del mes de noviembre"  #for example 4
  

  What modifications should I make to get these results? I think the problem is in this part of the regex (\b\d{1}\b)

  ---

  Edited code

  import re
  
  input_text = "hay alrededor de 4 coches el dia 7 del mes 5" #example 1
  #input_text = "Serian 4 unidades de mermelada para el dia 04 del 8 de este año 2023"  #example 2
  input_text = "Hay 10 unidades para el dia 15 del 12 y seran 9 ya para el 7 de noviembre"  #example 3
  #input_text = "Hay 10 unidades para el 15 del 1º mes del año y seran alrededor de 9 para el 7º dia del mes de noviembre"  #example 4
  
  #for days
  standard_number_of_digits_re_1 = r"(?:del dia|de el dia|el dia|del|de el|el)[\s|]*(\b\d)"
  standard_number_of_digits_re_2 = r"(\b\d)[\s|]*º[\s|]*dia"
  
  #for months
  standard_number_of_digits_re_3 = r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d)"
  standard_number_of_digits_re_4 = r"(?:del mes|de el mes|del|de el)[\s|]*(\b\d)[\s|]*(?:º[\s|]*mes del año|º[\s|]*mes)"
  
  
  #replacement with this conditions, and put '0' in front a day number only if it is one number and not two(or more numbers)
  #  example: '1' --> '01'  or  '10' --> '10'
  input_text = re.sub(standard_number_of_digits_re_3, r"0\1", input_text)
  
  print(repr(input_text)) #output
  
  input_text = re.sub(standard_number_of_digits_re_4, r"0\1", input_text)
  
  print(repr(input_text)) #output
  
  input_text = re.sub(standard_number_of_digits_re_1, r"0\1", input_text)
  
  print(repr(input_text)) #output
  
  input_text = re.sub(standard_number_of_digits_re_2, r"0\1", input_text)
  
  print(repr(input_text)) #output
  

  The problem is for example in the example 3 :

  'Hay 10 unidades para 015 012 y seran 9 ya para 07 de noviembre'
  

  And the correct output is:

  'Hay 10 unidades para 15 12 y seran 9 ya para 07 de noviembre'
  

  CodePudding user response：

  Get rid of \b after \d{1}. Python considers º to be a word character, so there's no word boundary between 7 and º. You don't need to match a word boundary, since the regexp only allows spaces or | between the number and º.

  There's also no need for {1}. All patterns match 1 time unless they're quantified, so {1} is redundant.

  (\b\d)[\s|]*º[\s|]*dia
  

  https://regex101.com/r/84QDbp/1