Get number present after a particular pattern of a

问题:

I want to get all the matching numbers(only numbers example '0012--22') or numbers which contain some text (example 'RF332') corresponding to it which matches with a list of strings provided("my_list" in the code). The format in which the text with number will be present is like separated by a space or two. Providing sample input file for reference.

This is the input file:

$cat input_file
some text before Expedien: 1-21-212-16-26 some random text
Reference RE9833 of all sentences.
abc
123
456
something blah blah Ref.: 
tramite  1234567
Ref.:
some junk Expedien N° 18-00777 # some new content
some text Expedien N°18-0022995 # some garbled content

The script till now is attached below: It is currently only identifying one element which is {'tramite': '1234567'}

import re
import glob
import os

my_list = ['Ref.:', 'Reference', 'tramite', 'Expediente', 'Expediente No', 'Expedien N°', 'Exp.No', 'Expedien']

#open the file as input
with open('garb.txt','r') as infile:
  res = dict()
  for line in infile:  
    elems = re.split('(?::)?\s+', line)
    #print(elems)
    if len(elems) >= 2 :
      contains = False
      tmp = ''
      for elem in elems:  
        if contains:
          res.update({tmp : elem})
          print(res)
          contains = False
          break
        if elem in my_list:
          contains = True
          tmp = elem
  #print(res)

This is the expected output:

Sample output:

{'Expedien N°': '18-0022995'}
{'Expedien N°': '18-0022995'}
{'Expedien': '1-21-212-16-26'}
{'Reference' : 'RE9833'}

etc etc.

回答1:

You may use

(?<!\w)(your|escaped|keywords|here)\W*([A-Z]*\d+(?:-+[A-Z]*\d+)*)

See the regex demo.

Pattern details

• (?<!\w) - left word boundary (unambiguous, \b meaning is context dependent and if the next char is a non-word char, it will require a word char on the left, and that is not something users usually expect)
• (your|escaped|keywords|here) - Capturing group 1: your list of keywords, it can be easily built using '|'.join(map(re.escape,my_list)) (note re.escape is necessary to escape special regex metacharacters like ., +, (, [, etc.)
• \W* - 0+ non-word chars (chars other than letters, digits or _)
• ([A-Z]*\d+(?:-+[A-Z]*\d+)*) - Capturing group 2:
  • [A-Z]* - zero or more uppercase ASCII letters
  • \d+ - 1 or more digits
  • (?:-+[A-Z]*\d+)* - 0 or more repetitions of
    • -+ - one or more hyphens
    • [A-Z]*\d+ - zero or more uppercase ASCII letters, 1 or more digits

See the Python demo:

import re
s="""your_text_here"""
my_list = ['Ref.:', 'Reference', 'tramite', 'Expediente', 'Expediente No', 'Expedien N°', 'Exp.No', 'Expedien']
rx = r'(?<!\w)({})\W*([A-Z]*\d+(?:-+[A-Z]*\d+)*)'.format('|'.join(map(re.escape,my_list)))
print(re.findall(rx, s))

Output:

[('Expedien', '1-21-212-16-26'), ('Reference', 'RE9833'), ('tramite', '1234567'), ('Expedien N°', '18-00777'), ('Expedien N°', '18-0022995')]

回答2:

There really needs to be something that allows users with less than 50+ rep points to comment, because this thread is one I'm really curious about and want to fork off of, but didn't want to have to give a full fledged answer on, because the answer I'm giving involves finite situations and isn't flexible.

@Wiktor Stribiżew

Your solution misses the "Ref." portion of the output based on your demo. It looks like he wants to skip "tramite"

@checkmate

In your desired output you need to edit it because "UV1234" does't show up anywhere in the string you posted

.

Anyway, I found a solution but am really hoping someone can improve upon this.

>>> import re

>>> string = '''some text before Expedien: 1-21-212-16-26 some random text
Reference RE9833 of all sentences.
abc
123
456
something blah blah Ref.: 
tramite  1234567
Ref.:
some junk Expedien N° 18-00777 # some new content
some text Expedien N°18-0022995 # some garbled content'''

>>> re.findall('(?:(Expedien[\s]+N\S|Ref\.(?!:[\S\s]{,11}Expedien)|Reference|Expedien))[\S\s]*?([A-Z\-]*(?:[\d]+)[\S]*)', string)

[('Expedien', '1-21-212-16-26'), ('Reference', 'RE9833'), ('Ref.', '1234567'), ('Expedien N\xb0', '18-00777'), ('Expedien N\xb0', '18-0022995')]

The Flaws:

• To capture correctly it relies in part on "Ref.(?!:[\S\s]{,11}Expedien)"
• First of all that "11" needs to be edited to account for other lengths of info that may be present so it is not flexible
• Secondly, if it is instead followed by "Reference" as opposed to then the third "Ref." will be captured incorrectly