Smart Way to do Regex

Fri, 06 Oct 2023 23:23:10 +0800

Below is a sample code that I found to get complex regex pattern instead of write it manually good example to make smaller parts first the combine to a very complex pattern.


import re

# convert the {name} test to rules in predefined format in macros
def regex_expand(macros, pattern, guarded = True):
 output = []
 pos = 0
 size = len(pattern)
 while pos < size:
 ch = pattern[pos]
 if ch == '\\':
 output.append(pattern[pos:pos + 2])
 pos += 2
 continue
 elif ch != '{':
 output.append(ch)
 pos += 1
 continue
 p2 = pattern.find('}', pos)
 if p2 < 0:
 output.append(ch)
 pos += 1
 continue
 p3 = p2 + 1
 name = pattern[pos + 1:p2].strip('\r\n\t ')
 if name == '':
 output.append(pattern[pos:p3])
 pos = p3
 continue
 elif name[0].isdigit():
 output.append(pattern[pos:p3])
 pos = p3
 continue
 elif ('<' in name) or ('>' in name):
 raise ValueError('invalid pattern name "%s"'%name)
 if name not in macros:
 raise ValueError('{%s} is undefined'%name)
 if guarded:
 output.append('(?:' + macros[name] + ')')
 else:
 output.append(macros[name])
 pos = p3
 return ''.join(output)

# given a set of rules, create rules dictionary
def regex_build(code, macros = None, capture = True):
 defined = {}
 if macros is not None:
 for k, v in macros.items():
 defined[k] = v
 line_num = 0
 for line in code.split('\n'):
 line_num += 1
 line = line.strip('\r\n\t ')
 if (not line) or line.startswith('#'):
 continue
 pos = line.find('=')
 if pos < 0:
 raise ValueError('%d: not a valid rule'%line_num)
 head = line[:pos].strip('\r\n\t ')
 body = line[pos + 1:].strip('\r\n\t ')
 if (not head):
 raise ValueError('%d: empty rule name'%line_num)
 elif head[0].isdigit():
 raise ValueError('%d: invalid rule name "%s"'%(line_num, head))
 elif ('<' in head) or ('>' in head):
 raise ValueError('%d: invalid rule name "%s"'%(line_num, head))
 try:
 pattern = regex_expand(defined, body, guarded = not capture)
 except ValueError as e:
 raise ValueError('%d: %s'%(line_num, str(e)))
 try:
 re.compile(pattern)
 except re.error:
 raise ValueError('%d: invalid pattern "%s"'%(line_num, pattern))
 if not capture:
 defined[head] = pattern
 else:
 defined[head] = '(?P<%s>%s)'%(head, pattern)
 return defined

# define a set of rules
rules = r'''
 protocol = http|https
 login_name = [^:@\r\n\t ]+
 login_pass = [^@\r\n\t ]+
 login = {login_name}(:{login_pass})?
 host = [^:/@\r\n\t ]+
 port = \d+
 optional_port = (?:[:]{port})?
 path = /[^\r\n\t ]*
 url = {protocol}://({login}[@])?{host}{optional_port}{path}?
'''

# convert above rules to dictionary
m = regex_build(rules, capture = True)

# print dictionary info
for k, v in m.items():
 print(k, '=', v)

print()

# use final rule "url" to regex a text
pattern = m['url']
s = re.match(pattern, 'https://name:pass@www.test.com:8080/abcdefg')

# print the complete list of matches
print('matched: "%s"'%s.group(0))
print()

# print the matched items based on rule defined names
for name in ('url', 'login_name', 'login_pass', 'host', 'port', 'path'):
 print('subgroup:', name, '=', s.group(name))

Output:

Python_Regex

Tue, 12 Oct 2021 20:59:46 +0800

Some Advanced usage on regex

get float number:

import re
target = 'total growth is : 8%-8.5%'
rate = re.findall('(\d+(?:\.\d+)?)', target)
print(rate)
['8', '8.5']
#\d is any int, + is one to many
#(?:...)
#A non-capturing group allows you to apply quantifiers to part of your regex but does not capture/assign an ID.
#so it means: find int or, if int follows a "." and any int afterwards, combine the int with ".int"
#For example, repeating 1-3 digits and a period 3 times can be done like #this: /(?:\d{1,3}\.){3}\d{1,3}/
#above regex can match string like : 219.254.232.123 or 232.212.222.3434 (can only go until 232.212.222.343), 
#can not match 232.2122.222.3434 or 232.212.22a.3434

#`(\d+(?:\.\d+)?)` is a group, so result will be put in a list. 

#if you just want to match any int or .xxx or xxxx.xxx,just use

rate = re.findall('(\d+(\.\d+)?)', target)
print(rate)
[('8', ''), ('8.5', '.5')]
#which will return a list to you

#if you just want to find a float, use below: 

target = 'total growth is : 8%-8.5% or 9.546%'
rate = re.findall('(\d+\.\d+)', target)
print(rate)
['8.5', '9.546']

A bit deeper:

A(?=B) ：Match String A ends with String B

Regex on Council of Elrond

Smart Way to do Regex

Python_Regex

Some Advanced usage on regex

get float number:

A bit deeper: