python regex match optional square brackets

I got all of them to match using this (You'll need to add the case-insensitive flag):

( ^ [a - z][a - z\ '&\(\) ]+\bv\b[a-z&\'\(\) ]+(?:.*?) \[?\d+ \w+ \d{4}\]?)

Suggestion : 2

Last Updated : 22 Jul, 2022

Any other string would not match the pattern.\d\ d\ d - \d\ d\ d - \d\ d\ d\ d

\
d {
   3
} - \d {
   3
} - \d {
   4
}

1._

Any other string would not match the pattern.\d\ d\ d - \d\ d\ d - \d\ d\ d\ d

2._

\
d {
   3
} - \d {
   3
} - \d {
   4
}

All the regex functions in Python are in the re module

import re

Output:

Phone number found: 415 - 555 - 4242

OUTPUT:

'415'

OUTPUT:

'415'

OUTPUT:

('415', '555-4242')

Suggestion : 3

I have a quick question on regex, I have a anycodings_regex certain string to match. It is shown below:,You can use re.sub to replace the first anycodings_python portion of the sentence if it starts anycodings_python with (square or round) brackets, with an anycodings_python empty string. No if statement is needed:,You can optionally match the (...) or anycodings_python [...] part before the first capturing anycodings_python group, as [])] only matches the optional anycodings_python closing one.,Is there a way to deal with this issue in anycodings_regex pure regex?

I have a quick question on regex, I have a anycodings_regex certain string to match. It is shown below:

"[someword] This Is My Name 2010"
or
   "This Is My Name 2010"
or
   "(someword) This Is My Name 2010"

What I have now, which I will use result = anycodings_regex re.search and then result.group() to get the anycodings_regex answer:

'[\]\)]? (.+) ([0-9]{4})\D'

Then you can capture all that follows in anycodings_python group 1, followed by matching the last 4 anycodings_python digits in group 2 and add a word anycodings_python boundary.

( ? : \([ ^ ()\ n] * \) | \[
   [ ^ ][\n] * \
]) ? (. + )([0 - 9] {
   4
})\ b

You can use re.sub to replace the first anycodings_python portion of the sentence if it starts anycodings_python with (square or round) brackets, with an anycodings_python empty string. No if statement is needed:

import re

s1 = "[someword] This Is My Name 2010"
s2 = "This Is My Name 2010"
s3 = "(someword) This Is My Name 2010"

reg = '\[.*?\] |\(.*?\) '

res1 = re.sub(reg, '', s1)
print(res1)

res2 = re.sub(reg, '', s2)
print(res2)

res3 = re.sub(reg, '', s3)
print(res3)

OUTPUT

This Is My Name 2010
This Is My Name 2010
This Is My Name 2010

Suggestion : 4

These sequences can be included inside a character class. For example, [\s,.] is a character class that will match any whitespace character, or ',' or '.'.,Here’s a complete list of the metacharacters; their meanings will be discussed in the rest of this HOWTO.,The first metacharacters we’ll look at are [ and ]. They’re used for specifying a character class, which is a set of characters that you wish to match. Characters can be listed individually, or a range of characters can be indicated by giving two characters and separating them by a '-'. For example, [abc] will match any of the characters a, b, or c; this is the same as [a-c], which uses a range to express the same set of characters. If you wanted to match only lowercase letters, your RE would be [a-z].,Metacharacters (except \) are not active inside classes. For example, [akm$] will match any of the characters 'a', 'k', 'm', or '$'; '$' is usually a metacharacter, but inside a character class it’s stripped of its special nature.

. ^ $ * + ? {} []\ | ()

>>>
import re
   >>>
   p = re.compile('ab*') >>>
   p
re.compile('ab*')

>>> p = re.compile('ab*', re.IGNORECASE)

>>>
import re
   >>>
   p = re.compile('[a-z]+') >>>
   p
re.compile('[a-z]+')

>>> p.match("") >>>
   print(p.match(""))
None

>>> m = p.match('tempo')
>>> m
<re.Match object; span=(0, 5), match='tempo'>