You've got a few problems there.

First, matches are case-sensitive unless you use the IGNORECASE/I flag to ignore case. So, 'AND' doesn't match 'and'.

Also, unless you use the VERBOSE/X flag, those spaces are part of the pattern. So, you're checking for 'AND ', not 'AND'. If you wanted that, you probably wanted spaces on each side, not just those sides (otherwise, 'band leader' is going to match…), and really, you probably wanted \b, not a space (otherwise a sentence starting with 'And another thing' isn't going to match).

Finally, if you think you need .* before and after your pattern and $ and ^ around it, there's a good chance you wanted to use search, findall, or finditer, rather than match.

So:

>>> s = "These are oranges and apples and pears, but not pinapples or .."
>>> r = re.compile(r'\bAND\b | \bOR\b | \bNOT\b', flags=re.I | re.X)
>>> r.findall(s)
['and', 'and', 'not', 'or']

Debuggex Demo

How about

^[(Administrators)(LOCAL SERVICE)(NETWORK SERVICE),\s]+$

or

^(Administrators|LOCAL SERVICE|NETWORK SERVICE|[,\s])+$

see working at https://regex101.com/r/oJF0aW/2/tests

both versions basically say that the whole string must contain only the specified user names as well as commas and whitespace.

You can use

(?=.*test)(?=.*long)

Source: MySQL SELECT LIKE or REGEXP to match multiple words in one record

You may leverage a non-capturing alternation group to match either VERSION or FREQ (optionally preceded with a word boundary, just check if it meets your requirements):

\b(?:VERSION|FREQ)="(.*?)"

See the regex demo

Details

• \b - a leading word boundary
• (?:VERSION|FREQ) - either VERSION or FREQ
• =" - a =" substring
• (.*?) - Group 1 (the actual output of findall): any 0+ chars other than line break chars, as few as possible
• " - a double quote.

See the Python demo:

import re
line='VERSION="OTHER" POWER="LOW" FREQ="OFF" MAXTUN="BLER"'
print(re.findall(r'\b(?:VERSION|FREQ)="(.*?)"', line))
# => ['OTHER', 'OFF']

A better idea, perhaps, is to capture key-value pairs and map them to a dictionary:

import re
line = 'VERSION="OTHER" POWER="LOW" FREQ="OFF" MAXTUN="BLER"'
results = re.findall(r'(VERSION|FREQ)="(.*?)"', line)
print(dict(results))
# => {'FREQ': 'OFF', 'VERSION': 'OTHER'}

See the Python demo.

You may use this regex for search using case insensitive switch:

\b(?:apples|pears|bananas)(?:(?:\s+(?:or|and)\s+|\s*,\s*)+(?:apples|pears|bananas)\b)*

and replace this with: [\g<0>]

RegEx Demo

Code:

import re

regex = r"\b(?:apples|pears|bananas)(?:(?:\s+(?:or|and)\s+|\s*,\s*)+(?:apples|pears|bananas))*"

subst = r"[\g<0>]"

result = re.sub(regex, subst, test_str, 0, re.IGNORECASE)

You could have tried looking for password validating regex, the site has a lot of them ;)

That said, you can use positive lookaheads to do that:

re.search(r"(?=.*[a-z])(?=.*[A-Z])(?=.*\d)", "1Az")

And to actually match the string...

re.search(r"(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{3}", "1Az")

And now, to ensure that the password is 3 chars long:

re.search(r"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{3}$", "1Az")

A positive lookahead (?= ... ) makes sure the expression inside is present in the string to be tested. So, the string has to have a lowercase character ((?=.*[a-z])), an uppercase character ((?=.*[A-Z])) and a digit ((?=.*\d)) for the regex to 'pass'.