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Python regex if match for more than one criteria in the same regex string

I am currently learning python and do some excercices and have the following problem. I take user input for Password which should be at least 8 chars long, have capital letter, small letter and a special char.
What I would like to understand is, can I combine all of the above in one regex as below, or I need to list each and every case separately (see below).
Using only one:
whole_check = re.compile(r'''(
[A-Z] #Check for capital letter
\d #Check for number
\W #check for special character)''', re.VERBOSE)
So how can I do a multiple if match here. As example:
if not [A-Z]:
do something
if not \d:
do something
The only other option is if i define each category in a separate variable:
cap_letter = re.compile(r'[A-Z]')
small_letter = re.compile(r'[a-z]')
Thanks for clearing this for me.

See Regex for password policy. Generally the answer is: yes, you could put it into one regex, but you should consider not doing that, as it will be much easier to maintain and read/understand in a week if you don't do that :)

Python Regex: Find specific phrase in any form in text (including if followed by . or ,)

I'm trying to find when a specific product name is mentioned in customer notes (i.e. un-standardized, messy text). The product name is "Lending QB." Within the text, the product name can appear in any of the follow ways:
str1 ='Lending QB is a great product.'
str2 ='lending qb is great.'
str3 ='I don't think lendingqb is great.'
str4 ='I like Lending QB, but not always.'
str5 ='The best product is Lending qb.'
Here is the regex that mostly works:
df['lendingQB'] = df['Text'].str.findall('(?i)(?<!\S)lending\s?qb(?!\S)', re.IGNORECASE)
Using regex101.com to test, and confirming within my Python program, I can capture the product name in strings (str) 1-3, but not 4 and 5; which makes me believe the issue is with not finding the product name when it's followed by a punctuation mark.
My understanding is the \S would include commas and periods.
I tried adding |[,.] to the regex but then nothing matches:
'(?i)(?<!\S)lending\s?qb(?!\S|[,.])'
(I realize the IGNORECASE is redundant, but to test with regex101.com, I added the "(?i)")
Any suggestions?
AC

The pattern (?!\S) uses a negative lookahead to check what follows is not a non whitespace character.
What you could so is replace the (?!\S) with a word boundary \b to let it not be part of a larger match:
(?i)(?<!\S)lending\s?qb\b
Regex demo
Another way could be to use a positive lookahead to check for a whitespace character or ., or the end of the string using (?=[\s,.]|$)
For example:
str5 ="The best product is Lending qb."
print(re.findall(r'(?<!\S)lending\s?qb(?=[\s,.]|$)', str5, re.IGNORECASE)) # ['Lending qb']

You have correctly identified one issue in the regex (punctuation immediately after QB), but there is a second edge case to consider given that the input is messy -- what if there are multiple spaces in Lending QB?.
I believe the most robust solution to your problem is:
(?i)(?<!\S)lending\s*qb\b
\b enforces that QB occur at the end of a word, automatically considering punctuation.
\s? was replaced with \s* to allow any amount of whitespace to be
a match, rather than just zero-to-one whitespaces.
PS. Another point to consider is that \b terminates on all punctuation, (?=\s|[,.]) will only terminate on the given punctuation: , or . in this case. Given the wide range of possible punctuation (colon, semicolon, dash, hyphen, emdash...) I would strongly recommend \b over (?=\s|[,.]). Unless you want precise control over allowable terminating punctuation of course...
PPS. further test cases to illustrate my points
str6 ='Lending Qb: simply the best'
str7 ='I'm a fan of lending QB'

This (?!\S) is a forward whitespace boundary.
It is really this (?![^\s]) a negative of a negative
with the added benefit of it matching at the EOS (end of string).
What that means is you can use the negative class form to add characters
that qualify as a boundary.
So, just put the period and comma in with the whitespace.
(?i)(?<![^\s,.])lending\s?qb(?![^\s,.])
https://regex101.com/r/BrOj2J/1
As a tutorial point, this concept encapsulates multiple assertions
and is basic engine Boolean class logic which speeds up the engine
by a ten fold factor by comparison.

Thank you "The fourth bird", "sln", and "Mark_Anderson". Your answers provided solutions and also were very educational. I went with Mark's answer since it seemed to be the most robust, which is where I'm trying to get to. Ideally, I do want to capture all cases when the product name is mentioned, no matter how messy it's typed.
I changed my code to this:
df['lendingQB'] = df['Text'].str.findall(r'(?i)(?<!\S)lending\s*qb\b', re.IGNORECASE)

Regular expression pattern questions?

I am having a hard time understanding regular expression pattern. Could someone help me regular expression pattern to match all words ending in s. And start with a and end with a (like ana).
How do I write ending?

Word boundaries are given by \b so the following regex matches words ending with ing or s: "\b(\w+?(?:ing|s))\b" where as \b is a word boundary, \w+ is one or more "word character" and (?:ing|s) is an uncaptured group of either ing or s.
As you asked "how to develop a regex":
First: Don't use regex for complex tasks. They are hard to read, write and maintain. For example there is a regex that validates email addresses - but its computer generated and nothing you should use in practice.
Start simple and add edge cases. At the beginning plan what characters you need to use: You said you need words ending with s or ing. So you probably need something to represent a word, endings of words and the literal characters s and ing. What is a word? This might change from case to case, but at least every alphabetical character. Looking up in the python documentation on regexes you can find \w which is [a-zA-Z0-9_], which fits my impression of a word character. There you can also find \b which is a word boundary.
So the "first pseudo code try" is something like \b\w...\w\b which matches a word. We still need to "formalize" ... which we want to have the meaning of "one ore more characters", which directly translates to \b\w+\b. We can now match a word! We still need the s or ing. | translates to or, so how is the following: \b\w+ing|s\b? If you test this, you'll see that it will match confusing things like ingest which should not match our regex. What is happening? As you probably already saw the | can't know "which part it should or", so we need to introduce parenthesis: \b\w+(ing|s)\b. Congratulations, you have now arrived at a working regex!
Why (and how) does this differ from the example I gave first? First I wrote \w+? instead of \w+, the ? turns the + into a non-greedy version. If you know what the difference between greedy and non greedy is, skip this paragraph. Consider the following: AaAAbA and we want to match the things enclosed with big letter A. A naive try: A\w+A, so one or more word characters enclosed with A. This matches AaA, but also AaAAbA, A is still something that can be matched by \w. Without further config the *+? quantifier all try to match as much as possible. Sometimes, like in the A example, you don't want that, you can then use a ? after the quantifier to signal you want a non-greedy version, a version that matches as little as possible.
But in our case this isn't needed, the words are well seperated by whitespaces, which are not part of \w. So in fact you can just let + be greedy and everything will be alright. If you use . (any character) you often need to be careful not to match to much.
The other difference is using (?:s|ing) instead of (s|ing). What does the ?: do here? It changes a capturing group to a non capturing group. Generally you don't want to get "everything" from the regex. Consider the following regex: I want to go to \w+. You are not interested in the whole sentence, but only in the \w+, so you can capture it in a group: I want to go to (\w+). This means that you are interested in this specific piece of information and want to retrieve it later. Sometimes (like when using |) you need to group expressions together, but are not interested in their content, you can then declare it as non capturing. Otherwise you will get the group (s or ing) but not the actual word!
So to summarize:
* start small
* add one case after another
* always test with examples
In fact I just tried re.findall(\b\w+(?:ing|s)\b, "fishing words") and it didn't work. \w+(?:ing|s) works. I've no idea why, maybe someone else can explain that. Regex are an arcane thing, only use them for easy and easy to test tasks.

Generally speaking I'd use \b to match "word boundaries" with \w which matches word components (short cut for [A-Za-z0-9_]). Then you can do an or grouping to match "s" or "ing". Result is:
/\b\w+(s|ing)\b/

What is the capitol of X? Regex

What I am trying to do is very simple, I think but I can't seem to get it to work.
My regex is:
"(?wW)hat is the Capital of (\w*?\s?\w*?)\?"
Which I am hoping will allow in things like "Russia" and "Costa Rica" to be in the capture group. Basically, I want to read in a question such as "what is the capitol of Argentina" and then be able to grab the word "Argentina" even if the sentence has a bunch of other stuff in it.
But I tried it and I entered "what is the Capital of russia?" and it said that string didn't match.

I think you are looking for this:
[wW]hat is the capitol of ([\w\s]*)\?
Your fundamental mistake is the mixing up of character classes and capture groups.
To look for a mixture of characters (like w or W) you want to use a character class like [wW]. This means when we are looking for word characters (\w = [a-zA-Z0-9_]) or whitespace characters (\s = [\r\n\t\f ]), we can simple say [\w\s].
The final issue would be your use of ? and * (repetition). First of all, they have no special meaning in the character classes so I removed them. * repeats 0+ characters (+ checks 1+), and ? makes the previous key optional. This means \w*? is unnecessary, since it is saying optionally 0+ matches.
Note, I used a capturing group (...) around the capitol name meaning we can reference the capitol from capture group 1.
Finally, we can use the i modifier to make our matches case-insensitive..the final expression may be simpler to understand:
/what is the capitol of ([a-z ]+)\?/i

This should match:
[wW]hat is the capitol of ([^?]+)\?

How to add tags to negated words in strings that follow "not", "no" and "never"

How do I add the tag NEG_ to all words that follow not, no and never until the next punctuation mark in a string(used for sentiment analysis)? I assume that regular expressions could be used, but I'm not sure how.
Input:It was never going to work, he thought. He did not play so well, so he had to practice some more.
Desired output:It was never NEG_going NEG_to NEG_work, he thought. He did not NEG_play NEG_so NEG_well, so he had to practice some more.
Any idea how to solve this?

To make up for Python's re regex engine's lack of some Perl abilities, you can use a lambda expression in a re.sub function to create a dynamic replacement:
import re
string = "It was never going to work, he thought. He did not play so well, so he had to practice some more. Not foobar !"
transformed = re.sub(r'\b(?:not|never|no)\b[\w\s]+[^\w\s]',
lambda match: re.sub(r'(\s+)(\w+)', r'\1NEG_\2', match.group(0)),
string,
flags=re.IGNORECASE)
Will print (demo here)
It was never NEG_going NEG_to NEG_work, he thought. He did not NEG_play NEG_so NEG_well, so he had to practice some more. Not NEG_foobar !
Explanation
The first step is to select the parts of your string you're interested in. This is done with
\b(?:not|never|no)\b[\w\s]+[^\w\s]
Your negative keyword (\b is a word boundary, (?:...) a non capturing group), followed by alpahnum and spaces (\w is [0-9a-zA-Z_], \s is all kind of whitespaces), up until something that's neither an alphanum nor a space (acting as punctuation).
Note that the punctuation is mandatory here, but you could safely remove [^\w\s] to match end of string as well.
Now you're dealing with never going to work, kind of strings. Just select the words preceded by spaces with
(\s+)(\w+)
And replace them with what you want
\1NEG_\2

I would not do this with regexp. Rather I would;
Split the input on punctuation characters.
For each fragment do
Set negation counter to 0
Split input into words
For each word
Add negation counter number of NEG_ to the word. (Or mod 2, or 1 if greater than 0)
If original word is in {No,Never,Not} increase negation counter by one.

You will need to do this in several steps (at least in Python - .NET languages can use a regex engine that has more capabilities):
First, match a part of a string starting with not, no or never. The regex \b(?:not?|never)\b([^.,:;!?]+) would be a good starting point. You might need to add more punctuation characters to that list if they occur in your texts.
Then, use the match result's group 1 as the target of your second step: Find all words (for example by splitting on whitespace and/or punctuation) and prepend NEG_ to them.
Join the string together again and insert the result in your original string in the place of the first regex's match.