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Filter out words depending on surrounding punctuation

Objective:
I'm looking for a way to match or skip words based on whether or not they are surrounded by quotations marks ' ', guillemets « » or parentheses ( ).
Examples of desired results:
len(re.findall("my word", "blablabla 'my word' blablabla")) should return 0 because linguistically speaking my word =/= 'my word' and hence shouldn't be matched;
len(re.findall("'my word'", "blablabla 'my word' blablabla")) should return 1 because linguistically speaking 'my word' = 'my word' and hence should be matched;
But here's the catch — both len(re.findall("my word", "blablabla «my word» blablabla")) and len(re.findall("my word", "blablabla (my word) blablabla")) should return 1.
My attempt:
I have the following expression (correct me if I'm wrong) at my disposal but am clueless as to how to implement it: (?<!\w)'[^ ].*?\w*?[^ ]'
I wish to make the following code len(re.findall(r'(?<!\w)'+re.escape(myword)+r'(?!\w)', sentence)) – whose aim is to strip out punctuation marks I believe – take into account all of the aforementioned situations.
For now, my code detects my word inside of 'my word' which is not what I want.
Thanks in advance!

I think one of the strategies is to use negative look-ahead feature:
my_word = "word"
r"(?!'" + my_word + "')[^']" + "my_word"
This should do the job as you can check here.
Since negative look-ahead does not consume characters, to prevent a match you need to use [^'] to ensure the quotation mark ' is not an allowed character preceding your my_word. The ^ starting an enumeration of characters means precisely that.
If you want to expand the list of quotation marks that should cause the word not to be counted as found it is enough that you change ' into a list of disallowed characters:
r"(?!['`]" + my_word + "['`])[^'`]my_word"
It is worth noting that the example from #Prasanna question is going to be impossible to match using regex. You would need to use a proper parser - e.g. pyparsing - to handle such situations because regular expressions are not able to handle a match that requires two arbitrary counts of characters to match (e.g. any number of 'a' followed by the same number of 'b' letters) and it will not be possible to create a generic regular expression with a look-ahead that handles n words then myword and at the same time skips n words if they are preceded by a quotation mark).

Replacing specific characters after a string match

I'm looking to replace specific characters in numbers I'm extracting but I cannot figure out how to do so.
Here "," is the float separator and (' or .) are thousands separators. I can match this way :
>>> myString = "I buy 456'123,45 then 45.654 then 123. I'm 30."
>>> re.findall(r"(?:\d+)(?:['|.]\d+)+(?:[,]\d+)?", myString)
["456'123,45", '45.654']
I want to replace in my string all thousands separator to get this :
>>> newString
"I buy 456123,45 then 45654 then 123. I'm 30."
I'm pretty sure I need to use groups and subgroups in order to replace what I want but I don't know how to deal with groups when "()+" is present, the length of the number can also be very long
(e.g : 123'456'789'123'456'789,123)
Thanks

You may use re.sub with
(?<=\d)['.](?=\d)
and replace with an empty string. See the regex demo.
Details
(?<=\d) - (positive lookbehind) a digit must appear immediately to the left of the current location
['.] - a single quote or a dot
(?=\d) - (positive lookahead) a digit must appear immediately to the right of the current location.
Python:
re.sub(r"(?<=\d)['.](?=\d)", "", myString)

How to group inside "or" matching in a regex?

I have two kinds of documents to parse:
1545994641 INFO: ...
and
'{"deliveryDate":"1545994641","error"..."}'
I want to extract the timestamp 1545994641 from each of them.
So, I decided to write a regex to match both cases:
(\d{10}\s|\"\d{10}\")
In the 1st kind of document, it matches the timestamp and groups it, using the first expression in the "or" above (\d{10}\s):
>>> regex = re.compile("(\d{10}\s|\"\d{10}\")")
>>> msg="1545994641 INFO: ..."
>>> regex.search(msg).group(0)
'1545994641 '
(So far so good.)
However, in the 2nd kind, using the second expression in the "or" (\"\d{10}\") it matches the timestamp and quotation marks, grouping them. But I just want the timestamp, not the "":
>>> regex = re.compile("(\d{10}\s|\"\d{10}\")")
>>> msg='{"deliveryDate":"1545994641","error"..."}'
>>> regex.search(msg).group(0)
'"1545994641"'
What I tried:
I decided to use a non-capturing group for the quotation marks:
(\d{10}\s|(?:\")\d{10}(?:\"))
but it doesn't work as the outer group catches them.
I also removed the outer group, but the result is the same.
Unwanted ways to solve:
I can surpass this by creating a group for each expression in the or,
but I just want it to output a single group (to abstract the code
from the regex).
I could also use a 2nd step of regex to capture the timestamp from
the group that has the quotation marks, but again that would break
the code abstraction.
I could omit the "" in the regex but that would match a timestamp in the middle of the message , as I want it to be objective to capture the timestamp as a value of a key or in the beginning of the document, followed by a space.
Is there a way I can match both cases above but, in the case it matches the second case, return only the timestamp? Or is it impossible?
EDIT:
As noticed by #Amit Bhardwaj, the first case also returns a space after the timestamp. It's another problem (I didn't figure out) with the same solution, probably!

You may use lookarounds if your code can only access the whole match:
^\d{10}(?=\s)|(?<=")\d{10}(?=")
See the regex demo.
In Python, declare it as
rx = r'^\d{10}(?=\s)|(?<=")\d{10}(?=")'
Pattern details
^\d{10}(?=\s):
^ - string start
\d{10} - ten digits
(?=\s) - a positive lookahead that requires a whitespace char immediately to the right of the current location
| - or
(?<=")\d{10}(?="):
(?<=") - a " char
\d{10} - ten digits
(?=") - a positive lookahead that requires a double quotation mark immediately to the right of the current location.

You could use lookarounds, but I think this solution is simpler, if you can just get the group:
"?(\d{10})(?:\"|\s)
EDIT:
Considering if there is a first " there must be a ", try this:
(^\d{10}\s|(?<=\")\d{10}(?=\"))
EDIT 2:
To also remove the trailing space in the end, use a lookahead too:
(^\d{10}(?=\s)|(?<=\")\d{10}(?=\"))

Extract text between double square brackets in Python

If I have a string that may look like this:
"[[Category:Political culture]]\n\n [[Category:Political ideologies]]\n\n"
How do I extract the categories and put them into a list?
I'm having a hard time getting the regular expression to work.

To expand on the explanation of the regex used by Avinash in his answer:
Category:([^\[\]]*) consists of several parts:
Category: which matches the text "Category:"
(...) is a capture group meaning roughly "the expression inside this group is a block that I want to extract"
[^...] is a negated set which means "do not match any characters in this set".
\[ and \] match "[" and "]" in the text respectively.
* means "match zero or more of the preceding regex defined items"
Where I have used ... to indicate that I removed some characters that were not important for the explanation.
So putting it all together, the regex does this:
Finds "Category:" and then matches any number (including zero) characters after that that are not the excluded characters "[" or "]". When it hits an excluded character it stops and the text matched by the regex inside the (...) part is returned. So the regex does not actually look for "[[" or "]]" as you might expect and so will match even if they are left out. You could force it to look for the double square brackets at the beginning and end by changing it to \[\[Category:([^\[\]]*)\]\].
For the second regex, Category:[^\[\]]*, the capture group (...) is excluded, so Python returns everything matched which includes "Category:".

Seems like you want something like this,
>>> str = "[[Category:Political culture]]\n\n [[Category:Political ideologies]]\n\n"
>>> re.findall(r'Category:([^\[\]]*)', str)
['Political culture', 'Political ideologies']
>>> re.findall(r'Category:[^\[\]]*', str)
['Category:Political culture', 'Category:Political ideologies']
By default re.findall will print only the strings which are matched by the pattern present inside a capturing group. If no capturing group was present, then only the findall function would return the matches in list. So in our case , this Category: matches the string category: and this ([^\[\]]*) would capture any character but not of [ or ] zero or more times. Now the findall function would return the characters which are present inside the group index 1.

Python code:
s = "[[Category:Political culture]]\n\n [[Category:Political ideologies]]\n\n"
cats = [line.strip().strip("[").strip("]") for line in s.splitlines() if line]
print(cats)
Output:
['Category:Political culture', 'Category:Political ideologies']

Match regex with \\n in it

I have the following string:
>>> repr(s)
" NBCUniversal\\n63 VOLGAFILM, INC VOLGAFILMINC\\n64 Video Service Corp
I want to match the string before the \\n -- everything before a whitespace character. The output should be:
['NBCUniversal', 'VOLGAFILMINC']
Here is what I have so far:
re.findall(r'[^s].+\\n\d{1,2}', s)
What would be the correct regex for this?

EDIT: sorry I haven't read carefully your question
If you want to find all groups of letters immediatly before a literal \n, re.findall is appropriate. You can obtain the result you want with:
>>> import re
>>> s = " NBCUniversal\\n63 VOLGAFILM, INC VOLGAFILMINC\\n64 Video Service Corp "
>>> re.findall(r'(?i)[a-z]+(?=\\n)', s)
['NBCUniversal', 'VOLGAFILMINC']
OLD ANSWER:
re.findall is not the appropriate method since you only need one result (that is a pair of strings). Here the re.search method is more appropriate:
>>> import re
>>> s = " NBCUniversal\\n63 VOLGAFILM, INC VOLGAFILMINC\\n64 Video Service Corp "
>>> res = re.search(r'^(?i)[^a-z\\]*([a-z]+)\\n[^a-z\\]*([a-z]+)', s)
>>> res.groups()
('NBCUniversal', 'VOLGAFILM')
Note: I have assumed that there are no other characters between the first word and the literal \n, but if it isn't the case, you can add [^a-z\\]* before the \\n in the pattern.

If you want to fix your existing code instead of replace it, you're on the right track, you've just got a few minor problems.
Let's start with your pattern:
>>> re.findall(r'[^s].+\\n\d{1,2}', s)
[' NBCUniversal\\n63 VOLGAFILM, INC VOLGAFILMINC\\n64']
The first problem is that .+ will match everything that it can, all the way up to the very last \\n\d{1,2}, rather than just to the next \\n\d{1,2}. To fix that, add a ? to make it non-greedy:
>>> re.findall(r'[^s].+?\\n\d{1,2}', s)
[' NBCUniversal\\n63', ' VOLGAFILM, INC VOLGAFILMINC\\n64']
Notice that we now have two strings, as we should. The problem is, those strings don't just have whatever matched the .+?, they have whatever matched the entire pattern. To fix that, wrap the part you want to capture in () to make it a capturing group:
>>> re.findall(r'[^s](.+?)\\n\d{1,2}', s)
[' NBCUniversal', ' VOLGAFILM, INC VOLGAFILMINC']
That's nicer, but it still has a bunch of extra stuff on the left end. Why? Well, you're capturing everything after [^s]. That means any character except the letter s. You almost certainly meant [\s], meaning any character in the whitespace class. (Note that \s is already the whitespace class, so [\s], meaning the class consisting of the whitespace class, is unnecessary.) That's better, but that's still only going to match one space, not all the spaces. And it will match the earliest space it can that still leaves .+? something to match, not the latest. So if you want to suck all all the excess spaces, you need to repeat it:
re.findall(r'\s+(.+?)\\n\d{1,2}', s)
['NBCUniversal', 'VOLGAFILM, INC VOLGAFILMINC']
Getting closer… but the .+? matches anything, including the space between VOLGAFILM and VOLGAFILMINC, and again, the \s+ is going to match the first run of spaces it can, leaving the .+? to match everything after that.
You could fiddle with the prefix , but there's an easier solution. If you don't want spaces in your capture group, just capture a run of nonspaces instead of a run of anything, using \S:
re.findall(r'\s+(\S+?)\\n\d{1,2}', s)
['NBCUniversal', 'VOLGAFILMINC']
And notice that once you've done that, the \s+ isn't really doing anything anymore, so let's just drop it:
re.findall(r'(\S+?)\\n\d{1,2}', s)
['NBCUniversal', 'VOLGAFILMINC']
I've obviously made some assumptions above that are correct for your sample input, but may not be correct for real data. For example, if you had a string like Weyland-Yutani\\n…, I'm assuming you want Weyland-Yutani, not just Yutani. If you have a different rule, like only letters, just change the part in parentheses to whatever fits that rule, like (\w+?) or ([A-Za-z]+?).

Assuming that the input actually has the sequence \n (backslash followed by letter 'n') and not a newline, this will work:
>>> re.findall(r'(\S+)\\n', s)
['NBCUniversal', 'VOLGAFILMINC']
If the string actually contains newlines then replace \\n with \n in the regular expression.