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Regex to parse queries with quoted substrings and return nested lists of individual words

I'm trying to write a regex that takes in a string of words containing quoted substrings like "green lizards" like to sit "in the sun", tokenizes it into words and quoted substrings (using either single or double quotes) separated by spaces, and then returns a list [['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']] where the list items are either single words or nested lists of words where a quoted substrings was encountered.
I am new to regex, and was able to find a solution that captures the quoted parts: re.findall('"([^"]*)"', '"green lizards" like to sit "in the sun"') ... which returns: ['green lizards', 'in the sun']
But this doesn't capture the individual words, and also doesn't tokenize them (returning a single string instead of list of words, which requires me to split() them each separately.
How would I make a regex that correctly returns the type of list I'm wanting? Also, I'm open to better methods/tools than regex for parsing these sorts of strings if anyone has suggestions.
Thanks!

With re.findall() function and built-in str methods:
import re
s = '"green lizards" like to sit "in the sun"'
result = [i.replace('"', "").split() if i.startswith('"') else i
for i in re.findall(r'"[^"]+"|\S+', s)]
print(result)
The output:
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']]

Another approach (supporting both single and double quotes):
import re
sentence = """"green lizards" like to sit "in the sun" and 'single quotes' remain alone"""
rx = re.compile(r"""(['"])(.*?)\1|\S+""")
tokens = [m.group(2).split()
if m.group(2) else m.group(0)
for m in rx.finditer(sentence)]
print(tokens)
Yielding
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun'], 'and', ['single', 'quotes'], 'remain', 'alone']
The idea here is:
(['"]) # capture a single or a double quote
(.*?) # 0+ characters lazily
\1 # up to the same type of quote previously captured
| # ...or...
\S+ # not a whitespace
In the list comprehension we check which condition was met.

You can use re.split and then a final str.split:
import re
s = '"green lizards" like to sit "in the sun"'
new_s = [[i[1:-1].split()] if i.startswith('"') else i.split() for i in re.split('(?<=")\s|\s(?=")', s)]
last_result = [i for b in new_s for i in b]
Output:
[['green', 'lizards'], 'like', 'to', 'sit', ['in', 'the', 'sun']]

Regex in Python to match words with special characters

I have this code
import re
str1 = "These should be counted as a single-word, b**m !?"
match_pattern = re.findall(r'\w{1,15}', str1)
print(match_pattern)
I want the output to be:
['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']
The output should exclude non-words such as the "!?" what are the other validation should I use to match and achieve the desired output?

I would use word boundaries (\b) filled with 1 or more non-space:
match_pattern = re.findall(r'\b\S+\b', str1)
result:
['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']
!? is skipped thanks to word boundary magic, which don't consider that as a word at all either.

Probably you want something like [^\s.!?] instead of \w but what exactly you want is not evident from a single example. [^...] matches a single character which is not one of those between the brackets and \s matches whitespace characters (space, tab, newline, etc).

You can also achieve a similar result not using RegEx:
string = "These should be counted as a single-word, b**m !?"
replacements = ['.',',','?','!']
for replacement in replacements:
if replacement in string:
string = string.replace(replacement, "");
print string.split()
>>> ['These', 'should', 'be', 'counted', 'as', 'a', 'single-word', 'b**m']

Regex parsing text and get relevant words / characters

I want to parse a file, that contains some programming language. I want to get a list of all symbols etc.
I tried a few patterns and decided that this is the most successful yet:
pattern = "\b(\w+|\W+)\b"
Using this on my text, that is something like:
string = "the quick brown(fox).jumps(over + the) = lazy[dog];"
re.findall(pattern, string)
will result in my required output, but I have some chars that I don't want and some unwanted formatting:
['the', ' ', 'quick', ' ', 'brown', '(', 'fox', ').', 'jumps', 'over',
' + ', 'the', ') = ', 'lazy', '[', 'dog']
My list contains some whitespace that I would like to get rid of and some double symbols, like (., that I would like to have as single chars. Of course I have to modify the \W+ to get this done, but I need a little help.
The other is that my regex doesn't match the ending ];, which I also need.

Why use \W+ for one or more, if you want single non-word characters in output? Additionally exclude whitespace by use of a negated class. Also it seems like you could drop the word boundaries.
re.findall(r"\w+|[^\w\s]", str)
This matches
\w+ one or more word characters
|[^\w\s] or one character, that is neither a word character nor a whitespace
See Ideone demo

python: splitting strings where numbers and letters meet (1234abcd-->1234, abcd)

I have a string composed of numbers and letters: string = 'this1234is5678it', and I would like the string.split() output to give me a list like ['this', '1234', 'is', '5678', 'it'], splitting at where numbers and letter meet. Is there an easy way to do this?

You can use Regex for this.
import re
s = 'this1234is5678it'
re.split('(\d+)',s)
Running example http://ideone.com/JsSScE
Outputs ['this', '1234', 'is', '5678', 'it']
Update
Steve Rumbalski mentioned in the comment the importance of the parenthesis in the regex. He quotes from the documentation:
If capturing parentheses are used in pattern, then the text of all
groups in the pattern are also returned as part of the resulting
list." Without the parenthesis the result would be ['this', 'is',
'it'].

python split a text file function

I wrote a tokenize function that basically reads a string representation and splits it into list of words.
My code:
def tokenize(document):
x = document.lower()
return re.findall(r'\w+', x)
My output:
tokenize("Hi there. What's going on? first-class")
['hi', 'there', 'what', 's', 'going', 'on', 'first', 'class']
Desired Output:
['hi', 'there', "what's", 'going', 'on', 'first-class']
Basically I want the apostrophed words and hypen words to remain as a single word in list along with double quotes. How can i change my function to get the desired output.

\w+ matches one or more word characters; this does not include apostrophes or hyphens.
You need to use a character set here to tell Python exactly what you want to match:
>>> import re
>>> def tokenize(document):
... return re.findall("[A-Za-z'-]+", document)
...
>>> tokenize("Hi there. What's going on? first-class")
['hi', 'there', "what's", 'going', 'on', 'first-class']
>>>
You'll notice too that I removed the x = document.lower() line. This is no longer necessary since we can match uppercase characters by simply adding A-Z to the character set.