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Combining positive and negative lookahead in python

I'm trying to extract tokens that satisfy many conditions out of which, I'm using lookahead to implement the following two conditions:
The tokens must be either numeric/alphanumeric (i.e, they must have at least one digit). They can contain few special characters like - '-','/','\','.','_' etc.,
I want to match strings like: 165271, agya678, yah#123, kj*12-
The tokens can't have consecutive special characters like: ajh12-&
I don't want to match strings like: ajh12-&, 671%&i^
I'm using a positive lookahead for the first condition: (?=\w*\d\w*) and a negative lookahead for the second condition: (?!=[\_\.\:\;\-\\\/\#\+]{2})
I'm not sure how to combine these two look-ahead conditions.
Any suggestions would be helpful. Thanks in advance.
Edit 1 :
I would like to extract complete tokens that are part of a larger string too (i.e., They may be present in middle of the string).
I would like to match all the tokens in the string:
165271 agya678 yah#123 kj*12-
and none of the tokens (not even a part of a token) in the string: ajh12-& 671%&i^
In order to force the regex to consider the whole string I've also used \b in the above regexs : (?=\b\w*\d\w*\b) and (?!=\b[\_\.\:\;\-\\\/\#\+]{2}\b)

You can use
^(?!=.*[_.:;\-\\\/#+*]{2})(?=[^\d\n]*\d)[\w.:;\-\\\/#+*]+$
Regex demo
The negative lookahead (?=[^\d\n]*\d) matches any char except a digit or a newline use a negated character class, and then match a digit.
Note that you also have to add * and that most characters don't have to be escaped in the character class.
Using contrast, you could also turn the first .* into a negated character class to prevent some backtracking
^(?!=[^_.:;\-\\\/#+*\n][_.:;\-\\\/#+*]{2})(?=[^\d\n]*\d)[\w.:;\-\\\/#+*]+$
Edit
Without the anchors, you can use whitespace boundaries to the left (?<!\S) and to the right (?!\S)
(?<!\S)(?!=\S*[_.:;\-\\\/#+*]{2})(?=[^\d\s]*\d)[\w.:;\-\\\/#+*]+(?!\S)
Regex demo

You can use multiple look ahead assertions to only capture strings that
(?!.*(?:\W|_){2,}.*) - doesn't have consecutive special characters and
(?=.*\d.*) - has at least 1 digit
^(?!.*(?:\W|_){2,}.*)(?=.*\d.*).*$

Regex to match (French) numbers

I'm trying to find a simple (not perfect) pattern to recognise French numbers in a French text. French numbers use comma for the Anglo-Saxon decimal, and use dot or space for the thousand separator. \u00A0 is non-breaking space, also often used in French documents for the thousand separator.
So my first attempt is:
number_pattern = re.compile(r'\d[\d\., \u00A0]*\d', flags=re.UNICODE)
... but the trouble is that this doesn't then match a single digit.
But if I do this
number_pattern = re.compile(r'\d[\d\., \u00A0]*\d?', flags=re.UNICODE)
it then picks up trailing space (or NBS) characters (or for that matter a trailing comma or full stop).
The thing is, the pattern must both START and END with a digit, but it is possible that these may be the SAME character.
How might I achieve this? I considered a two-stage process where you try to see whether this is in fact a single-digit number... but that in itself is not trivial: if followed by a space, NBS, comma or dot, you then have to see whether the character after that, if there is one, is or is not a digit.
Obviously I'm hoping to find a solution which involves only one regex: if there is only one regex, it is then possible to do something like:
doubled_dollars_plain_text = plain_text.replace('$', '$$')
substituted_plain_text = re.sub(number_pattern, '$number', doubled_dollars_plain_text)
... having to use a two-stage process would make this much more lengthy and fiddly.
Edit
I tried to see whether I could implement ThierryLathuille's idea, so I tried:
re.compile(r'(\d(?:[\d\., \u00A0]*\d)?)', flags=re.UNICODE)
... this seems to work pretty well. Unlike JvdV's solution it doesn't attempt to check that thousand separators are followed by 3 digits, and for that matter you could have a succession of commas and spaces in the middle and it would still pass, which is quite problematic when you have a list of numbers separated by ", ". But it's acceptable for certain purposes... until something more sophisticated can be found.
I wonder if there's a way of saying "any non-digit in this pattern must be on its own" (i.e. must be bracketed between two digits)?

What about:
\d{1,3}(?:[\s.]?\d{3})*(?:,\d+)?(?!\d)
See an online demo
\d{1,3} - 1-3 digits.
(?: - Open 1st non-capture group:
[\s.]? - An optional whitespace or literal dot. Note that with unicode \s should match \p{Z} to include the non-breaking whitespace.
\d{3} - Three digits.
)* - Close 1st non-capture group and match 0+ times.
(?:,\d+)? - A 2nd optional non-capture group to match a comma followed by at least 1 digit.
(?!\d) - A negative lookahead to prevent trailing digits.

Very much inspired by JvdV's answer, I suggest this:
number_pattern = re.compile(r'(\d{1,3}(?:(?:[. \u00A0])?\d{3})*(?:,\d+)?(?!\d))', flags=re.UNICODE)
... this makes the thousand separator optional, and also makes thousand groups optional. It restricts the thousand-separator to 3 possible characters: dot, space and NBS, which is necessary for French numbers as found in practice.
PS I just found today that in fact Swiss French-speakers appear sometimes to use an apostrophe (of which there are several candidates in the vastness of Unicode) as a thousand separator.

Negative lookahead not working after character range with plus quantifier

I am trying to implement a regex which includes all the strings which have any number of words but cannot be followed by a : and ignore the match if it does. I decided to use a negative look ahead for it.
/([a-zA-Z]+)(?!:)/gm
string: lame:joker
since i am using a character range it is matching one character at a time and only ignoring the last character before the : .
How do i ignore the entire match in this case?
Link to regex101: https://regex101.com/r/DlEmC9/1

The issue is related to backtracking: once your [a-zA-Z]+ comes to a :, the engine steps back from the failing position, re-checks the lookahead match and finds a match whenver there are at least two letters before a colon, returning the one that is not immediately followed by :. See your regex demo: c in c:real is not matched as there is no position to backtrack to, and rea in real:c is matched because a is not immediately followed with :.
Adding implicit requirement to the negative lookahead
Since you only need to match a sequence of letters not followed with a colon, you can explicitly add one more condition that is implied: and not followed with another letter:
[A-Za-z]+(?![A-Za-z]|:)
[A-Za-z]+(?![A-Za-z:])
See the regex demo. Since both [A-Za-z] and : match a single character, it makes sense to put them into a single character class, so, [A-Za-z]+(?![A-Za-z:]) is better.
Preventing backtracking into a word-like pattern by using a word boundary
As #scnerd suggests, word boundaries can also help in these situations, but there is always a catch: word boundary meaning is context dependent (see a number of ifs in the word boundary explanation).
[A-Za-z]+\b(?!:)
is a valid solution here, because the input implies the words end with non-word chars (i.e. end of string, or chars other than letter, digits and underscore). See the regex demo.
When does a word boundary fail?
\b will not be the right choice when the main consuming pattern is supposed to match even if glued to other word chars. The most common example is matching numbers:
\d+\b(?!:) matches 12 in 12,, but not in 12:, and also 12c and 12_
\d+(?![\d:]) matches 12 in 12, and 12c and 12_, not in 12: only.

Do a word boundary check \b after the + to require it to get to the end of the word.
([a-zA-Z]+\b)(?!:)
Here's an example run.

Python regex: using or statement

I may not being saying this right (I'm a total regex newbie). Here's the code I currently have:
bugs.append(re.compile("^(\d+)").match(line).group(1))
I'd like to add to the regex so it looks at either '\d+' (starts with digits) or that it starts with 2 capital letters and contains a '-' before the first whitespace. I have the regex for the capital letters:
^[A-Z]{2,}
but I'm not sure how to add the '-' and the make an OR with the \d+. Does this make sense? Thanks!

The way to do an OR in regexps is with the "alternation" or "pipe" operator, |.
For example, to match either one or more digits, or two or more capital letter:
^(\d+|[A-Z]{2,})
Debuggex Demo
You may or may not sometimes need to add/remove/move parentheses to get the precedence right. The way I've written it, you've got one group that captures either the digit string or the capitals. While you're learning the rules (in fact, even after you've learned the rules) it's helpful to look at a regular expression visualizer/debugger like the one I used.
Your rule is slightly more complicated: you want 2 or more capital letters, and a hyphen before the first space. That's a bit hard to write as is, but if you change it to two or more capital letters, zero or more non-space characters, and a hyphen, that's easy:
^(\d+|[A-Z]{2,}\S*?-)
Debuggex Demo
(Notice the \S*?—that means we're going to match as few characters as possible, instead of as many as possible, so we'll only match up to the first hyphen in THIS-IS-A-TEST instead of up to the last. If you want the other one, just drop the ?.)

Write | for "or". For a sequence of zero or more non-whitespace characters, write \S*.
re.compile('^(\d+|[A-Z][A-Z]\S*-\s)')

re.compile(r"""
^ # beginning of the line
(?: # non-capturing group; do not return this group in .group()
(\d+) # one or more digits, captured as a group
| # Or
[A-Z]{2} # Exactly two uppercase letters
\S* # Any number of non-whitespace characters
- # the dash you wanted
) # end of the non-capturing group
""",
re.X) # enable comments in the regex

Looking for a regular expression including alphanumeric + "&" and ";"

Here's the problem:
split=re.compile('\\W*')
This regular expression works fine when dealing with regular words, but there are occasions where I need the expression to include words like käyttäj&aml;auml;.
What should I add to the regex to include the & and ; characters?

I would treat the entities as a unit (since they also can contain numerical character codes), resulting in the following regular expression:
(\w|&(#(x[0-9a-fA-F]+|[0-9]+)|[a-z]+);)+
This matches
either a word character (including “_”), or
an HTML entity consisting of
the character “&”,
the character “#”,
the character “x” followed by at least one hexadecimal digit, or
at least one decimal digit, or
at least one letter (= named entity),
a semicolon
at least once.
/EDIT: Thanks to ΤΖΩΤΖΙΟΥ for pointing out an error.

You probably want to take the problem reverse, i.e. finding all the character without the spaces:
[^ \t\n]*
Or you want to add the extra characters:
[a-zA-Z0-9&;]*
In case you want to match HTML entities, you should try something like:
(\w+|&\w+;)*

you should make a character class that would include the extra characters. For example:
split=re.compile('[\w&;]+')
This should do the trick. For your information
\w (lower case 'w') matches word characters (alphanumeric)
\W (capital W) is a negated character class (meaning it matches any non-alphanumeric character)
* matches 0 or more times and + matches one or more times, so * will match anything (even if there are no characters there).

Looks like this RegEx did the trick:
split=re.compile('(\\\W+&\\\W+;)*')
Thanks for the suggestions. Most of them worked fine on Reggy, but I don't quite understand why they failed with re.compile.