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re.sub for string starting with special character

Sorry if this question seems too similar to other's I have found. This is a variation of using re.sub to replace exact characters in a string.
I have a string that looks like:
C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
I would like to only replace, for example, the '*:1' with 'Ar'. My current attempt looks like this:
smiles_all='C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
print(smiles_all)
new_smiles=re.sub('[*:]1','Ar',smiles_all)
print(new_smiles)
C1([*:5])C([*:6])C2=NC1=C([*Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*Ar0])C(=N4)C([*:3])=C5C([*Ar1])=C([*Ar2])C(=C2([*:4]))N5
As you can see, this is still changing the values that were previously 10,11, etc. I've tried different variations where I select [*:1], but that is also incorrect. Any help here would be greatly appreciated. In my current output, the * also remains. That needs to be swapped so that *:1 becomes Ar
Here is an example of what the output should be
C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
*Edit:
At one point this question was flagged as answered by this question:
Escaping regex string
When I implement re.escape as suggested, I still get an error:
new_smiles=re.sub(re.escape('*:1'),'Ar',smiles_all)
C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5
C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([Ar0])C(=N4)C([*:3])=C5C([Ar1])=C([Ar2])C(=C2([*:4]))N5

Given:
smiles_all='C1([*:5])C([*:6])C2=NC1=C([*:1])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
desired='C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5'
You are trying to replace the literal string [*:1] with [Ar]. In a regex, the expression [*:1] is a character class that matches a single one of the characters inside the class with one match. If you add any regex repetition to a character class, it will match those characters in any order up to the repetition limit.
The easiest way to to replace the literal [*:1] with [Ar] is to use Python's string methods:
>>> smiles_all.replace('[*:1]','[Ar]')==desired
True
If you want to use a regex, you need to escape those metacharaters to get a literal string:
>>> re.sub(r'\[\*:1\]', "[Ar]", smiles_all)==desired
True
Or let Python do the escaping for you:
>>> re.sub(re.escape(r'[*:1]'), "[Ar]", smiles_all)==desired
True

You can try:
re.sub(r"[*:]+1(?=])", "Ar", smiles_all)
Difference from yours is to allow 1+ repetitions of literal * and : followed by 1 which is also ensured to be followed by a ] via the ?=, i.e., positive lookahead.
to get
"C1([*:5])C([*:6])C2=NC1=C([Ar])C3=C([*:7])C([*:8])=C(N3)C([*:2])=C4C([*:9])=C([*:10])C(=N4)C([*:3])=C5C([*:11])=C([*:12])C(=C2([*:4]))N5"

Dynamically Removing string with regex python

I am currently having trouble removing the end of strings using regex. I have tried using .partition with unsuccessful results. I am now trying to use regex unsuccessfully. All the strings follow the format of some random words **X*.* Some more words. Where * is a digit and X is a literal X. For Example 21X2.5. Everything after this dynamic string should be removed. I am trying to use re.sub('\d\d\X\d.\d', string). Can someone point me in the right direction with regex and how to split the string?
The expected output should read:
some random words 21X2.5
Thanks!

Use following regex:
re.search("(.*?\d\dX\d\.\d)", "some random words 21X2.5 Some more words").groups()[0]
Output:
'some random words 21X2.5'

Your regex is not correct. The biggest problem is that you need to escape the period. Otherwise, the regex treats the period as a match to any character. To match just that pattern, you can use something like:
re.findall('[\d]{2}X\d\.\d', 'asb12X4.4abc')
[\d]{2} matches a sequence of two integers, X matches the literal X, \d matches a single integer, \. matches the literal ., and \d matches the final integer.
This will match and return only 12X4.4.
It sounds like you instead want to remove everything after the matched expression. To get your desired output, you can do something like:
re.split('(.*?[\d]{2}X\d\.\d)', 'some random words 21X2.5 Some more words')[1]
which will return some random words 21X2.5. This expression pulls everything before and including the matched regex and returns it, discarding the end.
Let me know if this works.

To remove everything after the pattern, i.e do exactly as you say...:
s = re.sub(r'(\d\dX\d\.\d).*', r'\1', s)
Of course, if you mean something else than what you said, something different will be needed! E.g if you want to also remove the pattern itself, not just (as you said) what's after it:
s = re.sub(r'\d\dX\d\.\d.*', r'', s)
and so forth, depending on what, exactly, are your specs!-)

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.

Python multiple repeat Error

I'm trying to determine whether a term appears in a string.
Before and after the term must appear a space, and a standard suffix is also allowed.
Example:
term: google
string: "I love google!!! "
result: found
term: dog
string: "I love dogs "
result: found
I'm trying the following code:
regexPart1 = "\s"
regexPart2 = "(?:s|'s|!+|,|.|;|:|\(|\)|\"|\?+)?\s"
p = re.compile(regexPart1 + term + regexPart2 , re.IGNORECASE)
and get the error:
raise error("multiple repeat")
sre_constants.error: multiple repeat
Update
Real code that fails:
term = 'lg incite" OR author:"http++www.dealitem.com" OR "for sale'
regexPart1 = r"\s"
regexPart2 = r"(?:s|'s|!+|,|.|;|:|\(|\)|\"|\?+)?\s"
p = re.compile(regexPart1 + term + regexPart2 , re.IGNORECASE)
On the other hand, the following term passes smoothly (+ instead of ++)
term = 'lg incite" OR author:"http+www.dealitem.com" OR "for sale'

The problem is that, in a non-raw string, \" is ".
You get lucky with all of your other unescaped backslashes—\s is the same as \\s, not s; \( is the same as \\(, not (, and so on. But you should never rely on getting lucky, or assuming that you know the whole list of Python escape sequences by heart.
Either print out your string and escape the backslashes that get lost (bad), escape all of your backslashes (OK), or just use raw strings in the first place (best).
That being said, your regexp as posted won't match some expressions that it should, but it will never raise that "multiple repeat" error. Clearly, your actual code is different from the code you've shown us, and it's impossible to debug code we can't see.
Now that you've shown a real reproducible test case, that's a separate problem.
You're searching for terms that may have special regexp characters in them, like this:
term = 'lg incite" OR author:"http++www.dealitem.com" OR "for sale'
That p++ in the middle of a regexp means "1 or more of 1 or more of the letter p" (in the others, the same as "1 or more of the letter p") in some regexp languages, "always fail" in others, and "raise an exception" in others. Python's re falls into the last group. In fact, you can test this in isolation:
>>> re.compile('p++')
error: multiple repeat
If you want to put random strings into a regexp, you need to call re.escape on them.
One more problem (thanks to Ωmega):
. in a regexp means "any character". So, ,|.|;|:" (I've just extracted a short fragment of your longer alternation chain) means "a comma, or any character, or a semicolon, or a colon"… which is the same as "any character". You probably wanted to escape the ..
Putting all three fixes together:
term = 'lg incite" OR author:"http++www.dealitem.com" OR "for sale'
regexPart1 = r"\s"
regexPart2 = r"(?:s|'s|!+|,|\.|;|:|\(|\)|\"|\?+)?\s"
p = re.compile(regexPart1 + re.escape(term) + regexPart2 , re.IGNORECASE)
As Ωmega also pointed out in a comment, you don't need to use a chain of alternations if they're all one character long; a character class will do just as well, more concisely and more readably.
And I'm sure there are other ways this could be improved.

The other answer is great, but I would like to point out that using regular expressions to find strings in other strings is not the best way to go about it. In python simply write:
if term in string:
#do whatever

Also make sure that your arguments are in the correct order!
I was trying to run a regular expression on some html code. I kept getting the multiple repeat error, even with very simple patterns of just a few letters.
Turns out I had the pattern and the html mixed up. I tried re.findall(html, pattern) instead of re.findall(pattern, html).

i have an example_str = "i love you c++" when using regex get error multiple repeat Error. The error I'm getting here is because the string contains "++" which is equivalent to the special characters used in the regex. my fix was to use re.escape(example_str ), here is my code.
example_str = "i love you c++"
regex_word = re.search(rf'\b{re.escape(word_filter)}\b', word_en)

A general solution to "multiple repeat" is using re.escape to match the literal pattern.
Example:
>>>> re.compile(re.escape("c++"))
re.compile('c\\+\\+')
However if you want to match a literal word with space before and after try out this example:
>>>> re.findall(rf"\s{re.escape('c++')}\s", "i love c++ you c++")
[' c++ ']

REGEX: Parsing n digits with non numeric word boundaries

I hope this message finds you in good spirits. I am trying to find a quick tutorial on the \b expression (apologies if there is a better term). I am writing a script at the moment to parse some xml files, but have ran into a bit of a speed bump. I will show an example of my xml:
<....></...><...></...><OrderId>123456</OrderId><...></...>
<CustomerId>44444444</CustomerId><...></...><...></...>
<...> is unimportant and non relevant xml code. Focus primarily on the CustomerID and OrderId.
My issue lies in parsing a string, similar to the above statement. I have a regexParse definition that works perfectly. However it is not intuitive. I need to match only the part of the string that contains 44444444.
My Current setup is:
searchPattern = '>\d{8}</CustomerId'
Great! It works, but I want to do it the right way. My thinking is 1) find 8 digits 2) if the some word boundary is non numeric after that matches CustomerId return it.
Idea:
searchPattern = '\bd{16}\b'
My issue in my tests is incorporating the search for CustomerId somewhere before and after the digits. I was wondering if any of you can either help me out with my issue, or point me in the right path (in words of a guide or something along the lines). Any help is appreciated.
Mods if this is in the wrong area apologies, I wanted to post this in the Python discussion because I am not sure if Python regex supports this functionality.
Thanks again all,
darcmasta

txt = """
<....></...><...></...><OrderId>123456</OrderId><...></...>
<CustomerId>44444444</CustomerId><...></...><...></...>
"""
import re
pattern = "<(\w+)>(\d+)<"
print re.findall(pattern,txt)
#output [('OrderId', '123456'), ('CustomerId', '44444444')]

You might consider using a look-back operator in your regex to make it easy for a human to read:
import re
a = re.compile("(?<=OrderId>)\\d{6}")
a.findall("<....></...><...></...><OrderId>123456</OrderId><...></...><CustomerId>44444444</CustomerId><...></...><...></...>")
['123456']
b = re.compile("(?<=CustomerId>)\\d{8}")
b.findall("<....></...><...></...><OrderId>123456</OrderId><...></...><CustomerId>44444444</CustomerId><...></...><...></...>")
['44444444']

You should be using raw string literals:
searchPattern = r'\b\d{16}\b'
The escape sequence \b in a plain (non-raw) string literal represents the backspace character, so that's what the re module would be receiving (unrecognised escape sequences such as \d get passed on as-is, i.e. backslash followed by 'd').