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Remove characters after matching two conditions

I have the Python code below and I would like the output to be a string: "P-1888" discarding all numbers after the 2nd "-" and removing the leading 0's after the 1st "-".
So far all I have been able to do in the following code is to remove the trailing 0's:
import re
docket_no = "P-01888-000"
doc_no_rgx1 = re.compile(r"^([^\-]+)\-(0+(.+))\-0[\d]+$")
massaged_dn1 = doc_no_rgx1.sub(r"\1-\2", docket_no)
print(massaged_dn1)

You can use the split() method to split the string on the "-" character and then use the join() method to join the first and second elements of the resulting list with a "-" character. Additionally, you can use the lstrip() method to remove the leading 0's after the 1st "-". Try this.
docket_no = "P-01888-000"
docket_no_list = docket_no.split("-")
docket_no_list[1] = docket_no_list[1].lstrip("0")
massaged_dn1 = "-".join(docket_no_list[:2])
print(massaged_dn1)

First way is to use capturing groups. You have already defined three of them using brackets. In your example the first capturing group will get "P", and the third capturing group will get numbers without leading zeros. You can get captured data by using re.match:
match = doc_no_rgx1.match(docket_no)
print(f'{match.group(1)}-{match.group(3)}') # Outputs 'P-1888'
Second way is to not use regex for such a simple task. You could split your string and reassemble it like this:
parts = docket_no.split('-')
print(f'{parts[0]}-{parts[1].lstrip("0")}')

It seems like a sledgehammer/nut situation but of you do want to use re then you could use:
doc_no_rgx1 = ''.join(re.findall('([A-Z]-)0+(\d+)-', docket_no)[0])

I don't think I'd use a regular expression for this purpose. Your usecase can be handled by standard string manipulation so using a regular expression would be overkill. Instead, consider doing this:
docket_nos = "P-01888-000".split('-')[:-1]
docket_nos[1] = docket_nos[1].lstrip('0')
docket_no = '-'.join(docket_nos)
print(docket_no) # P-1888
This might seem a little bit verbose but it does exactly what you're looking for. The first line splits docket_no by '-' characters, producing substrings P, 01888 and 000; and then discards the last substring. The second line strips leading zeros from the second substring. And the third line joins all these back together using '-' characters, producing your desired result of P-1888.

Functionally this is no different than other answers suggesting that you split on '-' and lstrip the zero(s), but personally I find my code more readable when I use multiple assignment to clarify intent vs. using indexes:
def convert_docket_no(docket_no):
letter, number, *_ = docket_no.split('-')
return f'{letter}-{number.lstrip("0")}'
_ is used here for a "throwaway" variable, and the * makes it accept all elements of the split list past the first two.

How I can use regex to remove repeated characters from string

I have a string as follows where I tried to remove similar consecutive characters.
import re
input = "abccbcbbb";
for i in input :
input = re.sub("(.)\\1+", "",input);
print(input)
Now I need to let the user specify the value of k.
I am using the following python code to do it, but I got the error message TypeError: can only concatenate str (not "int") to str
import re
input = "abccbcbbb";
k=3
for i in input :
input= re.sub("(.)\\1+{"+(k-1)+"}", "",input)
print(input)

The for i in input : does not do what you need. i is each character in the input string, and your re.sub is supposed to take the whole input as a char sequence.
If you plan to match a specific amount of chars you should get rid of the + quantifier after \1. The limiting {min,} / {min,max} quantifier should be placed right after the pattern it modifies.
Also, it is more convenient to use raw string literals when defining regexps.
You can use
import re
input_text = "abccbcbbb";
k=3
input_text = re.sub(fr"(.)\1{{{k-1}}}", "", input_text)
print(input_text)
# => abccbc
See this Python demo.
The fr"(.)\1{{{k-1}}}" raw f-string literal will translate into (.)\1{2} pattern. In f-strings, you need to double curly braces to denote a literal curly brace and you needn't escape \1 again since it is a raw string literal.

If I were you, I would prefer to do it like suggested before. But since I've already spend time on answering this question here is my handmade solution.
The pattern described below creates a named group named "letter". This group updates iterative, so firstly it is a, then b, etc. Then it looks ahead for all the repetitions of the group "letter" (which updates for each letter).
So it finds all groups of repeated letters and replaces them with empty string.
import re
input = 'abccbcbbb'
result = 'abcbcb'
pattern = r'(?P<letter>[a-z])(?=(?P=letter)+)'
substituted = re.sub(pattern, '', input)
assert substituted == result

Just to make sure I have the question correct you mean to turn "abccbcbbb" into "abcbcb" only removing sequential duplicate characters. Is there a reason you need to use regex? you could likely do a simple list comprehension. I mean this is a really cut and dirty way to do it but you could just put
input = "abccbcbbb"
input = list(input)
previous = input.pop(0)
result = [previous]
for letter in input:
if letter != previous : result += letter
previous = letter
result = "".join(result)
and with a method like this, you could make it easier to read and faster with a bit of modification id assume.

Remove brackets and number inside from string Python

I've seen a lot of examples on how to remove brackets from a string in Python, but I've not seen any that allow me to remove the brackets and a number inside of the brackets from that string.
For example, suppose I've got a string such as "abc[1]". How can I remove the "[1]" from the string to return just "abc"?
I've tried the following:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[]')
but this only outputs the string without the final bracket
abc[1
I've also tried with a wildcard option:
stringTest = "abc[1]"
stringTestWithoutBrackets = str(stringTest).strip('[\w+\]')
but this also outputs the string without the final bracket
abc[1

You could use regular expressions for that, but I think the easiest way would be to use split:
>>> stringTest = "abc[1][2][3]"
>>> stringTest.split('[', maxsplit=1)[0]
'abc'

You can use regex but you need to use it with the re module:
re.sub(r'\[\d+\]', '', stringTest)
If the [<number>] part is always at the end of the string you can also strip via:
stringTest.rstrip('[0123456789]')
Though the latter version might strip beyond the [ if the previous character is in the strip list too. For example in "abc1[5]" the "1" would be stripped as well.

Assuming your string has the format "text[number]" and you only want to keep the "text", then you could do:
stringTest = "abc[1]"
bracketBegin = stringTest.find('[')
stringTestWithoutBrackets = stringTest[:bracketBegin]

Regular expression to retrieve string parts within parentheses separated by commas

I have a String from which I want to take the values within the parenthesis. Then, get the values that are separated from a comma.
Example: x(142,1,23ERWA31)
I would like to get:
142
1
23ERWA31
Is it possible to get everything with one regex?
I have found a method to do so, but it is ugly.
This is how I did it in python:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
secondResult = re.search("(?<=\()(.*?)(?=\))", firstResult.group(0))
finalResult = [x.strip() for x in secondResult.group(0).split(',')]
for i in finalResult:
print(i)
142
1
23ERWA31

This works for your example string:
import re
string = "x(142,1,23ERWA31)"
l = re.findall (r'([^(,)]+)(?!.*\()', string)
print (l)
Result: a plain list
['142', '1', '23ERWA31']
The expression matches a sequence of characters not in (,,,) and – to prevent the first x being picked up – may not be followed by a ( anywhere further in the string. This makes it also work if your preamble x consists of more than a single character.
findall rather than search makes sure all items are found, and as a bonus it returns a plain list of the results.

You can make this a lot simpler. You are running your first Regex but then not taking the result. You want .group(1) (inside the brackets), not .group(0) (the whole match). Once you have that you can just split it on ,:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
for e in firstResult.group(1).split(','):
print(e)

A little wonky looking, and also assuming there's always going to be a grouping of 3 values in the parenthesis - but try this regex
\((.*?),(.*?),(.*?)\)
To extract all the group matches to a single object - your code would then look like
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?),(.*?),(.*?)\)", string).groups()
You can then call the firstResult object like a list
>> print(firstResult[2])
23ERWA31

Getting rid of certain characters in a string in python

I have characters in the middle of a string that I want to get rid of. These characters are =, p,, and H. Since they are not the leftmost and the rightmost characters in the string, I cannot use strip(). Is there a function that gets rid of a certain character in any location in a string?

The usual tool for this job is str.translate
https://docs.python.org/2/library/stdtypes.html#str.translate
>>> 'hello=potato'.translate(None, '=p')
'hellootato'

Check the .replace() function:
> 'aaba'.replace('a','').replace('b','')
< ''

My usual tool for this is the regular expression.
>>> import re
>>> invalidCharacters = r'[=p H]'
>>> mystring = re.sub(invalidCharacters, '', ' poH==hHoPPp p')
'ohoPP'
If you need to constrain the number (i.e., the count) of characters you remove, see the count argument.