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Remove Characters From A String Until A Specific Format is Reached

So I have the following strings and I have been trying to figure out how to manipulate them in such a way that I get a specific format.
string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo
I want to be able to get rid of any of the last string so I am just left with the month and year, like below:
string1-itd_jan2021
string2itd_mar2021
string3itd_feb2021
string4-itd_mar2021
string5itd_jun2021
string6-itd_feb2021
I thought about using string.split on the - but then realized that for some strings this wouldn't work. I also thought about getting rid of a set amount of characters by putting it into a list and slicing but the end is varying characters length?
Is there anything I can do it with regex or any other python module?

Use str.rsplit with the appropriate maxsplit parameter:
s = s.rsplit("-", 1)[0]
You could also use str.split (even though this is clearly the worse choice):
s = "-".join(s.split("-")[:-1])
Or using regular expressions:
s = re.sub(r'-[^-]*$', '', s)
# "-[^-]*" a "-" followed by any number of non-"-"

With a regex:
import re
re.sub(r'([0-9]{4}).*$', r'\1', s)

Use re.sub like so:
import re
lines = '''string1-itd_jan2021-internal
string2itd_mar2021-space
string3itd_feb2021-internal
string4-itd_mar2021-moon
string5itd_jun2021-internal
string6-itd_feb2021-apollo'''
for old in lines.split('\n'):
new = re.sub(r'[-][^-]+$', '', old)
print('\t'.join([old, new]))
Prints:
string1-itd_jan2021-internal string1-itd_jan2021
string2itd_mar2021-space string2itd_mar2021
string3itd_feb2021-internal string3itd_feb2021
string4-itd_mar2021-moon string4-itd_mar2021
string5itd_jun2021-internal string5itd_jun2021
string6-itd_feb2021-apollo string6-itd_feb2021
Explanation:
r'[-][^-]+$' : Literal dash (-), followed by any character other than a dash ([^-]) repeated 1 or more times, followed by the end of the string ($).

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

Why is the split() returning list objects that are empty? [duplicate]

I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']

I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)

Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')

If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']

Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.

>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']

how to extract string inside single quotes using python script

Have a set of string as follows
text:u'MUC-EC-099_SC-Memory-01_TC-25'
text:u'MUC-EC-099_SC-Memory-01_TC-26'
text:u'MUC-EC-099_SC-Memory-01_TC-27'
These data i have extracted from a Xls file and converted to string,
now i have to Extract data which is inside single quotes and put them in a list.
expecting output like
[MUC-EC-099_SC-Memory-01_TC-25, MUC-EC-099_SC-Memory-01_TC-26,MUC-EC-099_SC-Memory-01_TC-27]
Thanks in advance.

Use re.findall:
>>> import re
>>> strs = """text:u'MUC-EC-099_SC-Memory-01_TC-25'
text:u'MUC-EC-099_SC-Memory-01_TC-26'
text:u'MUC-EC-099_SC-Memory-01_TC-27'"""
>>> re.findall(r"'(.*?)'", strs, re.DOTALL)
['MUC-EC-099_SC-Memory-01_TC-25',
'MUC-EC-099_SC-Memory-01_TC-26',
'MUC-EC-099_SC-Memory-01_TC-27'
]

You can use the following expression:
(?<=')[^']+(?=')
This matches zero or more characters that are not ' which are enclosed between ' and '.
Python Code:
quoted = re.compile("(?<=')[^']+(?=')")
for value in quoted.findall(str(row[1])):
i.append(value)
print i

That text: prefix seems a little familiar. Are you using xlrd to extract it? In that case, the reason you have the prefix is because you're getting the wrapped Cell object, not the value in the cell. For example, I think you're doing something like
>>> sheet.cell(2,2)
number:4.0
>>> sheet.cell(3,3)
text:u'C'
To get the unwrapped object, use .value:
>>> sheet.cell(3,3).value
u'C'
(Remember that the u here is simply telling you the string is unicode; it's not a problem.)