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How can I get the last name in a path next to the right of a \ character using RE on Python?

I have been trying to make a good regular expression that catches the last name (of a file or folder) for a given path, meaning that for instance, the following path:
C:\Users\ResetStoreX\Pictures
Can be reduced to:
Pictures
What I have achieved so far is the following RE: (?:[^\\]\\)[^\\]*$
Which reduces the previous path to:
X\Pictures
Any ideas?

You almost had it. I removed the \ inside of the parentheses, if you want all backslashes to be captured.
(?:[^\\])[^\\]*$

Trying to find a way to filter out parts of a string python

I'm trying to filter out strings in file names that appear in a for loop
if search == "List":
onlyfiles = [f for f in listdir("path") if isfile(join("path", f))]
for i in onlyfiles:
print(i)
now it will output all the filenames, as expected and wanted, but I want to filter out the .json at the end of the file as well as a few other elements in the name of the file so that I can just see the file name.
For example: filename-IDENTIFIER.json
I want to filter out "-IDENTIFIER.json" out from the for loop's output
Thanks for any help

There are a few approaches here, based on how much your data can vary:
So let's try to build a get_filename(f) method
Quick and dirty
If you know that f always ends in exactly the same way, then you can directly try to remove those characters. So here we have to remove the last 16 characters. It's useful to know that in Python, a string can be considered as an (immutable) array of characters, so you can use list indexing as well.
get_filename(f: str):
return f[:-16]
This will however fail if the Identifier or suffix changes in length.
Varying lenghts
If the suffix changes based on the length, then you should split the string on a fixed delimiter and return the relevant part. In this case you want to split on -.
get_filename(f: str):
return f.split("-")[0]
Note however that this will fail if the filename also contains a -.
You can fix that by dropping the last part and rejoining all the earlier pieces, in the following way.
get_filename(f: str):
return "-".join(f.split("-")[:-1])
Using regexes to match the format
The most general approach would be to use python regexes to select the relevant part. These allow you to very specifically target a specific pattern. The exact regex that you'll need will depend on the complexity of your strings.

Split the string on "-" and get the first element:
filename = f.split("-")[0]
This will get messed up case filename contains "-" though.

This should work:
i.split('-')[0].split('.')[0]
Case 1: filename-IDENTIFIER.json
It takes the substring before the dash, so output will become filename
Case 2: filename.json
There is no dash in the string, so the first split does nothing (full string will be in the 0th element), then it takes the substring before the point. Output will be filename
Case 3: filename
Nothing to split, output will be filename
If it's always .json and -IDENTIFIER, then it's safer to use this:
i.split('-IDENTIFIER')[0].split('.json')[0]
Case 4: filename-blabla.json
If the filename has an extra dash in it, it won't be a problem, output will be filename-blabla

Python regular expression for Windows file path

The problem, and it may not be easily solved with a regex, is that I want to be able to extract a Windows file path from an arbitrary string. The closest that I have been able to come (I've tried a bunch of others) is using the following regex:
[a-zA-Z]:\\([a-zA-Z0-9() ]*\\)*\w*.*\w*
Which picks up the start of the file and is designed to look at patterns (after the initial drive letter) of strings followed by a backslash and ending with a file name, optional dot, and optional extension.
The difficulty is what happens, next. Since the maximum path length is 260 characters, I only need to count 260 characters beyond the start. But since spaces (and other characters) are allowed in file names I would need to make sure that there are no additional backslashes that could indicate that the prior characters are the name of a folder and that what follows isn't the file name, itself.
I am pretty certain that there isn't a perfect solition (the perfect being the enemy of the good) but I wondered if anyone could suggest a "best possible" solution?

Here's the expression I got, based on yours, that allow me to get the path on windows : [a-zA-Z]:\\((?:[a-zA-Z0-9() ]*\\)*).* . An example of it being used is available here : https://regex101.com/r/SXUlVX/1
First, I changed the capture group from ([a-zA-Z0-9() ]*\\)* to ((?:[a-zA-Z0-9() ]*\\)*).
Your original expression captures each XXX\ one after another (eg : Users\ the Users\).
Mine matches (?:[a-zA-Z0-9() ]*\\)*. This allows me to capture the concatenation of XXX\YYYY\ZZZ\ before capturing. As such, it allows me to get the full path.
The second change I made is related to the filename : I'll just match any group of character that does not contain \ (the capture group being greedy). This allows me to take care of strange file names.
Another regex that would work would be : [a-zA-Z]:\\((?:.*?\\)*).* as shown in this example : https://regex101.com/r/SXUlVX/2
This time, I used .*?\\ to match the XXX\ parts of the path.
.*? will match in a non-greedy way : thus, .*?\\ will match the bare minimum of text followed by a back-slash.
Do not hesitate if you have any question regarding the expressions.
I'd also encourage you to try to see how well your expression works using : https://regex101.com . This also has a list of the different tokens you can use in your regex.
Edit : As my previous answer did not work (though I'll need to spend some times to find out exactly why), I looked for another way to do what you want. And I managed to do so using string splitting and joining.
The command is "\\".join(TARGETSTRING.split("\\")[1:-1]).
How does this work : Is plit the original string into a list of substrings, based. I then remove the first and last part ([1:-1]from 2nd element to the one before the last) and transform the resulting list back into a string.
This works, whether the value given is a path or the full address of a file.
Program Files (x86)\\Adobe\\Acrobat Distiller\\acrbd.exe fred is a file path
Program Files (x86)\\Adobe\\Acrobat Distiller\\acrbd.exe fred\ is a directory path

How to extract string between key substring and "/" with regex?

I have a string that's
/path/to/file?_subject_ID_SOMEOTHERSTRING
the path/to/file part changes depends on situation, and subject_ID is always there. I try to write a regex that extract only file part of the string. Using ?subject_ID is definite, but I don't know how to safely get the file
My current regex looks like (.*[\/]).*\?_subject_ID
url = '/path/to/file?_subject_ID_SOMEOTHERSTRING'
file_re = re.compile('(.*[\/]).*\?_subject_ID')
file_re.search(url)
this will find the right string, but I still can't extract the file name
printing _.group(1) will get me /path/to/. What's the next step that gets me the actual file name?

As for your '(.*[\/]).*\?_subject_ID' regex approach, you just need to add a capturing group around the second .*. You could use r'(.*/)(.*)\?_subject_ID' (then, there will be .group(1) and .group(2) parts captured), but it is not the most appropriate way to parse URLs in Python.
You may use the non-regex approach here, here is a snippet showing how to leverage urlparse and os.path to parse the URL like yours:
import urlparse
path = urlparse.urlparse('/path/to/file?_subject_ID_SOMEOTHERSTRING').path
import os.path
print(os.path.split(path)[1]) # => file
print(os.path.split(path)[0]) # => /path/to
See the IDEONE demo

It's pretty simple, really. Just match a / before and ?subject_ID after:
([^/?]*)\?subject_ID
The [^/?]* (as opposed to .*) is because otherwise it'd match the part before, too. The ? in the character class
If you want to get both the path and the file, you can do much the same thing, but also grab the part before the /:
([^?]*)([^/?]*)\?subject_ID
It's basically the same as the one before but with the first bit captured instead of ignored.

Compare & manipulate strings with python

I've written an XML parser in Python and have just added functionality to read a further script from a different directory.
I've got two args, first is the path where I'm parsing XML. Second is a string in another XML file which I want to match with the first path;
arg1 = \work\parser\main\tools\app\shared\xml\calculators\2012\example\calculator
path = calculators/2012/example/calculator
How can I compare the two strings to match identify that they're both referencing the same thing and also, how can I strip calculator from either string so I can store that & use it?
edit
Just had a thought. I have used a Regex to get the year out of the path already with year = re.findall(r"\.(\d{4})\.", path) following a problem Python has with numbers when converting the path to an import statement.
I could obviously split the strings and use a regex to match the path as a pattern in arg1 but this seems a long way round. Surely there's a better method?

Here I am assuming you are actually talking about strings, and not file paths - for which #mgilson's suggestion is better
How can I compare the two strings to match identify that they're both
referencing the same thing
Well first you need to identify what you mean by "the same thing"
At first glance it seems that if the the second string ends with the first string with the reversed slash, you have a match.
arg1 = r'\work\parser\main\tools\app\shared\xml\calculators\2012\example\calculator'
arg2 = r'calculators/2012/example/calculator'
>>> arg1.endswith(arg2.replace('/','\\'))
True
and also, how can I strip calculator from
either string so I can store that & use it?
You also need to decide if you want to strip the first calculator, the last calculator or any occurance of calculator in the string.
If you just want to remove the last string after the separator, then its simply:
>>> arg2.split('/')[-1]
'calculator'
Now to get the orignal string back, without the last bit:
>>> '/'.join(arg2.split('/')[:-1])
'calculators/2012/example'

check out os.path.samefile:
http://docs.python.org/library/os.path.html#os.path.samefile
and os.path.dirname:
http://docs.python.org/library/os.path.html#os.path.dirname
or maybe os.path.basename (I'm not sure what part of the string you want to keep).

Here, try this:
arg1 = "\work\parser\main\tools\app\shared\xml\calculators\2012\example\calculator"
path = "calculators/2012/example/calculator"
arg1=arg1.replace("/","\\")
path=path.replace("/","\\")
if str(arg1).endswith(str(path)) or str(path).endswith(str(arg1)):
print "Match"
That should work for your needs. Cheers :)