You have the right idea with escaping the backslashes, but despite how it looks, your input string doesn't actually have any backslashes in it. You need to escape them in the input, too!

>>> a = "1\\2\\3\\4"  # Note the doubled backslashes here!
>>> print(a.split('\\'))  # Split on '\\'
['1', '2', '3', '4']

You could also use a raw string literal for the input, if it's likely to have many backslashes. This notation is much cleaner to look at (IMO), but it does have some limitations: read the docs!

>>> a = r"1\2\3\4"
>>> print(a.split('\\'))
['1', '2', '3', '4']

If you're getting a elsewhere, and a.split('\\') doesn't appropriately split on the visible backslashes, that means you've got something else in there instead of real backslashes. Try print(repr(a)) to see what the "literal" string actually looks like.

>>> a = '1\2\3\4'
>>> print(a)
1☻♥♦
>>> print(repr(a))
'1\x02\x03\x04'

>>> b = '1\\2\\3\\4'
>>> print(b)
1\2\3\4
>>> print(repr(b))
'1\\2\\3\\4'

Assuming you have a correct path, then:

import os

# Note that we're using the **r** prefix to make it a raw string - backslashes don't escape  
path = r'C:\Users\Desktop\TEST\Excel_Reports\1837.xlsx'
print os.path.split(path)[1]
# 1837.xlsx

You could also then further split to just get the base of the filename, eg:

print os.path.splitext(os.path.split(path)[1])[0]
# 1837

I think you're trying harder than you need to. Python has an os module which has a function to do exactly what you want, independent of platform. Here, you just need to use

fileName=os.path.basename(targetDirectory)

Also, if you're using Python you should seriously consider moving away from camel case (fileName) to the more Pythonic snake case (file_name).

Your first backslash is escaping the second at the level of the string literal. But the regex engine needs that backslash escaped also, since it's a special character for regex too.

Use a "raw" string literal (e.g. r' |/|\\') or quadruple backslash.

To capture a delimiter, it's easier to use findall instead of split:

re.findall(r'[^\\/]+|[\\/]', string)

[^\\/]+ would find 1 or more occurrences of sub-strings that do not contain forward or backward slash. | works as an or operator. Finally, [\\/] will match with the occurrences of forward and backward slash. The result would provide separate sub-strings for the occurrences of forward and backward slash and string matches where they do not occur.

As for why your code didn't work, your expression is (\\/). When Python interpreter parses this, it sees an escaped slash and creates a string of four characters: ( \ / ). Then, this string is sent to the regex engine, which also does escaping. It sees a slash followed by a backslash, and since backslash is not special, it "escapes" to itself, so the final expression is just (/). Finally, re applies this expression, splits by a backslash and captures it - exactly what you're observing.

The correct command for your approach would be re.split('([\\\/])',string) due to double escaping.

The moral of the story: always use raw literals r"..." with regexes to avoid double escaping issues.

Declare a variable with the separator to avoid the quotation&escaping alchemy. Use Single-Quoted Style; backslash '\' can be used freely. For example, the task below

  - debug:
      msg: "{{ dirs[1].split(separator)[-1] }}"
    vars:
      separator: '\'
      dirs: ['Web\this','Web\that','Web']

gives

  msg: that