Extract file names from a given directory with regex

Question

I am pretty weak in regex. I'm looking for some help with how to extract the .sav file name from the following string:

C:\Users...\Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20\AutumnHi-20180531-183047-34-SystemNormal\AutumnHi-20180531-183047-34-SystemNormal.sav

Currently I am using this code:

re.findall(r'\\(.+).sav',txt)

but it only finds

['Users\\...\\Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20\\AutumnHi-20180531-183047-34-SystemNormal\AutumnHi-20180531-183047-34-SystemNormal.sav was']

I'm trying to find "AutumnHi-20180531-183047-34-SystemNormal.sav"

I am using Python 3.7.

Does this answer your question? [Extract file name from path, no matter what the os/path format](https://stackoverflow.com/questions/8384737/extract-file-name-from-path-no-matter-what-the-os-path-format) — Tomerikoo, Mar 05 '21 at 13:50

score 1 · Answer 1 · answered Jun 29 '19 at 08:09

You could match a backslash and then capture in a group matching not a backslash using a negated character class. Then match a dot and sav.

You might use a negative lookahead to assert what is directly on the right is not a non whitespace char.

\\([^\\]+\.sav)(?!\S)

Regex demo

score 0 · Answer 2 · answered Jun 29 '19 at 05:50

Regex101 (link):

txt = r'''C:\Users\\...\\Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20\\WinterLo-20180729-043047-34-SystemNormal\\WinterLo-20180729-043047-34-SystemNormal.sav'''

import re

print(re.findall(r'(?<=\\)[^\\]+sav',txt)[0])

Prints:

WinterLo-20180729-043047-34-SystemNormal.sav

You could achieve the same without re module:

print(txt.split('\\')[-1])

SanV · Answer 3 · 2019-06-29T06:13:57.467

0

The following pattern should match the filename:
(?=[^\\]*$).*\.sav

Regex101 Demo

The above pattern asserts (?= is positive lookahead) that no other character up to the end of the string is a backslash. So essentially it finds the last backslash and then matches the desired text. For other details, see "EXPLANATION" on the right side of the regex101 demo at the link above.

edited Jun 29 '19 at 06:13

answered Jun 29 '19 at 06:00

SanV

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score 0 · Answer 4 · answered Jun 29 '19 at 08:27

I am assuming you are not learning about regex but want to know how to handle parsing filenames.

I would use the pathlib module to handle parsing the filename.

C:\Users\barry>py -3.7
Python 3.7.2 (tags/v3.7.2:9a3ffc0492, Dec 23 2018, 23:09:28) [MSC v.1916 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import pathlib
>>> filename = r'C:\Users\...\Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20\WinterLo-20180729-043047-34-SystemNormal\WinterLo-20180729-043047-34-SystemNormal.sav'
>>> path = pathlib.Path(filename)
>>> path.name
'WinterLo-20180729-043047-34-SystemNormal.sav'
>>> path.parent
WindowsPath('C:/Users/.../Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20/WinterLo-20180729-043047-34-SystemNormal')
>>>

score 0 · Answer 5 · answered Jun 29 '19 at 15:37

I'm guessing that these expressions:

[^\\]+\.sav
([^\\]+\.sav)

or some similar derivative of those might likely extract what we might want here.

Test

import re

print(re.findall(r"([^\\]+\.sav)", "C:\\Users...\\Standard Loadflows Seq and Dyn PSSEv34 - 2019-02-20\\AutumnHi-20180531-183047-34-SystemNormal\\AutumnHi-20180531-183047-34-SystemNormal.sav"))

Output

['AutumnHi-20180531-183047-34-SystemNormal.sav']

Extract file names from a given directory with regex

5 Answers5

Test

Output

Demo