Yet another regex string match

Question

I've been at it, for ~4 hours now, and I've failed completely, so I humbly ask for help

I've got a string with the following structure

a197 8101 aaa/bbb/ccc/ddd.doc

I need a regex that will give me ddd.doc. Obviously ddd is not always ddd, might be 'potato', might contain numerals etc. Basically I want regex that will give my anything between the last '/' and up until (including) .doc

Edit: \/(.*\.html) this is the closest I've got but it will return /bbb/ccc/ddd.doc

Edit2: I'm not looking to split, maybe I misspoke. I just want to match

https://frightanic.com/software-development/regex-match-last-occurrence/ — Robert Harvey, Feb 26 '20 at 20:03
Can you please show what 4 hours of trying has yielded so we can better teach? — Sterling Archer, Feb 26 '20 at 20:04
Would the following link be [helpfull](https://stackoverflow.com/q/6169324/9758194) — JvdV, Feb 26 '20 at 20:06
@JvdV thanks but no, I'm not looking to split anything, and doing so on a '/' would be suicide as I have a LOT of them — , Feb 26 '20 at 20:08
@RahulKhanna thank you very much, for your wisdom, but I don't need to split anything :) — , Feb 26 '20 at 20:09
I'm pretty sure that `result = yourstring.rsplit('/',maxsplit=1)[1]` will actually be faster than any regexp-based solution. — Błotosmętek, Feb 26 '20 at 20:18

score 2 · Accepted Answer · answered Feb 26 '20 at 20:57

import re
pattern = re.compile(r"/([^/\\]+.doc)")
print(pattern.search("a197 8101 aaa/bbb/ccc/ddd.doc").group(1))
print(pattern.search("a197 8101 aaa/bbb/ccc/potato.doc").group(1))
print(pattern.search("a197 8101 aaa/bbb/ccc/01_-2,,.3.doc").group(1))

output:

ddd.doc
potato.doc
01_-2,,.3.doc

score 0 · Answer 2 · answered Feb 26 '20 at 20:09

0

This should work

import re

string="a197 8101 aaa/bbb/ccc/ddd.doc"
result = re.findall(r'\w+\.\w+$', string)
print(result)

answered Feb 26 '20 at 20:09

Eraklon

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Depending on what is included in the "etc" in the question, \w may or may not suffice. – Mark Smith Feb 26 '20 at 20:26

WGriffing · Answer 3 · 2020-02-26T20:31:48.293

import re

string = "a197 8101 aaa/bbb/ccc/ddd.doc"
# parenthesis forms regex groups
# (group 1 matches start of line to and including /, greedily) = (^.*\/)
# (group 2 matches any character, non-greedy, until end of line) = (.*?$)
result = re.search(r'(^.*\/)(.*?$)', string)
print(result.group(0))
print(result.group(1))
print(result.group(2))

output = result.group(2)

will yield:

a197 8101 aaa/bbb/ccc/ddd.doc
a197 8101 aaa/bbb/ccc/
ddd.doc

Note that group 0 will always be the whole match.

Yet another regex string match

3 Answers3