How to get String Before Last occurrence of substring?

Question

I want to get String before last occurrence of my given sub string.

My String was,

path = D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov

my substring, 1001-1010 which will occurred twice. all i want is get string before its last occurrence.

Note: My substring is dynamic with different padding but only number.

I want,

D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v

I have done using regex and slicing,

>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> q = re.findall("\d*-\d*",p)
>>> q[-1].join(p.split(q[-1])[:-1])
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v'
>>>

Is their any better way to do by purely using regex?

Please Note I have tried so many eg:

I got answer by using regex with slicing but i want to achieve by using regex alone..

score 8 · Answer 1 · answered Apr 25 '19 at 14:42

8

Why use regex. Just use built in string methods:

path = "D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov"
index = path.rfind("1001-1010")
print(path[:index])

answered Apr 25 '19 at 14:42

Error - Syntactical Remorse

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looks good.. nice example without regex.. but problem is i dont know substring.. substring will be some number with padding.. – Mohideen bin Mohammed Apr 25 '19 at 14:55
2

Yes this approach will work only with a static pattern but if keyword is unknown and a pattern search is required then regex is must. – anubhava Apr 25 '19 at 14:58

anubhava · Accepted Answer · 2019-04-25T14:55:41.290

3

You can use a simple greedy match and a capture group:

(.*)1001-1010

Your match is in capture group #1

Since .* is greedy by nature, it will match longest match before matching your keyword 1001-1010.

RegEx Demo

As per comments below if keyword is not a static string then you may use this regex:

r'(.*\D)\d+-\d+'

Python Code:

>>> p = 'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v1001-1010.mov'
>>> print (re.findall(r'(.*\D)\d+-\d+', p))
['D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v']

edited Apr 25 '19 at 14:55

answered Apr 25 '19 at 14:42

anubhava

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great i have slightly modified your answer.. `(.*)\d*-\d*` number also not static so i made it like this.. padding will differ .. thanks i have tried .*(\d*-\d*)\/ instead. thats why i didnt found – Mohideen bin Mohammed Apr 25 '19 at 14:47
Yes you may also use: `r'(.*\D)\d+-\d+'` – anubhava Apr 25 '19 at 14:49
1

thanks @anubhava (.*\D)\d+-\d+ this is what i want.. previous one gave me string along with substring but this has both. – Mohideen bin Mohammed Apr 25 '19 at 14:54
1

ok answer updated accordingly. Please mention this in your question as well so readers would better understand why a regex approach is needed and string search using `rfind` won't work. – anubhava Apr 25 '19 at 14:56
1

accepted +1 rfind can be useful for static but mine is dynamic.. thats i found my solution – Mohideen bin Mohammed Apr 25 '19 at 15:00

Mohideen bin Mohammed · Answer 3 · 2019-04-25T14:58:38.817

Thanks @anubhava,

My first regex was,

.*(\d*-\d*)\/

Now i have corrected mine..

.*(\d*-\d*)

or

(.*)(\d*-\d*)

which gives me,

>>> q = re.search('.+(\d*-\d*)', p)
>>> q.group()
'D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v0001-1001'
>>> 

(.*\D)\d+-\d+

this gives me exactly what i want...

>>> q = re.search('(.*\D)\d+-\d+', p)
>>> q.groups()
('D:/me/vol101/Prod/cent/2019_04_23_01/image/AVEN_000_3400_img_pic_p1001-1010/pxy/AVEN_000_3400_img-mp4_to_MOV_v',)
>>>

How to get String Before Last occurrence of substring?

3 Answers3