Python regex: Remove a pattern at the end of string

Question

Input: blah.(2/2)

Desired Output: blah

Input can be "blah.(n/n)" where n can be any single digit number.

How do I use regex to achieve "blah"? This is the regex I currently have which doesn't work:

m = re.sub('[.[0-9] /\ [0-9]]{6}$', '', m)

Solution 1

You need to use regex \.\(\d/\d\)$

>>> import re
>>> str = "blah.(2/2)"
>>> re.sub(r'\.\(\d/\d\)$', '',str)
'blah'

Regex explanation here

Solution 2

I really like the simplicity of the @idjaw's approach. If you were to solve it with regular expressions:

In [1]: import re

In [2]: s = "blah.(2/2)"

In [3]: re.sub(r"\.\(\d/\d\)$", "", s)
Out[3]: 'blah'

Here, \.\(\d/\d\)$ would match a single dot followed by an opening parenthesis, followed by a single digit, followed by a slash, followed by a single digit, followed by a closing parenthesis at the end of a string.

Solution 3

Just do this since you will always be looking to split around the .

s = "stuff.(asdfasdfasdf)"
m = s.split('.')[0]

Solution 4

Just match what you do not want and remove it. In this case, non letters at the end of the string:

>>> re.sub(r'[^a-zA-Z]+$', '', "blah.(1/2)")
'blah'

View more solutions

Comments

[rebeling]

blah. blah. boom. What about '.'.join(s.split('.')[:-1])

[idjaw]

I don't see that it is necessary to convert your list to a string when you can simply just extract the string you need after your split right away.

[rebeling]

blah. blah. boom.(4/2) What do you expect to get?

[idjaw]

Apologies, I don't understand what you are asking.

[rebeling]

No worries. But sometimes there is another dot. Try print "blah. blah. boom.(4/2)".split('.')[0]

[idjaw]

Oh I understand now, thanks. The intention of this was more to fall with the pattern as exactly specified by the OP and the shortest solution path to that was to just split on the ..