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split string on any special character using python

currently I can have many dynamic separators in string like
new_123_12313131
new$123$12313131
new#123#12313131
etc etc . I just want to check if there is a special character in string then just get value after last separator like in this example just want 12313131

This is a good use case for isdigit():
l = [
'new_123_12313131',
'new$123$12313131',
'new#123#12313131',
]
output = []
for s in l:
temp = ''
for char in s:
if char.isdigit():
temp += char
output.append(temp)
print(output)
Result: ['12312313131', '12312313131', '12312313131']

Assuming you define 'special character' as anything thats not alphanumeric, you can use the str.isalnum() function to determine the first special character and leverage it something like this:
def split_non_special(input) -> str:
"""
Find first special character starting from the end and get the last piece
"""
for i in reversed(input):
if not i.isalnum():
return input.split(i)[-1] # return as soon as a separator is found
return '' # no separator found
# inputs = ['new_123_12313131', 'new$123$12313131', 'new#123#12313131', 'eefwfwrfwfwf3243']
# outputs = [split_non_special(input) for input in inputs]
# ['12313131', '12313131', '12313131', ''] # outputs

just get value after last separator
the more obvious way is using re.findall:
from re import findall
findall(r'\d+$',text) # ['12313131']

Python supplies what seems to be what you consider "special" characters using the string library as string.punctuation. Which are these characters:
!"#$%&'()*+,-./:;<=>?#[\]^_`{|}~
Using that in conjunction with the re module you can do this:
from string import punctuation
import re
re.split(f"[{punctuation}]", my_string)
my_string being the string you want to split.
Results for your examples
['new', '123', '12313131']
To get just digits you can use:
re.split("\d", my_string)
Results:
['123', '12313131']

How to remove text before a particular character or string in multi-line text?

I want to remove all the text before and including */ in a string.
For example, consider:
string = ''' something
other things
etc. */ extra text.
'''
Here I want extra text. as the output.
I tried:
string = re.sub("^(.*)(?=*/)", "", string)
I also tried:
string = re.sub(re.compile(r"^.\*/", re.DOTALL), "", string)
But when I print string, it did not perform the operation I wanted and the whole string is printing.

I suppose you're fine without regular expressions:
string[string.index("*/ ")+3:]
And if you want to strip that newline:
string[string.index("*/ ")+3:].rstrip()

The problem with your first regex is that . does not match newlines as you noticed. With your second one, you were closer but forgot the * that time. This would work:
string = re.sub(re.compile(r"^.*\*/", re.DOTALL), "", string)
You can also just get the part of the string that comes after your "*/":
string = re.search(r"(\*/)(.*)", string, re.DOTALL).group(2)

Update: After doing some research, I found that the pattern (\n|.) to match everything including newlines is inefficient. I've updated the answer to use [\s\S] instead as shown on the answer I linked.
The problem is that . in python regex matches everything except newlines. For a regex solution, you can do the following:
import re
strng = ''' something
other things
etc. */ extra text.
'''
print(re.sub("[\s\S]+\*/", "", strng))
# extra text.
Add in a .strip() if you want to remove that remaining leading whitespace.

to keep text until that symbol you can do:
split_str = string.split(' ')
boundary = split_str.index('*/')
new = ' '.join(split_str[0:boundary])
print(new)
which gives you:
something
other things
etc.

string_list = string.split('*/')[1:]
string = '*/'.join(string_list)
print(string)
gives output as
' extra text. \n'

Function for Replace And Strip

I am trying to create a function to replace some values from a given string, but I'm receiving the following error : EOL while scanning single-quoted string.
No sure what I am doing wrong:
def DataClean(strToclean):
cleanedString = strToclean
cleanedString.strip()
cleanedString= cleanedString.replace("MMMM/", "").replace("/KKKK" ,"").replace("}","").replace(",","").replace("{","")
cleanedString = cleanedString.replace("/TTTT","")
if cleanedString[-1:] == "/":
cleanedString = cleanedString[:-1]
return str(cleanedString)

You can achieve that with a much simpler solution using the regex module. Define a pattern that will match any MMM/ or /TTT and replace it with ''.
import re
pattern = r'(MMM/)?(/TTT)?'
text = 'some text MMM/ and /TTT blabla'
re.sub(pattern, '', text)
# some text and blabla
In your function it would look like
import re
def DataClean(strToclean):
clean_str = strToclean.strip()
pattern = '(MMM/)?(KKKK)?'
new_str = re.sub(pattern, '', text)
return str(new_str.rstrip('/'))
The rstrip method will remove / at the end of the string, if there any. (remove the need for if).
Build the pattern with all the patterns you are searching in the string. Using (pattern)? you define the patterns as optional. You can state as many as you want.
It is more readable than concatenating string operations.
Note the rstrip method will remove all the trailing slashes, not just one. If you want to remove just the last char, you need an if statement:
if new_str[-1] == '/':
new_str = new_str[:-1]
The if statement use index access to the string, -1 means last char. The assignment happens with slicing, up to the last char.

How to remove words after certain character in a line in python [duplicate]

I have a string. How do I remove all text after a certain character? (In this case ...)
The text after will ... change so I that's why I want to remove all characters after a certain one.

Split on your separator at most once, and take the first piece:
sep = '...'
stripped = text.split(sep, 1)[0]
You didn't say what should happen if the separator isn't present. Both this and Alex's solution will return the entire string in that case.

Assuming your separator is '...', but it can be any string.
text = 'some string... this part will be removed.'
head, sep, tail = text.partition('...')
>>> print head
some string
If the separator is not found, head will contain all of the original string.
The partition function was added in Python 2.5.
S.partition(sep) -> (head, sep, tail)
Searches for the separator sep in S, and returns the part before it,
the separator itself, and the part after it. If the separator is not
found, returns S and two empty strings.

If you want to remove everything after the last occurrence of separator in a string I find this works well:
<separator>.join(string_to_split.split(<separator>)[:-1])
For example, if string_to_split is a path like root/location/child/too_far.exe and you only want the folder path, you can split by "/".join(string_to_split.split("/")[:-1]) and you'll get
root/location/child

Without a regular expression (which I assume is what you want):
def remafterellipsis(text):
where_ellipsis = text.find('...')
if where_ellipsis == -1:
return text
return text[:where_ellipsis + 3]
or, with a regular expression:
import re
def remwithre(text, there=re.compile(re.escape('...')+'.*')):
return there.sub('', text)

import re
test = "This is a test...we should not be able to see this"
res = re.sub(r'\.\.\..*',"",test)
print(res)
Output: "This is a test"

The method find will return the character position in a string. Then, if you want remove every thing from the character, do this:
mystring = "123⋯567"
mystring[ 0 : mystring.index("⋯")]
>> '123'
If you want to keep the character, add 1 to the character position.

From a file:
import re
sep = '...'
with open("requirements.txt") as file_in:
lines = []
for line in file_in:
res = line.split(sep, 1)[0]
print(res)

This is in python 3.7 working to me
In my case I need to remove after dot in my string variable fees
fees = 45.05
split_string = fees.split(".", 1)
substring = split_string[0]
print(substring)

Yet another way to remove all characters after the last occurrence of a character in a string (assume that you want to remove all characters after the final '/').
path = 'I/only/want/the/containing/directory/not/the/file.txt'
while path[-1] != '/':
path = path[:-1]

another easy way using re will be
import re, clr
text = 'some string... this part will be removed.'
text= re.search(r'(\A.*)\.\.\..+',url,re.DOTALL|re.IGNORECASE).group(1)
// text = some string

Strip in Python

I have a question regarding strip() in Python. I am trying to strip a semi-colon from a string, I know how to do this when the semi-colon is at the end of the string, but how would I do it if it is not the last element, but say the second to last element.
eg:
1;2;3;4;\n
I would like to strip that last semi-colon.

Strip the other characters as well.
>>> '1;2;3;4;\n'.strip('\n;')
'1;2;3;4'

>>> "".join("1;2;3;4;\n".rpartition(";")[::2])
'1;2;3;4\n'

how about replace?
string1='1;2;3;4;\n'
string2=string1.replace(";\n","\n")

>>> string = "1;2;3;4;\n"
>>> string.strip().strip(";")
"1;2;3;4"
This will first strip any leading or trailing white space, and then remove any leading or trailing semicolon.

Try this:
def remove_last(string):
index = string.rfind(';')
if index == -1:
# Semi-colon doesn't exist
return string
return string[:index] + string[index+1:]
This should be able to remove the last semicolon of the line, regardless of what characters come after it.
>>> remove_last('Test')
'Test'
>>> remove_last('Test;abc')
'Testabc'
>>> remove_last(';test;abc;foobar;\n')
';test;abc;foobar\n'
>>> remove_last(';asdf;asdf;asdf;asdf')
';asdf;asdf;asdfasdf'
The other answers provided are probably faster since they're tailored to your specific example, but this one is a bit more flexible.

You could split the string with semi colon and then join the non-empty parts back again using ; as separator
parts = '1;2;3;4;\n'.split(';')
non_empty_parts = []
for s in parts:
if s.strip() != "": non_empty_parts.append(s.strip())
print "".join(non_empty_parts, ';')

If you only want to use the strip function this is one method:
Using slice notation, you can limit the strip() function's scope to one part of the string and append the "\n" on at the end:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:8].strip(';') + str[8:]
Using the rfind() method(similar to Micheal0x2a's solution) you can make the statement applicable to many strings:
# create a var for later
str = "1;2;3;4;\n"
# format and assign to newstr
newstr = str[:str.rfind(';') + 1 ].strip(';') + str[str.rfind(';') + 1:]

re.sub(r';(\W*$)', r'\1', '1;2;3;4;\n') -> '1;2;3;4\n'