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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']

Python Removing non-alphabetical characters with exceptions

I am having a hard time doing Data Analysis on a large text that has lots of non-alphabetical chars. I tried using
string = filter(str.isalnum, string)
but I also have "#" in my text that I want to keep. How do I make an exception for a character like "#" ?

It is easier to use regular expressions:
string = re.sub("[^A-Za-z0-9#]", "", string)

You can use re.sub
re.sub(r'[^\w\s\d#]', '', string)
Example:
>>> re.sub(r'[^\w\s\d#]', '', 'This is # string 123 *$^%')
This is # string 123

One way to do this would be to create a function that returns True or False if an input character is valid.
import string
valid_characters = string.ascii_letters + string.digits + '#'
def is_valid_character(character):
return character in valid_characters
# Instead of using `filter`, we `join` all characters in the input string
# if `is_valid_character` is `True`.
def get_valid_characters(string):
return "".join(char for char in string if is_valid_character(char))
Some example output:
>>> print(valid_characters)
abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789#
>>> get_valid_characters("!Hello_#world?")
'Helloworld'
>>> get_valid_characters("user#example")
'user#example'
A simpler way to write it would be using regex. This will accomplish the same thing:
import re
def get_valid_characters(string):
return re.sub(r"[^\w\d#]", "", string)

You could use a lambda function to specify your allowed characters. But also note that filter returns a <filter object> which is an iterator over the returned values. So you will have to stich it back to a string:
string = "?filter_#->me3!"
extra_chars = "#!"
filtered_object = filter(lambda c: c.isalnum() or c in extra_chars, string)
string = "".join(filtered_object)
print(string)
Gives:
filter#me3!

Removing Punctuation and Replacing it with Whitespace using Replace in Python

trying to remove the following punctuation in python I need to use the replace methods to remove these punctuation characters and replace it with whitespace ,.:;'"-?!/
here is my code:
text_punct_removed = raw_text.replace(".", "")
text_punct_removed = raw_text.replace("!", "")
print("\ntext with punctuation characters removed:\n", text_punct_removed)
It only will remove the last one I try to replace, so I tried combining them
text_punct_removed = raw_text.replace(".", "" , "!", "")
print("\ntext with punctuation characters removed:\n", text_punct_removed)
but I get an error message, how do I remove multiple punctuation? Also there will be an issue if I put the " in quotes like this """ which will make a comment, is there a way around that? thanks

If you don't need to explicitly use replace:
exclude = set(",.:;'\"-?!/")
text = "".join([(ch if ch not in exclude else " ") for ch in text])

Here's a naive but working solution:
for sp in '.,"':
raw_text = raw_text.replace(sp, '')

If you need to replace all punctuations with space, you can use the built-in punctuation list to replace the string:
Python 3
import string
import re
my_string = "(I hope...this works!)"
translator = re.compile('[%s]' % re.escape(string.punctuation))
translator.sub(' ', my_string)
print(my_string)
# Result:
# I hope this works
After, if you want to remove double spaces inside string, you can make:
my_string = re.sub(' +',' ', my_string).strip()
print(my_string)
# Result:
# I hope this works

This works in Python3.5.3:
from string import punctuation
raw_text_with_punctuations = "text, with: punctuation; characters? all over ,.:;'\"-?!/"
print(raw_text_with_punctuations)
for char in punctuation:
raw_text_with_punctuations = raw_text_with_punctuations.replace(char, '')
print(raw_text_with_punctuations)

Either remove one character at a time:
raw_text.replace(".", "").replace("!", "")
Or, better, use regular expressions (re.sub()):
re.sub(r"\.|!", "", raw_text)

Splitting a string using re module of python

I have a string
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
#I have to capture only the field 'count_EVENT_GENRE'
field = re.split(r'[(==)(>=)(<=)(in)(like)]', s)[0].strip()
#o/p is 'cou'
# for s = 'sum_EVENT_GENRE in [1,2,3,4,5]' o/p = 'sum_EVENT_GENRE'
which is fine
My doubt is for any character in (in)(like) it is splitting the string s at that character and giving me first slice.(as after "cou" it finds one matching char i:e n). It's happening for any string that contains any character from (in)(like).
Ex : 'percentage_AMOUNT' o/p = 'p'
as it finds a matching char as 'e' after p.
So i want some advice how to treat (in)(like) as words not as characters , when splitting occurs/matters.
please suggest a syntax.

Answering your question, the [(==)(>=)(<=)(in)(like)] is a character class matching single characters you defined inside the class. To match sequences of characters, you need to remove [ and ] and use alternation:
r'==?|>=?|<=?|\b(?:in|like)\b'
or better:
r'[=><]=?|\b(?:in|like)\b'
You code would look like:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
field = re.split(r'[=><]=?|\b(?:in|like)\b', s)[0].strip()
print(field)
However, there might be other (easier, or safer - depending on the actual specifications) ways to get what you want (splitting with space and getting the first item, use re.match with r'\w+' or r'[a-z]+(?:_[A-Z]+)+', etc.)
If your value is at the start of the string and starts with lowercase ASCII letters, and then can have any amount of sequences of _ followed with uppercase ASCII letters, use:
re.match(r'[a-z]+(?:_[A-Z]+)*', s)
Full demo code:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
fieldObj = re.match(r'[a-z]+(?:_[A-Z]+)*', s)
if fieldObj:
print(fieldObj.group())

If you want only the first word of your string, then this should do the job:
import re
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
field = re.split(r'\W', s)[0]
# count_EVENT_GENRE

Is there anything wrong with using split?
>>> s = 'count_EVENT_GENRE in [1,2,3,4,5]'
>>> s.split(' ')[0]
'count_EVENT_GENRE'
>>> s = 'coint_EVENT_GENRE = "ROMANCE"'
>>> s.split(' ')[0]
'coint_EVENT_GENRE'
>>>

How to remove all characters after a specific character in python?

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