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How to get everything after string x in python

I have a string:
s3://tester/test.pdf
I want to exclude s3://tester/ so even if i have s3://tester/folder/anotherone/test.pdf I am getting the entire path after s3://tester/
I have attempted to use the split & partition method but I can't seem to get it.
Currently am trying:
string.partition('/')[3]
But i get an error saying that it out of index.
EDIT: I should have specified that the name of the bucket will not always be the same so I want to make sure that it is only grabbing anything after the 3rd '/'.

You can use str.split():
path = 's3://tester/test.pdf'
print(path.split('/', 3)[-1])
Output:
test.pdf
UPDATE: With regex:
import re
path = 's3://tester/test.pdf'
print(re.split('/',path,3)[-1])
Output:
test.pdf

Have you tried .replace?
You could do:
string = "s3://tester/test.pdf"
string = string.replace("s3://tester/", "")
print(string)
This will replace "s3://tester/" with the empty string ""
Alternatively, you could use .split rather than .partition
You could also try:
string = "s3://tester/test.pdf"
string = "/".join(string.split("/")[3:])
print(string)

To answer "How to get everything after x amount of characters in python"
string[x:]

PLEASE SEE UPDATE
ORIGINAL
Using the builtin re module.
p = re.search(r'(?<=s3:\/\/tester\/).+', s).group()
The pattern uses a lookbehind to skip over the part you wish to ignore and matches any and all characters following it until the entire string is consumed, returning the matched group to the p variable for further processing.
This code will work for any length path following the explicit s3://tester/ schema you provided in your question.
UPDATE
Just saw updates duh.
Got the wrong end of the stick on this one, my bad.
Below re method should work no matter S3 variable, returning all after third / in string.
p = ''.join(re.findall(r'\/[^\/]+', s)[1:])[1:]

Python regex to extract substring at start and end of string

I am looking for a regex that will extract everything up to the first . (period) in a string, and everything including and after the last . (period)
For example:
my_file.10.4.5.6.csv
myfile2.56.3.9.txt
Ideally the regex when run against these strings would return:
my_file.csv
myfile2.txt
The numeric stamp in the file will be different each time the script is run, so I am looking essentially to exclude it.
The following prints out the string up to the first . (period)
print re.search("^[^.]*", data_file).group(0)
I am having trouble though getting it to also return the the last period and string after it.
Sorry just to update this based upon feedback and comments below:
This does need to be a regex. The regex will be passed into the program from a configuration file. The user will not have access to the source code as it will be packaged.
The user may need to change the regex based upon some arbitrary criteria, so they will need to update the config file, rather than edit the application and re-build the package.
Thanks

You don’t need a regular expression!
parts = data_file.split(".")
print parts[0] + "." + parts[-1]

Instead of regular expressions, I would suggest using str.split. For example:
>>> data_file = 'my_file.10.4.5.6.csv'
>>> parts = data_file.split('.')
>>> print parts[0] + '.' + parts[-1]
my_file.csv
However if you insist on regular expressions, here is one approach:
>>> print re.sub(r'\..*\.', '.', data_file)
my_file.csv

You don't need a regex.
tokens = expanded_name.split('.')
compressed_name = '.'.join((tokens[0], tokens[-1]))
If you are concerned about performance, you could use a length limit and rsplit() to only chop up the string as much as you need.
compressed_name = expanded_name.split('.', 1)[0] + '.' + expanded_name.rsplit('.', 1)[1]

Do you need a regex here?
>>> address = "my_file.10.4.5.6.csv"
>>> split_by_periods = address.split(".")
>>> "{}.{}".format(address[0], address[-1])
>>> "my_file.csv"

Escape (Insert backslash) before brackets in a Python string

I need to format many strings that contain a similar structure:
u'LastName FirstName (Department / Subdepartment)'
My wish is to get the string to look like this:
u'LastName FirstName \(Department / Subdepartment\)'
Meaning I need to add a backslash to the opening bracket and to the closing bracket.
So far I am doing this in Python:
displayName = displayName.replace('(', '\(').replace(')', '\)').
Which seems OK, but I am just wondering:
Is there is a more Pythonic way to do it?
I did not find a proper way Python's String documentation, but maybe I am looking in the wrong place...

You've already found the most Pythonic way, regex provides a not so readable solution:
>>> import re
>>> s = u'LastName FirstName (Department / Subdepartment)'
>>> print re.sub(r'([()])', r'\\\1', s)
LastName FirstName \(Department / Subdepartment\)

you can use re.escape('string').
example:
import re
escaped = re.escape(u'LastName FirstName (Department / Subdepartment)')
Note:
This method will return the string with all non-alphanumerics backslashed which includes punctuation and white-space.
Though that may be useful for you.

Python Regular Expression findall with variable

I am trying to use re.findall with look-behind and look-forward to extract data. The regular expression works fine when I am not using a raw_input variable, but I need users to be able to input a variety of different search terms.
Here is the current code:
me = re.findall(r"(?<='(.+)'+variable+'(.+)')(.*?)(?='(.+)+variable+(.+)')", raw)
As you can see, I am trying to pull out strings between one search term.
However, each time I use this type of formatting, I get a fixed width error. Is there anyway around this?
I have also tried the following formats with no success.
variable = raw_input('Term? ')
'.*' + variable + '.*'
and
'.*%s.*' % (variable, )
and
'.*{0}.*'.format(variable)
and
'.*{variable}.*'.format(variable=variable)

I'm not sure if this is what you mean, but it may get you started. As far as I understand your question, you don't need lookaheads or lookbehinds. This is for Python 2.x (won't work with Python 3):
>>> import re
>>> string_to_search = 'fish, hook, swallowed, reeled, boat, fish'
>>> entered_by_user = 'fish'
>>> search_regex = r"{0}(.+){0}".format(entered_by_user)
>>> match = re.search(search_regex, string_to_search)
>>> if match:
... print "result:", match.group(1).strip(' ,')
...
result: hook, swallowed, reeled, boat
If you really want the last 'fish' in the result as in your comment above, then just remove the second {0} from the format() string.

This solution should work:
me = re.findall(rf"(?<='(.+)'+{variable}+'(.+)')(.*?)(?='(.+)+{variable}+(.+)')", raw)
You also can add many different variables as you wish.
Add rf for the regular expression and the desired variables in between {}
import re
text = "regex is the best"
var1 = "is the"
var2 = "best"
yes = re.findall(rf"regex {var1} {var2}", text)
print(yes)
['regex is the best']

The way lookbehind is usually implemented (including its Python implementation) has an inherent limitation that you are unfortunately running into: lookbehinds cannot be variable-length. The "Important Notes About Lookbehind" section here explains why. I think you should be able to do the regex without a lookbehind, though.

Python Regex, re.sub, replacing multiple parts of pattern?

I can't seem to find a good resource on this.. I am trying to do a simple re.place
I want to replace the part where its (.*?), but can't figure out the syntax on how to do this.. I know how to do it in PHP, so I've been messing around with what I think it could be based on that (which is why it has the $1 but I know that isn't correct in python).. I would appreciate if anyone can show the proper syntax, I'm not asking specifics for any certain string, just how I can replace something like this, or if it had more than 1 () area.. thanks
originalstring = 'fksf var:asfkj;'
pattern = '.*?var:(.*?);'
replacement_string='$1' + 'test'
replaced = re.sub(re.compile(pattern, re.MULTILINE), replacement_string, originalstring)

>>> import re
>>> originalstring = 'fksf var:asfkj;'
>>> pattern = '.*?var:(.*?);'
>>> pattern_obj = re.compile(pattern, re.MULTILINE)
>>> replacement_string="\\1" + 'test'
>>> pattern_obj.sub(replacement_string, originalstring)
'asfkjtest'
Edit: The Python Docs can be pretty useful reference.

>>> import re
>>> regex = re.compile(r".*?var:(.*?);")
>>> regex.sub(r"\1test", "fksf var:asfkj;")
'asfkjtest'

The python docs are online, and the one for the re module is here. http://docs.python.org/library/re.html
To answer your question though, Python uses \1 rather than $1 to refer to matched groups.