How do I remove a substring from the end of a string (remove a suffix of the string)?


I have the following code:

url = ''

I expected: abcdc

I got: abcd

Now I do

url.rsplit('.com', 1)

Is there a better way?

See How do the .strip/.rstrip/.lstrip string methods work in Python? for a specific explanation of what the first attempt is doing.

Asked By: Ramya



strip doesn’t mean "remove this substring". x.strip(y) treats y as a set of characters and strips any characters in that set from both ends of x.

On Python 3.9 and newer you can use the removeprefix and removesuffix methods to remove an entire substring from either side of the string:

url = ''
url.removesuffix('.com')    # Returns 'abcdc'
url.removeprefix('abcdc.')  # Returns 'com'

The relevant Python Enhancement Proposal is PEP-616.

On Python 3.8 and older you can use endswith and slicing:

url = ''
if url.endswith('.com'):
    url = url[:-4]

Or a regular expression:

import re
url = ''
url = re.sub('.com$', '', url)
Answered By: Steef

How about url[:-4]?

Answered By: Daren Thomas

This is a perfect use for regular expressions:

>>> import re
>>> re.match(r"(.*).com", "").group(1)
Answered By: Aaron Maenpaa

If you know it’s an extension, then

url = ''
url.rsplit('.', 1)[0]  # split at '.', starting from the right, maximum 1 split

This works equally well with or or abcdc.[anything] and is more extensible.

Answered By: JohnMetta

Depends on what you know about your url and exactly what you’re tryinh to do. If you know that it will always end in ‘.com’ (or ‘.net’ or ‘.org’) then


is the quickest solution. If it’s a more general URLs then you’re probably better of looking into the urlparse library that comes with python.

If you on the other hand you simply want to remove everything after the final ‘.’ in a string then


will work. Or if you want just want everything up to the first ‘.’ then try

Answered By: dagw
def strip_end(text, suffix):
    if suffix and text.endswith(suffix):
        return text[:-len(suffix)]
    return text
Answered By: yairchu

If you are sure that the string only appears at the end, then the simplest way would be to use ‘replace’:

url = ''
Answered By: Charles Collis

On Python 3.9+:


On any Python version:

def remove_suffix(text, suffix):
    return text[:-len(suffix)] if text.endswith(suffix) and len(suffix) != 0 else text

or the one-liner:

remove_suffix = lambda text, suffix: text[:-len(suffix)] if text.endswith(suffix) and len(suffix) != 0 else text
Answered By: David Foster

You can use split:

# 'abccomputer'
Answered By: Lucas

For urls (as it seems to be a part of the topic by the given example), one can do something like this:

import os
url = ''
name,ext = os.path.splitext(url)
print (name, ext)

ext = '.'+url.split('.')[-1]
name = url[:-len(ext)]
print (name, ext)

Both will output:
('http://www.stackoverflow', '.com')

This can also be combined with str.endswith(suffix) if you need to just split “.com”, or anything specific.

Answered By: b0bz

Since it seems like nobody has pointed this on out yet:

url = ""
new_url = url[:url.rfind(".")]

This should be more efficient than the methods using split() as no new list object is created, and this solution works for strings with several dots.

Answered By: user3129181

In my case I needed to raise an exception so I did:

class UnableToStripEnd(Exception):
    """A Exception type to indicate that the suffix cannot be removed from the text."""

    def get_exception(text, suffix):
        return UnableToStripEnd("Could not find suffix ({0}) on text: {1}."
                                .format(suffix, text))

def strip_end(text, suffix):
    """Removes the end of a string. Otherwise fails."""
    if not text.endswith(suffix):
        raise UnableToStripEnd.get_exception(text, suffix)
    return text[:len(text)-len(suffix)]
Answered By: juan Isaza
import re

def rm_suffix(url = '', suffix='.com'):
    return(re.sub(suffix+'$', '', url))

I want to repeat this answer as the most expressive way to do it. Of course, the following would take less CPU time:

def rm_dotcom(url = ''):
    return(url[:-4] if url.endswith('.com') else url)

However, if CPU is the bottle neck why write in Python?

When is CPU a bottle neck anyway? In drivers, maybe.

The advantages of using regular expression is code reusability. What if you next want to remove ‘.me’, which only has three characters?

Same code would do the trick:

>>> rm_sub('','.me')
Answered By: Ofer Rahat

DSCLAIMER This method has a critical flaw in that the partition is not anchored to the end of the url and may return spurious results. For example, the result for the URL "" is "www" (incorrect) instead of the expected "". This solution therefore is evil. Don’t use it unless you know what you are doing!


This is fairly easy to type and also correctly returns the original string (no error) when the suffix ‘.com’ is missing from url.

Answered By: winni2k

If you mean to only strip the extension:

# 'abcdc'

It works with any extension, with potential other dots existing in filename as well. It simply splits the string as a list on dots and joins it without the last element.

Answered By: Dcs

Here,i have a simplest code.


Assuming you want to remove the domain, no matter what it is (.com, .net, etc). I recommend finding the . and removing everything from that point on.

url = ''
dot_index = url.rfind('.')
url = url[:dot_index]

Here I’m using rfind to solve the problem of urls like which should be reduced to the name

If you’re also concerned about www.s, you should explicitly check for them:

if url.startswith("www."):
   url = url.replace("www.","", 1)

The 1 in replace is for strange edgecases like

If your url gets any wilder than that look at the regex answers people have responded with.

Answered By: Xavier Guay

If you need to strip some end of a string if it exists otherwise do nothing. My best solutions. You probably will want to use one of first 2 implementations however I have included the 3rd for completeness.

For a constant suffix:

def remove_suffix(v, s):
    return v[:-len(s)] if v.endswith(s) else v
remove_suffix("", ".com") == 'abc'
remove_suffix("abc", ".com") == 'abc'

For a regex:

def remove_suffix_compile(suffix_pattern):
    r = re.compile(f"(.*?)({suffix_pattern})?$")
    return lambda v: r.match(v)[1]
remove_domain = remove_suffix_compile(r".[a-zA-Z0-9]{3,}")
remove_domain("") == "abc"
remove_domain("") == ""
remove_domain("abc.") == "abc."
remove_domain("abc") == "abc"

For a collection of constant suffixes the asymptotically fastest way for a large number of calls:

def remove_suffix_preprocess(*suffixes):
    suffixes = set(suffixes)
    except KeyError:

    def helper(suffixes, pos):
        if len(suffixes) == 1:
            suf = suffixes[0]
            l = -len(suf)
            ls = slice(0, l)
            return lambda v: v[ls] if v.endswith(suf) else v
        si = iter(suffixes)
        ml = len(next(si))
        exact = False
        for suf in si:
            l = len(suf)
            if -l == pos:
                exact = True
                ml = min(len(suf), ml)
        ml = -ml
        suffix_dict = {}
        for suf in suffixes:
            sub = suf[ml:pos]
            if sub in suffix_dict:
                suffix_dict[sub] = [suf]
        if exact:
            del suffix_dict['']
            for key in suffix_dict:
                suffix_dict[key] = helper([s[:pos] for s in suffix_dict[key]], None)
            return lambda v: suffix_dict.get(v[ml:pos], lambda v: v)(v[:pos])
            for key in suffix_dict:
                suffix_dict[key] = helper(suffix_dict[key], ml)
            return lambda v: suffix_dict.get(v[ml:pos], lambda v: v)(v)
    return helper(tuple(suffixes), None)
domain_remove = remove_suffix_preprocess(".com", ".net", ".edu", ".uk", '.tv', '', '')

the final one is probably significantly faster in pypy then cpython. The regex variant is likely faster than this for virtually all cases that do not involve huge dictionaries of potential suffixes that cannot be easily represented as a regex at least in cPython.

In PyPy the regex variant is almost certainly slower for large number of calls or long strings even if the re module uses a DFA compiling regex engine as the vast majority of the overhead of the lambda’s will be optimized out by the JIT.

In cPython however the fact that your running c code for the regex compare almost certainly outweighs the algorithmic advantages of the suffix collection version in almost all cases.


Answered By: user1424589

Starting in Python 3.9, you can use removesuffix instead:

# 'abcdc'
Answered By: Xavier Guihot

I used the built-in rstrip function to do it like follow:

string = ""
suffix = ".com"
newstring = string.rstrip(suffix)
Answered By: Zioalex

A broader solution, adding the possibility to replace the suffix (you can remove by replacing with the empty string) and to set the maximum number of replacements:

def replacesuffix(s,old,new='',limit=1):
    String suffix replace; if the string ends with the suffix given by parameter `old`, such suffix is replaced with the string given by parameter `new`. The number of replacements is limited by parameter `limit`, unless `limit` is negative (meaning no limit).

    :param s: the input string
    :param old: the suffix to be replaced
    :param new: the replacement string. Default value the empty string (suffix is removed without replacement).
    :param limit: the maximum number of replacements allowed. Default value 1.
    :returns: the input string with a certain number (depending on parameter `limit`) of the rightmost occurrences of string given by parameter `old` replaced by string given by parameter `new`
    if s[len(s)-len(old):] == old and limit != 0:
        return replacesuffix(s[:len(s)-len(old)],old,new,limit-1) + new
        return s

In your case, given the default arguments, the desired result is obtained with:

>>> 'abcdc'

Some more general examples:

>>> 'whatever-qweNN'

>>> 'whatever-NNN'

replacesuffix('12.53000','0',' ',-1)
>>> '12.53   '
Answered By: mmj

Because this is a very popular question i add another, now available, solution. With python 3.9 ( the function removesuffix() will be added (and removeprefix()) and this function is exactly what was questioned here.

url = ''



PEP 616 ( shows how it will behave (it is not the real implementation):

def removeprefix(self: str, prefix: str, /) -> str:
    if self.startswith(prefix):
        return self[len(prefix):]
        return self[:]

and what benefits it has against self-implemented solutions:

  1. Less fragile:
    The code will not depend on the user to count the length of a literal.

  2. More performant:
    The code does not require a call to the Python built-in len function nor to the more expensive str.replace() method.

  3. More descriptive:
    The methods give a higher-level API for code readability as opposed to the traditional method of string slicing.

Answered By: D-E-N

Using replace and count

This might seems a little bit a hack but it ensures you a safe replace without using startswith and if statement, using the count arg of replace you can limit the replace to one:

mystring = ""



Suffix (you write the prefix reversed) .com becomes moc.:

Answered By: G M

Use the public suffix list hosted by Mozilla. It’s available as the tldextract python library.

import tldextract

url = ''

# Extract the domain and TLD
extracted = tldextract.extract(url)
domain, tld = extracted.domain, extracted.suffix

if tld and tld != 'localhost':
    url_without_tld = domain
    url_without_tld = url

Answered By: Anonymoose

Function To Remove a Suffix in Python 3.8 :

def removesuffix(text, suffix):
if text.endswith(suffix):
    return text[:-len(suffix)]
    return text
Answered By: alemol
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