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

url = 'abcdc.com' print(url.replace('.com','')) 

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

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

url = "www.example.com" 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.

Starting in Python 3.9, you can use removesuffix instead:

'abcdc.com'.removesuffix('.com') # 'abcdc' 

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

 url=url[:-4] 

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

url.rsplit('.',1)[0] 

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

url.split('.',1)[0] 

On Python 3.9+:

text.removesuffix(suffix) 

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 

If you know it's an extension, then

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

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

How about url[:-4]?

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 = 'http://www.stackoverflow.com' name,ext = os.path.splitext(url) print (name, ext) #Or: 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.

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 "www.comcast.net" is "www" (incorrect) instead of the expected "www.comcast.net". This solution therefore is evil. Don't use it unless you know what you are doing!

url.rpartition('.com')[0] 

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

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

Here I'm using rfind to solve the problem of urls like abcdc.com.net which should be reduced to the name abcdc.com.

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 www.net.www.com

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

If you mean to only strip the extension:

'.'.join('abcdc.com'.split('.')[:-1]) # '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.

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("abc.com", ".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.com") == "abc" remove_domain("sub.abc.net") == "sub.abc" 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) try: suffixes.remove('') except KeyError: pass 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 else: 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].append(suf) else: 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]) else: 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', '.co.uk', '.org.uk') 

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.

Edit: https://m.xkcd.com/859/

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

url = 'abcdc.com' print(url.removesuffix('.com')) 

output:

'abcdc' 

PEP 616 (https://www.python.org/dev/peps/pep-0616/) 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):] else: 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.

import re def rm_suffix(url = 'abcdc.com', 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 = 'abcdc.com'): 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('abcdc.me','.me') 'abcdc' 

Function To Remove a Suffix in Python 3.8 :

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

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.""" @staticmethod 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)] 

You can use split:

'abccomputer.com'.split('.com',1)[0] # 'abccomputer' 

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 else: return s 

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

replacesuffix('abcdc.com','.com') >>> 'abcdc' 

Some more general examples:

replacesuffix('whatever-qweqweqwe','qwe','N',2) >>> 'whatever-qweNN' replacesuffix('whatever-qweqweqwe','qwe','N',-1) >>> 'whatever-NNN' replacesuffix('12.53000','0',' ',-1) >>> '12.53 ' 

This is a perfect use for regular expressions:

>>> import re >>> re.match(r"(.*)\.com", "hello.com").group(1) 'hello' 

In case anybody else is having this issue, I realized that I was not reassigning the return value of strip() and incorrectly making the assumption that the original variable was being altered by reference.

MY ORIGINAL PROBLEM

og_str = "first second third" og_str.strip("second", "") print(og_str) . >> result: "first second third" >> !! I expected "first third" 

.
SOLUTION:

og_str = "first second third" og_str = og_str.strip("second", "") print(og_str) . >> result: "first third" >> That is what I expected <whew> 

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 = "www.comwww.com" 

Prefix:

print(mystring.replace("www.","",1)) 

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

print(mystring[::-1].replace("moc.","",1)[::-1]) 

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

import tldextract url = 'abcdc.com' # Extract the domain and TLD extracted = tldextract.extract(url) domain, tld = extracted.domain, extracted.suffix if tld and tld != 'localhost': url_without_tld = domain else: url_without_tld = url print(url_without_tld) 

Here,i have a simplest code.

url=url.split(".")[0]