get everything after a particular string in python using regular expression

Question

url="www.example.com/thedubaimall"

I want to save everything that comes aftercom/ that is I need only thedubaimall

How can i do it with regular expression??

Answer 1

Obligatory "what if you didn't use regex?" answer:

>>> url="www.example.com/thedubaimall"
>>> url.partition(".com/")[2]
'thedubaimall'

Answer 2

If you want to use regular expression, you use re.findall:

re.findall('(?<=com/).*$', "www.example.com/thedubaimall")
# ['thedubaimall']

Some speed tests with @DeepSpace's suggestion:

%timeit re.findall('(?<=com/).*$', "www.example.com/thedubaimall")
# The slowest run took 7.57 times longer than the fastest. This could mean that an intermediate result is being cached.
# 1000000 loops, best of 3: 1.29 µs per loop

%timeit re.findall('com/(.*)', "www.example.com/thedubaimall")
# The slowest run took 6.48 times longer than the fastest. This could mean that an intermediate result is being cached.
# 1000000 loops, best of 3: 992 ns per loop

%timeit "www.example.com/thedubaimall".partition(".com/")[2]
# The slowest run took 7.87 times longer than the fastest. This could mean that an intermediate result is being cached.
# 1000000 loops, best of 3: 204 ns per loop

Looks like @DeepSpace's suggestion is a bit faster, and @Kevin's answer is much faster.