Showing the domain of a URL

Question

I have such a method which takes a URL, such as http://twitter.com/ttt or twitter.com/dasd, and should show its domain with http: http://twitter.com and name twitter

def show_domain(url):
 sites = {
 'twitter': ['http://www.twitter.com/', 'twitter'],
 'facebook': ['http://www.facebook.com/', 'facebook'],
 'pininterest': ['http://www.pininterest.com/', 'pininterest']
 }
 if sites['twitter'][0] in url:
 link = sites['twitter'][0]
 brand = sites['twitter'][1]
 elif sites['facebook'][0] in url:
 link = sites['facebook'][0]
 brand = sites['facebook'][1]
 else:
 link = sites['pininterest'][0]
 brand = sites['pininterest'][1]
 return link, brand

Is there way to optimise this code?

• jcollado: Have you considered using the urlparse module?

Answer 1

You can use tuple unpacking/list unpacking to write :

if sites['twitter'][0] in url:
 link, brand = sites['twitter']
elif sites['facebook'][0] in url:
 link, brand = sites['facebook']
else:
 link, brand = sites['pininterest']
return link, brand

Now, the data structure you are using is a bit akward because you don't use the key at all. Instead of mapping string to list of strings, you could have a list of list of strings. Even better, you could use list of tuple of strings :

sites = [
 ('http://www.twitter.com/', 'twitter'),
 ('http://www.facebook.com/', 'facebook'),
 ('http://www.pininterest.com/', 'pininterest')
]
if sites[0][0] in url:
 link, brand = sites[0]
elif sites[1][0] in url:
 link, brand = sites[1]
else:
 link, brand = sites[2]
return link, brand

Now, this seems to be begging to be written with a loop

for (link, brand) in sites:
 if link in url:
 return link, brand
return None # or anything you like

• user2424174: but it local variables link, brand are unused, pycharm says
• Davidmh: @user2424174 then it is a bug in pycharm, because they are being returned, so used.