Python One-liner to get your site's Alexa Rank

alexa.py

#!/usr/bin/env python
import urllib, sys, bs4
print bs4.BeautifulSoup(urllib.urlopen("http://data.alexa.com/data?cli=10&dat=s&url="+ sys.argv[1]).read(), "xml").find("REACH")['RANK']

error is popping that index is out of bound...help!!!

:) @risheek20

import urllib.request, sys, re
import xmltodict, json

xml = urllib.request.urlopen('http://data.alexa.com/data?cli=10&dat=s&url={}'.format("www.example.com")).read()
 
result= xmltodict.parse(xml)
 
data = json.dumps(result).replace("@","")
data_tojson = json.loads(data)
url = data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["URL"]
rank= data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["TEXT"]
 
print("site {site}, rank {rank}".format(site=url,rank=rank))

I made one here: alexa_check.py

Not a one liner but it gets all ranks available, global and national. It will also show other countries rank (sometimes only one country will be shown on the website-besides global)...

For instance facebook.com will show only Global and US rank on the Alexa page. However my script finds that they are number 1 in several countries

:) @risheek20

import urllib.request, sys, re
import xmltodict, json

xml = urllib.request.urlopen('http://data.alexa.com/data?cli=10&dat=s&url={}'.format("www.example.com")).read()
 
result= xmltodict.parse(xml)
 
data = json.dumps(result).replace("@","")
data_tojson = json.loads(data)
url = data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["URL"]
rank= data_tojson["ALEXA"]["SD"][1]["POPULARITY"]["TEXT"]
 
print("site {site}, rank {rank}".format(site=url,rank=rank))

hey buddy, ther's problem with parser.Parse(xml_input, True) ..... can you please edited???
I would rate it very much
Thanks in advance!