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urlparse.urlparse takes a url of the format:
 <scheme>://<netloc>/<path>;<params>?<query>#<fragment>
And returns a 6-tuple of the format:
 (scheme, netloc, path, params, query, fragment).
An example from the library refrence takes:
 urlparse('http://www.cwi.nl:80/%7Eguido/Python.html')
And produces:
 ('http', 'www.cwi.nl:80', '/%7Eguido/Python.html', '',
'', '')
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Note that there isn't a field for the port number in the
6-tuple. Instead, it is included in the netloc. Urlparse
should handle your example as:
>>> urlparse.urlparse('1.2.3.4:80','http') 
('http', '1.2.3.4:80', '', '', '', '')
Instead, it gives the incorrect output as you indicated.