Python open: Invalid argument error [duplicate]

Question

I have the following two xml links, each of which I want to use as the file parameter to open:

table1 = 'https://www.sec.gov/Archives/edgar/data/1103804/000110380417000040/xslForm13F_X01/Form13fInfoTable.xml'
table2 = 'https://www.sec.gov/Archives/edgar/data/1103804/000110380417000040/Form13fInfoTable.xml'

What I have tried:

• Using raw strings (r'https:// ...)
• Excluding https:// in each of the path names (to get rid of the colon on a Windows system)
• Using 'r' within open(), which should be unnecessary anyway because it is the default

There are a number of similar SO questions, none of which offer solutions that get rid of this error. Although the following is seemingly innocuous I cannot get past the error. For instance,

d = open(table1, 'r')
---------------------------------------------------------------------------
FileNotFoundError                         Traceback (most recent call last)
<ipython-input-110-07d32326a11e> in <module>()
----> 1 d = open(table1, 'r')

FileNotFoundError: [Errno 2] No such file or directory: 'www.sec.gov/Archives/edgar/data/1103804/000110380417000040/xslForm13F_X01/Form13fInfoTable.xml'

Answer 1

This is the open function you want.

import urllib

urllib.urlopen('http://example.com') #python 2

urllib.request.urlopen('http://example.com') #python 3