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'

Is there a specific reason you need to use open? Since those are online resources, I would suggest you download them and work with the returned bytes/string. — Mike Scotty
– Mike Scotty, Commented Jun 18, 2017 at 19:58

Carl Shiles · Accepted Answer · 2017-06-18 20:07:35Z

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

edited Jun 18, 2017 at 20:07

answered Jun 18, 2017 at 19:59

Carl Shiles

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1 Comment

Brad Solomon Over a year ago

I was following along blindly with this walkthrough, which uses open. Thanks, and in 3.x this is urllib.request.urlopen

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