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  • Pysam installation problem in Red Hat Linux

    When installing the pysam Python module, version 0.7.4, it seems that the setup.py code first downloads "distribute" as below:

    Downloading http://pypi.python.org/packages/sour...-0.6.34.tar.gz

    On the server, it took a long time to respond with an error: connection timeout. Is there a way to get rid of this step? If there is another way of installing Pysam that would be great. Thanks!

  • #2
    Your local institutional firewall may be blocking access to the python.org site.

    Pysam is available here: http://code.google.com/p/pysam/downloads/list
    Last edited by GenoMax; 06-14-2013, 11:30 AM.

    Comment


    • #3
      Does this look like a proxy issue? Results from: $ python setup.py build

      [root@g300-149-b0 pysam-0.7.4]# python setup.py build
      Downloading http://pypi.python.org/packages/sour...-0.6.34.tar.gz
      Traceback (most recent call last):
      File "setup.py", line 127, in <module>
      use_setuptools()
      File "/app/pysam-0.7.4/distribute_setup.py", line 169, in use_setuptools
      download_delay)
      File "/app/pysam-0.7.4/distribute_setup.py", line 131, in _do_download
      to_dir, download_delay)
      File "/app/pysam-0.7.4/distribute_setup.py", line 201, in download_setuptools
      src = urlopen(url)
      File "/usr/lib64/python2.6/urllib2.py", line 126, in urlopen
      return _opener.open(url, data, timeout)
      File "/usr/lib64/python2.6/urllib2.py", line 397, in open
      response = meth(req, response)
      File "/usr/lib64/python2.6/urllib2.py", line 510, in http_response
      'http', request, response, code, msg, hdrs)
      File "/usr/lib64/python2.6/urllib2.py", line 429, in error
      result = self._call_chain(*args)
      File "/usr/lib64/python2.6/urllib2.py", line 369, in _call_chain
      result = func(*args)
      File "/usr/lib64/python2.6/urllib2.py", line 616, in http_error_302
      return self.parent.open(new, timeout=req.timeout)
      File "/usr/lib64/python2.6/urllib2.py", line 391, in open
      response = self._open(req, data)
      File "/usr/lib64/python2.6/urllib2.py", line 409, in _open
      '_open', req)
      File "/usr/lib64/python2.6/urllib2.py", line 369, in _call_chain
      result = func(*args)
      File "/usr/lib64/python2.6/urllib2.py", line 1198, in https_open
      return self.do_open(httplib.HTTPSConnection, req)
      File "/usr/lib64/python2.6/urllib2.py", line 1165, in do_open
      raise URLError(err)
      urllib2.URLError: <urlopen error [Errno -2] Name or service not known>

      Comment


      • #4
        Quite possible.

        If you are behind a proxy server, are you configured to go through it? What's the contents of your /etc/environment ?

        You can add the following lines to it ... obviously customising as appropriate:

        no_proxy=localhost,127.0.0.0/8,*.local
        NO_PROXY=localhost,127.0.0.0/8,*.local
        all_proxy=socks://proxy.example.com:8080/
        ALL_PROXY=socks://proxy.example.com:8080/
        http_proxy=http://proxy.example.com:8080
        HTTP_PROXY=http://proxy.example.com:8080
        ftp_proxy=http://proxy.example.com:8080
        FTP_PROXY=http://proxy.example.com:8080
        https_proxy=http://proxy.example.com:8080
        HTTPS_PROXY=http://proxy.example.com:8080

        Comment


        • #5
          It turns out I needed to have my hostname identified in /etc/hosts file for python to work with urlopen

          Comment

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