title: Why does os.path.abspath() return the path of cwd+file?
body: <p>If I have a structure like such:</p>
  
  <pre><code>root/
  -- group1/
  ---- names/
  ---- places/
  ------ foo.zip
  </code></pre>
  
  <p>Why is that when I call <code>os.path.abspath('foo.zip')</code> I get the file path of where the Python script is located plus the <code>foo.zip</code>? </p>
  
  <p>Looks like: <code>H:\Program\Scripts\foo.zip</code> </p>
  
  <p>Needs to be: <code>H:\Progran\Groups\group1\names\places\foo.zip</code></p>
  
  <p>This is the code I have for a function where the problem is resulting:</p>
  
  <pre><code>def unzip(in_dir):
      # in_dir is places passed to unzip()
      files = [f for f in os.listdir(os.path.abspath(in_dir)) if f.endswith('.zip')]
      for zip in files:
          # This prints the 'looks like' path above
          print os.path.abspath(zip)
  </code></pre>
  
  <p>Shouldn't <code>print os.path.abspath(zip)</code> give me the full path of each file that was found in the <code>os.listdir(os.path.abspath(in_dir))</code>?</p>