title: Compressing directory using shutil.make_archive() while preserving directory structure
body: <p>I'm trying to zip a directory called <code>test_dicoms</code> to a zip file named <code>test_dicoms.zip</code> using the following code:</p>
  <pre class="lang-py prettyprint-override"><code>shutil.make_archive('/home/code/test_dicoms', 'zip', '/home/code/test_dicoms')
  </code></pre>
  <p>The problem is that when I unzip it, all of the files that were in <code>/test_dicoms/</code> are extracted to <code>/home/code/</code> instead of the folder <code>/test_dicoms/</code> and all of it's contained files being extracted to <code>/home/code/</code>. So <code>/test_dicoms/</code> has a file called <code>foo.txt</code> and after I zip and unzip <code>foo.txt</code>'s path is <code>/home/code/foo.txt</code> as opposed to <code>/home/code/test_dicoms/foo.txt</code>. How do I fix this? Also, some of the directories I'm working with are very large. Will I need to add anything to my code to make it ZIP64 or is the function smart enough to do that automatically?</p>
  <p>Here's what's currently in the archive created:</p>
  <pre class="lang-none prettyprint-override"><code>[gwarner@jazz gwarner]$ unzip -l test_dicoms.zip
  Archive: test_dicoms.zip
  Length    Date       Time  Name
  --------- ---------- ----- ----
      93324 09-17-2015 16:05 AAscout_b_000070
      93332 09-17-2015 16:05 AAscout_b_000125
      93332 09-17-2015 16:05 AAscout_b_000248
  </code></pre>