 <p>Yes <a href="http://docs.python.org/library/stdtypes.html#dict.get" rel="nofollow">dict.get</a> is the correct (or at least, the simplest) way:</p> <pre><code>sorted(trial_list, key=trial_dict.get) </code></pre> <p>As Mark Amery commented, the equivalent explicit lambda:</p> <pre><code>sorted(trial_list, key=lambda x: trial_dict[x]) </code></pre> <p>might be better, for at least two reasons:</p> <ol> <li>the sort expression is visible and immediately editable</li> <li>it doesn't suppress errors (when the list contains something that is not in the dict). </li> </ol>