 <pre><code>dups = {} newlist = [] for x in biglist: if x['link'] not in dups: newlist.append(x) dups[x['link']] = None print newlist </code></pre> <p>produces</p> <pre><code>[{'link': 'u2.com', 'title': 'U2 Band'}, {'link': 'abc.com', 'title': 'ABC Station'}] </code></pre> <p>Note that here I used a dictionary. This makes the test <code>not in dups</code> much more efficient than using a list.</p>