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copied!<p>I'd just use <code>itertools.groupby</code>. Something like:</p> <pre><code>top_users = [(k, list(g)) for k,g in groupby(top_users, key=lambda x: x.score))] for u in top_users[0][1]: u.status = 'First prize' for u in top_users[1][1]: u.status = 'Second prize' for u in top_users[2][1]: u.status = 'Third prize' for score, users in top_users[3:]: for u in users: u.status = 'Highly recommended' </code></pre> <p>Or even better, use <code>itertools.count</code> instead of the 4 loops:</p> <pre><code>top_users = [(k, list(g)) for k,g in groupby(top_users, key=lambda x: x.score))] for c, (score, group) in zip(count(0), top_users): if c == 0: prize = 'First prize' elif c == 1: prize = 'Second prize' elif c == 2: prize = 'Third prize' else: prize = 'Highly recommended' map(lambda x: setattr(x, 'status', prize), group) </code></pre> <p>And the last refinement, maybe keep a prize list instead of the if statements.</p> <pre><code>top_users = [(k, list(g)) for k,g in groupby(top_users, key=lambda x: x.score))] prize_list = ['First prize', 'Second prize', 'Third prize', 'Highly recommended'] for c, (score, group) in zip(count(0), top_users): prize = prize_list[c] if c < len(prize_list) else prize_list[-1] map(lambda x: setattr(x, 'status', prize), group) </code></pre> <p>The caveat of this approach is that you're not doing the grouping in the database, but instead you'd be doing it in memory. This may be a problem if there are a lot of users. See <a href="https://stackoverflow.com/questions/629551/how-to-query-as-group-by-in-django">How to query as GROUP BY in django?</a> for some guidance on how to do this in the database.</p>

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