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copied!<p>First, I would stick with <code>itervalues</code>, which uses easy evaluation.</p> <p>Second, I would be wary of relying on using <code>set</code> since it performs looking up the value in the set on every iteration of going through dictionary. It's <code>O(1)</code> on the overage (and <code>O(n)</code> in the worse case which is <code>O(1)</code> in our case if all the length are the same, and <code>O(n)</code> if all the length are different) according to the <a href="http://wiki.python.org/moin/TimeComplexity" rel="nofollow">docs</a>. But it's difficult to asses the overhead of using set.</p> <p>I would use <code>all</code> in this case. <code>all</code> fails when it finds the first <code>False</code> value. So, the first mismatch of the length would stop the interating process. While, if using <code>set</code>, it would go through <strong>all the list</strong> to the end and only then compare its length to <code>1</code>.</p> <pre><code>>>> dct = {'a': [1, 2, 3], 'b': [1, 2, 3, 4], 'c': [1, 2]} >>> lenght_1 = len(dct.itervalues().next()) >>> all(len(value)==lenght_1 for value in dct.itervalues()) False >>> dct = {'a': [1, 2, 3], 'b': [1, 2, 4], 'c': [1, 2, 5]} >>> lenght_1 = len(dct.itervalues().next()) >>> all(len(value)==lenght_1 for value in dct.itervalues()) True </code></pre> <p>The code can be optimized by using the same iterator <code>it</code> which will not go through the first value twice:</p> <pre><code>>>> it = dct.itervalues() >>> length_1 = len(next(it)) >>> all(len(value)==l1 for value in it) True </code></pre>

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