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    <pre><code>In [6]: sorted(mydict.iteritems(), key=lambda (k,v): operator.itemgetter(1)(v)) Out[6]: [('a2', ['e', 2]), ('a4', ['s', 2]), ('a3', ['h', 3]), ('a1', ['g', 6]), ('a6', ['y', 7]), ('a5', ['j', 9])] </code></pre> <p>The key parameter is always a function that is fed one item from the iterable (<code>mydict.iteritems()</code>) at a time. In this case, an item could be something like </p> <pre><code>('a2',['e',2]) </code></pre> <p>So we need a function that can take <code>('a2',['e',2])</code> as input and return 2.</p> <p><code>lambda (k,v): ...</code> is an anonymous function which takes one argument -- a 2-tuple -- and unpacks it into <code>k</code> and <code>v</code>. So when the <code>lambda</code> function is applied to our item, <code>k</code> would be <code>'a2'</code> and <code>v</code> would be <code>['e',2]</code>. </p> <p><code>lambda (k,v): operator.itemgetter(1)(v)</code> applied to our item thus returns <code>operator.itemgetter(1)(['e',2])</code>, which "itemgets" the second item in <code>['e',2]</code>, which is 2.</p> <p>Note that <code>lambda (k,v): operator.itemgetter(1)(v)</code> is not a good way to code in Python. As gnibbler points out, <code>operator.itemgetter(1)</code> is recomputed <em>for each item</em>. That's inefficient. The point of using <code>operator.itemgetter(1)</code> is to create a function that can be applied many times. You don't want to re-create the function each time. <code>lambda (k,v): v[1]</code> is more readable, and faster:</p> <pre><code>In [15]: %timeit sorted(mydict.iteritems(), key=lambda (k,v): v[1]) 100000 loops, best of 3: 7.55 us per loop In [16]: %timeit sorted(mydict.iteritems(), key=lambda (k,v): operator.itemgetter(1)(v)) 100000 loops, best of 3: 11.2 us per loop </code></pre>
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    1. COCan you please explain this: `operator.itemgetter(1)(v))`?
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    2. CO@A A: `operator.itemgetter(1)(v)` is equivalent to `v[1]`. You can replace just any occurrence of `a[b]` with `operator.itemgetter(b)(a)`, but this is not how *I* did understand your question :)
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    3. COCreating an itemgetter is relatively expensive. This creates one for each key in the dict
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