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    <p>The other great answers and comments explain well the roles of <a href="http://docs.python.org/tutorial/controlflow.html#unpacking-argument-lists" rel="noreferrer"><strong>argument unpacking</strong></a> and <a href="http://docs.python.org/library/functions.html#zip" rel="noreferrer"><strong>zip()</strong></a>.</p> <p>As <a href="https://stackoverflow.com/questions/2233204/how-does-zipitersn-work-in-python/2233247#2233247">Ignacio</a> and <a href="https://stackoverflow.com/questions/2233204/how-does-zipitersn-work-in-python/2233289#2233289">ujukatzel</a> say, you pass to <code>zip()</code> three references to the same iterator and <code>zip()</code> makes 3-tuples of the integers—in order—from each reference to the iterator:</p> <pre><code>1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9 1,2,3,4,5,6,7,8,9 ^ ^ ^ ^ ^ ^ ^ ^ ^ </code></pre> <p>And since you ask for a more verbose code sample:</p> <pre><code>chunk_size = 3 L = [1,2,3,4,5,6,7,8,9] # iterate over L in steps of 3 for start in range(0,len(L),chunk_size): # xrange() in 2.x; range() in 3.x end = start + chunk_size print L[start:end] # three-item chunks </code></pre> <p>Following the values of <code>start</code> and <code>end</code>: </p> <pre><code>[0:3) #[1,2,3] [3:6) #[4,5,6] [6:9) #[7,8,9] </code></pre> <p>FWIW, you can get the same result with <code>map()</code> with an initial argument of <code>None</code>:</p> <pre><code>&gt;&gt;&gt; map(None,*[iter(s)]*3) [(1, 2, 3), (4, 5, 6), (7, 8, 9)] </code></pre> <p>For more on <code>zip()</code> and <code>map()</code>: <a href="http://muffinresearch.co.uk/archives/2007/10/16/python-transposing-lists-with-map-and-zip/" rel="noreferrer">http://muffinresearch.co.uk/archives/2007/10/16/python-transposing-lists-with-map-and-zip/</a></p>
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    1. POHow does zip(*[iter(s)]*n) work in Python?
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    1. COIt's not three copies of the same iterator, it's three times the same iterator object :)
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      1. USThomas Wouters
 

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