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copied!<p>I prefer the Gordian Knot approach:</p> <pre><code>print '\n'.join(['Jack', 'Kack', 'Lack', 'Mack', 'Nack', 'Ouack', 'Pack', 'Quack']) </code></pre> <p>Strictly speaking, that's just a (somewhat radical) modification.</p> <p>Stuff that, Perl guy.</p> <p>All kidding aside, you could do it as the others have mentioned by special casing the "O" and "Q". A (very) slightly more general approach is to think of the extra vowels not as a special case but rather as the only "visible" members of a general case:</p> <pre><code>suffixes = ['ack'] * 8 prefixes = 'JKLMNOPQ' infixes = ',,,,,u,,u'.split(',') </code></pre> <p>The split of the infixes string returns empty strings surrounding the consecutive commas, so the 'u's are the only non-empty elements and they fall in the right places for the final result.</p> <p>This makes three iterables (think lists, but strings count too) where all of the parts are lined up in the correct slots. This is enough of a shift that you can trade the old loop for a zip, which is probably not the intent of the exercise but is still a pythonic technique. Now you just need to run a zipper down the three, which is what the zip function is for, of course.</p> <p>The only wrinkle is that zip gives you a out a list of string pairs and we want whole strings, which is what a ''.join is for. They are processed by a list comprehension, which you will definitely want to learn if you don't already know:</p> <pre><code>result = [''.join(pair) for pair in zip(infixes, suffixes)] result = [''.join(pair) for pair in zip(prefixes, result)] </code></pre> <p>All that is left is to print:</p> <pre><code>print '\n'.join(result) </code></pre> <p>To paraphrase a great man, my disclaimer: In theory, I can prove the correctness of this code, but I have not run it, so there may be bugs.</p>

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