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copied!<p>If you're just looking to iterate over a flattened version of the data structure and don't need an indexable sequence, consider <a href="http://docs.python.org/library/itertools.html#itertools.chain" rel="noreferrer">itertools.chain and company</a>.</p> <pre><code>>>> list_of_menuitems = [['image00', 'image01'], ['image10'], []] >>> import itertools >>> chain = itertools.chain(*list_of_menuitems) >>> print(list(chain)) ['image00', 'image01', 'image10'] </code></pre> <p>It will work on anything that's iterable, which should include Django's iterable <code>QuerySet</code>s, which it appears that you're using in the question.</p> <p><strong>Edit:</strong> This is probably as good as a reduce anyway, because reduce will have the same overhead copying the items into the list that's being extended. <code>chain</code> will only incur this (same) overhead if you run <code>list(chain)</code> at the end.</p> <p><strong>Meta-Edit:</strong> Actually, it's less overhead than the question's proposed solution, because you throw away the temporary lists you create when you extend the original with the temporary.</p> <p><strong>Edit:</strong> As <a href="https://stackoverflow.com/questions/406121/flattening-a-shallow-list-in-python#411548">J.F. Sebastian says</a> <code>itertools.chain.from_iterable</code> avoids the unpacking and you should use that to avoid <code>*</code> magic, but <a href="https://stackoverflow.com/questions/406121/flattening-a-shallow-list-in-python#408281">the timeit app</a> shows negligible performance difference.</p>

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