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copied!<p>One solution would be to store current index and/or depth information and use it to traverse the nested list. But that seems like a solution that would do a lot of complicated forking -- testing for ends of lists, and so on. Instead, I came up with a compromise. Instead of flattening the list of lists, I created a generator that creates a flat list of indices into the list of lists:</p> <pre><code>def enumerate_nested(nested, indices): for i, item in enumerate(nested): if isinstance(item, collections.Iterable) and not isinstance(item, basestring): for new_indices in enumerate_nested(item, indices + (i,)): yield new_indices else: yield indices + (i,) </code></pre> <p>Then a simple function that extracts an innermost item from the list of lists based on an index tuple:</p> <pre><code>def tuple_index(nested_list, index_tuple): for i in index_tuple: nested_list = nested_list[i] return nested_list </code></pre> <p>Now all you have to do is traverse the flat index list, in whatever way you like.</p> <pre><code>>>> indices = list(enumerate_nested(l, tuple())) >>> print l [1, 2, 3, ['a', 'b', 'c'], 4, ['d', 'e', [100, 200, 300]], 5, ['a', 'b', 'c'], 6] >>> for i in indices: ... print tuple_index(l, i), ... 1 2 3 a b c 4 d e 100 200 300 5 a b c 6 </code></pre> <hr> <p>Since this answer was accepted for the stack-based solution that I posted on ideone in the comments, and since it's preferable not to use external pastebins for answer code, please note that <a href="https://stackoverflow.com/questions/6558365/convert-list-of-positions-4-1-2-of-arbitrary-length-to-an-index-for-a-nested/6618179#6618179">this answer</a> also contains my stack-based solution. </p>

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