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copied!<p>This is not a csv problem, since you don't have a csv. (In fact you confused me for some time because I was trying to imagine what the csv file looked like!)</p> <ol> <li>Establish a "test" of whether a list needs normalization (<code>is_tuple_string_row</code> function below).</li> <li>Use <a href="http://docs.python.org/2/library/ast.html#ast.literal_eval" rel="nofollow"><code>ast.literal_eval</code></a> to safely convert the string-of-tuple into an actual tuple.</li> <li>Use <a href="http://docs.python.org/2/library/itertools.html#itertools.chain.from_iterable" rel="nofollow"><code>chain.from_iterable</code></a> to combine all the tuples into a single sequence.</li> </ol> <p>Full code below implemented as the generator <code>normalized_list()</code>.</p> <pre><code>l = [["('N', '', 'SHOWERS', '', 'RD', '11734', '11734', '', '')", "('61060', '61060')"], ['', '', 'NORA', '', 'RD', '10301', '10999', '10400', '10998', '61089', '61089']] l2 = [['N', '', 'SHOWERS', '', 'RD', '11734', '11734', '', '', '61060', '61060'], ['', '', 'NORA', '', 'RD', '10301', '10999', '10400', '10998', '61089', '61089']] import ast from itertools import imap, chain def is_tuple_string_row(L): """Return whether list L is a list of strings containing tuples""" return len(L)==2 and all(s.startswith("('") and s.endswith("')") for s in L) def normalized_list(L): """Yield lists in sequence L as lists of strings Normalizes a list of strings of tuples down to a simple list of strings """ for item in L: if is_tuple_string_row(item): yield list(chain.from_iterable(imap(ast.literal_eval, item))) else: yield item # You can iterate efficiently for item in normalized_list(l): print item # or you can put the whole thing to a single new list aslist = list(normalized_list(l)) # verify results assert aslist == l2 </code></pre>

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