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    <p>This solution uses only an Ordered Dictionary.<br> <strong>deepcopy()</strong> is necessary if one wants the original copy to remain unchanged.</p> <pre><code>from collections import OrderedDict from copy import deepcopy def treat(passed_list): L = deepcopy(passed_list) dic = OrderedDict() for subl in L: for x in subl: if x not in dic: dic[x] = subl print 'dic at start' print '\n'.join('%-3s : %s' % (a,dic[a]) for a in dic) + '\n' for sublist in L: short = [] short.extend(el for el in sublist if el not in short) seen = [] for k,val in dic.iteritems(): if val is sublist: break if k in short: if val not in seen: seen.append(val) sumseen = [] for elseen in seen: for y in elseen: sumseen.append(y) dic[y] = sumseen if seen: for el in sublist: if el not in sumseen: sumseen.append(el) dic[el] = sumseen sublist[:] = short cumul = [] cumul.extend(lu for lu in dic.itervalues() if lu not in cumul) return cumul plus = [[1,2,3,2,1],[10,5,5,5,10], [8,5,3,3,5],[45,50,12,45,40,12]] lst = [[1,2,3], [10,5], [3,8,5]] for one_list in (plus,lst): print 'one_list before == %r\n' % one_list print 'treat(one_list) == %r\n' % treat(one_list) print 'one_list after == %r\n' % one_list print '====================================' </code></pre> <p>result</p> <pre><code>one_list before == [[1, 2, 3, 2, 1], [10, 5, 5, 5, 10], [8, 5, 3, 3, 5], [45, 50, 12, 45, 40, 12]] dic at start 1 : [1, 2, 3, 2, 1] 2 : [1, 2, 3, 2, 1] 3 : [1, 2, 3, 2, 1] 10 : [10, 5, 5, 5, 10] 5 : [10, 5, 5, 5, 10] 8 : [8, 5, 3, 3, 5] 45 : [45, 50, 12, 45, 40, 12] 50 : [45, 50, 12, 45, 40, 12] 12 : [45, 50, 12, 45, 40, 12] 40 : [45, 50, 12, 45, 40, 12] treat(one_list) == [[1, 2, 3, 10, 5, 8], [45, 50, 12, 40]] one_list after == [[1, 2, 3, 2, 1], [10, 5, 5, 5, 10], [8, 5, 3, 3, 5], [45, 50, 12, 45, 40, 12]] ==================================== one_list before == [[1, 2, 3], [10, 5], [3, 8, 5]] dic at start 1 : [1, 2, 3] 2 : [1, 2, 3] 3 : [1, 2, 3] 10 : [10, 5] 5 : [10, 5] 8 : [3, 8, 5] treat(one_list) == [[1, 2, 3, 10, 5, 8]] one_list after == [[1, 2, 3], [10, 5], [3, 8, 5]] ==================================== </code></pre> <p>This solution has an inconvenience: it is 2 to 3 times slower than the J.F. Sebastian's solution. </p>
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