[EMAIL PROTECTED] wrote: > hi everyone > there is a way, using re, to test (for es) in > a=[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14] if a list b is > composed by three "sublists" of a separated or not by elements. > > if b=[a2,a3,a4,a7,a8,a12,a13] gives true because in a > we have [....,a2,a3,a3,...,a7,a8,...,a12,a13,...] > or b=[a1,a2,a5,a14] gives true because in a we have > [a1,a2,....,a5,...,a14] and so on...
difflib.SequenceMatcher can do this for you: >>> a = [1,2,3,4,5,6,7,8,9,10,11,12,13,14] >>> b = [2,3,4,7,8,12,13] >>> import difflib >>> sm = difflib.SequenceMatcher(None, a, b) >>> sm.get_matching_blocks() [(1, 0, 3), (6, 3, 2), (11, 5, 2), (14, 7, 0)] >>> b = [1, 2, 5, 14] >>> sm = difflib.SequenceMatcher(None, a, b) >>> sm.get_matching_blocks() [(0, 0, 2), (4, 2, 1), (13, 3, 1), (14, 4, 0)] You should test for len(sm.get_matching_blocks()) == 4 (the last element is a dummy) Kent -- http://mail.python.org/mailman/listinfo/python-list
