Neal Becker wrote: > Any suggestions for transforming the sequence: > > [1, 2, 3, 4...] > Where 1,2,3.. are it the ith item in an arbitrary sequence > > into a succession of tuples: > > [(1, 2), (3, 4)...] > > In other words, given a seq and an integer that specifies the size of tuple > to return, then for example:
>>> trf = lambda iterable,n : zip(*[iter(iterable)]*n) >>> trf(range(15),3) [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, 14)] note though that it will silently drop any remainder. hth, bb -- http://mail.python.org/mailman/listinfo/python-list
