[EMAIL PROTECTED] wrote: > But a really fast approach is to use a dictionary or other structure > that turns the inner loop into a fast lookup, not a slow loop through > the 'Customers' list.
Another approach is to sort both sequences, loop over both in one loop and just update the index for the smaller item. Something like this (below, I'm assuming no duplicates in the lists, and I don't know if that's true for your [2]): >>> l1 = (1,2,4,6,7,8,9) >>> l2 = (2,5,6,7,9,10) >>> i1 = i2 = 0 >>> while 1: ... if l1[i1]==l2[i2]: ... print l1[i1], 'is in both' ... i1 += 1; i2 += 1 ... elif l1[i1]<l2[i2]: ... i1 += 1 ... else: ... i2 += 1 ... 2 is in both 6 is in both 7 is in both 9 is in both Traceback (most recent call last): File "<stdin>", line 2, in ? IndexError: tuple index out of range Unless I'm terribly confused, this will lead to m+n iterations in the worst case. (Of course, the sort operations are something like O(n log n) I guess.) By the way, you might be able to use sets too: How fast is Set.intersection? -- http://mail.python.org/mailman/listinfo/python-list
