On 12/29/2012 4:40 PM, Mitya Sirenef wrote:
On 12/29/2012 03:15 PM, Joel Goldstick wrote:
Would this help: >>> w = [1,2,3,1,2,4,4,5,6,1] >>> s = set(w) >>> s set([1, 2, 3, 4, 5, 6]) >>> {x:w.count(x) for x in s} {1: 3, 2: 2, 3: 1, 4: 2, 5: 1, 6: 1} >>>Indeed, this is much better -- I didn't think of it..
It still turns an O(n) problem into an O(k*n) problem, where k is the number of distinct items.
-- Terry Jan Reedy -- http://mail.python.org/mailman/listinfo/python-list
