Steven D'Aprano <ste...@remove.this.cybersource.com.au> wrote: > But that's the wrong solution to the problem. The OP wants the largest > (or smallest) item, which he expects to get by sorting, then grabbing > the first element: > > sorted(alist)[0] > > That requires sorting the entire list, only to throw away everything > except the first item. A better solution is to use min() and max(), > neither of which sort the list, so they are much faster. > And if they want the n smallest or largest items (where n is significantly less than the length of the list[*]) they might consider using heapq.nsmallest() and heapq.nlargest()
I find it interesting that the heapq functions tell you in the documentation that they aren't suitable for use where n==1 or where n is near the total size of the sequence whereas random.sample() chooses what it thinks is the best algorithm based on the input. At the very least I would have thought the heapq functions could check for n==1 and call min/max when it is. [*] Some quick playing with timeit seems to indicate that with a list of 200 integers nsmallest is fastest (by a very small margin) when n==2 and n==3 but is beaten by sorted[:n] when n==4, so I guess you need a reasonably long list to make it worth not sorting it: with list of 20,000 integers it isn't worth sorting unless you want more than about 700 items. -- Duncan Booth http://kupuguy.blogspot.com -- http://mail.python.org/mailman/listinfo/python-list
