million thanks. So the default compare funcion of heapq also do it like sort ?
The size of the list is not very large but has the potential of being run many times(web apps). So I believe second one should be faster(from the app perspective) as it goes into the optimized code quickly without all the overheads in the merge case. Alex Martelli wrote: > You can do it either way -- simplest, and pretty fast, is to concatenate > them all and sort the result (the sort method is very good at taking > advantage of any sorting that may already be present in some parts of > the list it's sorting), but you can also try a merging approach. E.g.: > > > import heapq > > def merge_by_heap(*lists): > sources = [[s.next(), i, s.next] > for i, s in enumerate(map(iter,lists))] > heapq.heapify(sources) > while sources: > best_source = sources[0] > yield best_source[0] > try: best_source[0] = best_source[-1]() > except StopIteration: heapq.heappop(sources) > else: heapq.heapreplace(sources, best_source) > > Now, L=list(merge_by_heap(l1, l2, l3)) should work, just as well as, > say, L = sorted(l1+l2+l3). I suspect the second approach may be faster, > as well as simpler, but it's best to _measure_ (use the timeit.py module > from the standard library) if this code is highly speed-critical for > your overall application. > > > Alex -- http://mail.python.org/mailman/listinfo/python-list
