Yes, this is what he's saying. Its not "broken," just a bit different. After all, you dont have a problem with: lst = [1, 2, 3] ptr = lst lst.append(4) # this changes ptr
And a "view" of the dictionary is orders faster than creating a copy of it (which is required to keep k0 from changing in your example). If you're LUCKY, copying a dictionary is O(n), and i would expect it to take longer because hashmaps usually have a lot of empty space in them (anyone with python specific experiance know if its diffent for python hashmaps?), and each empty space takes time to iterate across. So a dictionary is over O(n), while a view takes O(1) time. considering how often this operation is used, it makes sense to do this optimization And if you really want an independant list, you can always say k0 = list(d.keys()) Paul Rubin wrote: > > Wait a sec, you're saying > > k0 = d.keys() > d['whee'] = 'parrot' # this can change k0??? > > That sounds broken. -- http://mail.python.org/mailman/listinfo/python-list
