Paul Rubin wrote: > How about something like: > > kv_sorted = sorted(kv, key=lambda x: (id(x[0]), x[1]))
You mean like this? #sort by id and then value kv_sorted = sorted(kv, key=lambda x: (id(x[0]),x[1])) #groupby: first element in each group is object and its min value d =dict( g.next() for k,g in groupby( kv_sorted, key=lambda x: x[0] ) ) Yes, that appears to be fastest and is pretty easy to read. Thanks, Alan -- http://mail.python.org/mailman/listinfo/python-list
