order independent hash?

[88888 Dihedral]

On Monday, December 5, 2011 1:50:08 PM UTC+8, Dan Stromberg wrote:
> Two methods:
> 1) If you need your hash only once in an infrequent while, then save
> the elements in a list, appending as needed, and sort prior to
> hashing, as needed
> 
> 2) If you need your hash more often, you could keep your elements in a
> treap or red-black tree; these will maintain sortedness throughout the
> life of the datastructure.
> 
> 3) If A bunch of log(n) or n or nlog(n) operations doesn't sound
> appealing, then you might try this one: Create some sort of mapping
> from your elements to the integers.  Then just use a sum.  This won't
> scatter things nearly as well as a cryptographic hash, but it's very
> fast, because you don't need to reevaluate some of your members as you
> go.
> 
> HTH
> 
A sorted list can behave like a hash table.  This is of O(log(n)) in accesses 
of n items in theory. 

I agree with you a hash key computation method too slow than a list of n items  in accesses for a range of n items  should be reloadable.

But this is not supported in Python yet.  

For tedious trivial jobs of non-heavy computing , I'll opt for easy use.