Could you use a Super CF?

Super Col name is the Integer, and the Col Names are the UUID. Not sure what your col values are or your key. 
There are some limitations to Super CF but I do not think they would apply in this case http://wiki.apache.org/cassandra/CassandraLimitations

You can the slice the super col names (your integers) and get back the super col and all it's columns. 

Or you could also use a two CF solution...

Index CF where you integer is the column name, not sure what your key is. The column value is not important.  
Item CF where the row key is the Integer, col names are the UUID not sure what the col value is. 

Some things to consider...
- is there a natural grouping to your integers ? e.g. every day
- what is the column value ? Will this make for big rows?

Hope that helps. 
Aaron


On 02 Dec, 2010,at 04:56 AM, Daniel Lundin <dln@eintr.org> wrote:

Unless I misunderstand the Q, composing the column names with the row
keys and merging the resulting would yield something useful.

keyA => (1, uuid), (2, uuid), (3, uid)
keyB => (1, uuid), (2, uuid), (3, uid)

Should be transformed into:

(1, keyA, uuid),
(1, keyB, uuid),
(2, keyA, uuid),
(2, keyB, uuid),
(3, keyA, uuid),
(3, keyB, uuid)

map + merge to the rescue.

On Wed, Dec 1, 2010 at 3:33 PM, Benjamin Waldher <lgbr@laserbunny.net> wrote:
> I have a fairly simple problem that might require a complicated solution.
>
> I need to store Integer -> UUID in a column family, and be able to query
> (and then paginate) the rows ordered by the integer in descending order.
> This is simple enough if no two rows have the same integer, as the integer
> could be a column name which can easily be sorted. However, in my scenario,
> two rows may have the same Integer value. As such, I would need to use the
> integer as the key in the column family. However, this means I must use
> OrderPreservingPartitioner, which is going to cause a huge load imbalance on
> one of my nodes.
>
> How can I have a sorted set of rows of Integer -> UUID where the integer may
> exist many times?
>