Your reverse index of "which rows contain a column named X" will have very wide rows. You could look at cassandra's secondary indexing, or possibly look at a solandra/solr approach. Another option is you can shift the problem slightly, "which rows have column X that was added between time y and time z". Remember with few distinct column names that reverse index of column to row is going to be a very big list.

On Thu, Apr 4, 2013 at 5:45 PM, Drew Kutcharian <> wrote:
Hi Edward,

I anticipate that the column names will be reused a lot. For example, key1 will be in many rows. So I think the number of distinct column names will be much much smaller than the number of rows.Is there a way to have a separate CF that keeps track of the column names?

What I was thinking was to have a separate CF that I write only the column name with a null value in there every time I write a key/value to the main CF. In this case if that column name exist, then it will just be overridden. Now if I wanted to get all the column names, then I can just query that CF. Not sure if that's the best approach at high load (100k inserts a second).

-- Drew

On Apr 4, 2013, at 12:02 PM, Edward Capriolo <> wrote:

You can not get only the column name (which you are calling a key) you can use get_range_slice which returns all the columns. When you specify an empty byte array (new byte[0]{}) as the start and finish you get back all the columns. From there you can return only the columns to the user in a format that you like.

On Thu, Apr 4, 2013 at 2:18 PM, Drew Kutcharian <> wrote:
Hey Guys,

I'm working on a project and one of the requirements is to have a schema free CF where end users can insert arbitrary key/value pairs per row. What would be the best way to know what are all the "keys" that were inserted (preferably w/o any locking). For example,

Row1 => key1 -> XXX, key2 -> XXX
Row2 => key1 -> XXX, key3 -> XXX
Row3 => key4 -> XXX, key5 -> XXX
Row4 => key2 -> XXX, key5 -> XXX

The query would be give me all the inserted keys and the response would be {key1, key2, key3, key4, key5}