Yeah, I intentionally didn't mention the expected data set size, hoping I could
find a more elegant solution that would work both in the small N and large N
cases. In any case, I appreciate the recommendations.
When I get some time I am interested in looking at the source and figuring out
wheth
I was thinking it was going to be a lot more than that, you might want to
consider just storing them all as a single serialized array of timestamps
and uuids. By my math, you could fit up to 40 uuid/timestamp pairs for
under 1K. Then you'd just store something like this:
// Row key is userId
123
That is a good question, because realistically I see N being under 10, and
there are no current plans to make use of a large historical record. I could
have the update process pull all columns and issue deletes as necessary such
that only M (M >= N) are kept.
Thanks for the inspiration.
On Ma
Sorry, missed that. I'm not sure if there's a cleaner way than using the
approaches you've looked at, hopefully someone else has an answer. How big
is N and do you need to keep more than N around?
On Sat, May 8, 2010 at 10:26 AM, William Ashley wrote:
> This would be a solution if I wanted to
This would be a solution if I wanted to get the N most recently CREATED guids,
but I'm interested in the most recently SEEN guids.
On May 8, 2010, at 10:00 AM, Ed Anuff wrote:
> Is there a reason you can't use time-based guids? Those would be sorted the
> way you wanted.
>
> On Fri, May 7, 20
Is there a reason you can't use time-based guids? Those would be sorted the
way you wanted.
On Fri, May 7, 2010 at 8:31 PM, William Ashley wrote:
>
> Hopefully I’ve sufficiently explained what I’m trying to do. Now on to
> solving this problem in Cassandra. I’ve been trying to find a way that
>
Daniel,
Partitioning applies to row keys, not column sorting. You could take both of my
Cassandra solutions and refactor them to use row keys containing userId:time or
userId:guid, but you ultimately wind up with the same compromises on update or
retrieval efficiency, plus then you have to use a
Query : Why are you sorting AFAIK cassandra sorts the keys by itself if you
are using ordered partitioning. And how do you store data pertaining to
single user but having several GUID's to attach with.
___
Vineet Daniel
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