about sorted set

[Lin]

Hello guys,

I am learning Sorted Set from here (http://redis.io/topics/data-types), I am confused by the following statements, especially confused by the statement "Since elements are taken in order and not ordered afterwards". I am wondering what means "taken in order and not ordered afterwards"? Appreciate if anyone could show me an example? Thanks.

With sorted sets you can add, remove, or update elements in a very fast way (in a time proportional to the logarithm of the number of elements). Since elements are taken in order and not ordered afterwards, you can also get ranges by score or by rank (position) in a very fast way. Accessing the middle of a sorted set is also very fast, so you can use Sorted Sets as a smart list of non repeating elements where you can quickly access everything you need: elements in order, fast existence test, fast access to elements in the middle!

regards,
Lin

[Didier Spezia]

Hi Lin,

it just means the data structure used to represent a sorted set (i.e. a dictionary plus a skip list),

does enforce the order by keeping a sorted view of the items (in the skip list).

So when you need to retrieve the item by score and value order, there is no further processing

to do except walking through the skip list. In other words, order is enforced at insert time,

not at retrieve time.

Regards,

Didier.

[Lin]

Hi Didier, thanks for the detailed reply. For sorted set, I think if we could have a skip list to store sorted values by score, it should be enough. Why we need another dictionary?

regards,
Lin

[Didier Spezia]

Because a sorted set is also a set, and therefore we should be able to quickly

check for membership, delete an item per value, increase the score for a given

value, etc ... For these operations, the score is not provided in the input query,

so if we only had a data structure indexed with the score, it would result in

linear searches.

http://redis.io/commands/zrem

http://redis.io/commands/zincrby

http://redis.io/commands/zrank

http://redis.io/commands/zscore

Regards,

Didier.

[Lin]

Thanks Didier, my question is answered.

regards,
Lin