Redis Capped Sorted Set, List, or Queue?

问题:

Has anyone implemented a capped data-structure of any kind in Redis? I'm working on building something like a news feed. The feed will wind up being manipulated and read from very frequently, and holding it in a sorted set in Redis would be cheap and perfect for my use case. The only issue is I only ever need n items per feed, and I'm worried about memory overflow, so I'd like to ensure each feed never gets above n items. It seems pretty trivial to make a capped sorted collection in Redis with Lua:

redis-cli EVAL "$(cat update_feed.lua)" 1 feeds:some_feed "thing_to_add", n

Where update_feed.lua looks something like (without testing it):

redis.call('ZADD', KEYS[1], os.time(), ARGV[1])
local num = redis.call('ZCARD', KEYS[1])
if num > ARGV[2]:
    redis.call('ZREMRANGEBYRANK', KEYS[1], -n, -inf)

That's not bad at all, and pretty cheap, but it seems like such a basic thing that could be doable much more cheaply by instantiating the sorted set with only n buckets to begin with. I can't find a way to do that in redis, so I guess my question is: did I miss something, and if I didn't, why is there no structure for this in redis, even if it just ran the basic Lua script I described, it seems like it would be a typical enough use-case that it ought to be implemented as an option for redis data structures?

回答1:

You can use LTRIM if it is a list.

Excerpt from the documentation.

LPUSH mylist someelement
LTRIM mylist 0 99

This pair of commands will push a new element on the list, while making sure that the list will not grow larger than 100 elements. This is very useful when using Redis to store logs for example. It is important to note that when used in this way LTRIM is an O(1) operation because in the average case just one element is removed from the tail of the list.

回答2:

I use sorted sets myself for this. I too thought about using lists, but then I found that manipulating the INSIDE of a list is fairly expensive -- O(n) -- while manipulating the inside of a sorted set is O(log n).

That's what sealed the deal for me--will you ever be manipulating the inside of the set? If so, stick with sorted sets and just flush the oldest whenever you have to, just like you were thinking.