问题
What is the difference in space between sorted sets and lists in redis? My guess is that sorted sets are some kind of balanced binary tree, and lists are a linked list. This means that on top of the three values that I'm encoding for each of them, key, score, value, although I'll munge together score and value for the linkedlist, the overhead is that the linkedlist needs to keep track of one other node, and the binary tree needs to keep track of two, so that the space overhead to using a sorted set is O(N).
If my value, and score are both longs, and the pointers to the other nodes are also longs, it seems like the space overhead of a single node goes from 3 longs to 4 longs on a 64-bit computer, which is a 33% increase in space.
Is this true?
回答1:
It is much more than your estimation. Let's suppose ziplists are not used (i.e. you have a significant number of items).
A Redis list is a classical double-linked list: 3 pointers (prev,next,value) per item.
A sorted set is a dictionary plus a skip list. In the dictionary, items will be stored with 3 pointers as well (key,value,next). The skip list memory footprint is more complex to evaluate: each node takes 1 double (score), 2 pointers (obj,backward), plus n couples (pointer,span value) with n between 1 and 32. Most items will take only 1 or 2 couples.
In other words, when it is not represented as a ziplist, a sorted set is by far the Redis data structure with the most overhead. Compared to a list, the memory overhead is more than 200% (i.e. 3 times).
Note: the best way to evaluate memory consumption with Redis is to try to build a big list or sorted set with pseudo-data and use INFO to get the memory footprint.
来源：https://stackoverflow.com/questions/12269429/redis-data-structure-space-requirements