MapReduce的Shuffle和Sort阶段复制操作:何时达到m*r次最大值?
Great question—this cuts to the core of how partitioning works in MapReduce's Shuffle phase. Let’s break this down plainly:
The Exact Scenario for m*r Copies
The number of copy operations hits the maximum m*r when every single mapper produces intermediate output that gets distributed across all r reducers. Put another way: for each of the m mappers, its output has at least one key-value pair assigned to every one of the r reducer partitions.
This typically happens when:
- The partitioning function (most commonly the default
HashPartitioner) spreads the mapper’s keys evenly across allrpartitions. - No reducer partition is empty for any mapper’s output—every reducer receives at least some data from every mapper.
Concrete Example
Let’s use a small, tangible example to make this real:
- Let’s say we have
m=2mappers andr=3reducers. - We’re using the standard
HashPartitioner, which assigns keys to partitions viakey.hashCode() % r. - Mapper 1 outputs keys:
apple,banana,cherryapple.hashCode() % 3 = 0→ sent to reducer 0banana.hashCode() % 3 = 1→ sent to reducer 1cherry.hashCode() % 3 = 2→ sent to reducer 2
- Mapper 2 outputs keys:
date,elderberry,figdate.hashCode() % 3 = 0→ sent to reducer 0elderberry.hashCode() % 3 = 1→ sent to reducer 1fig.hashCode() % 3 = 2→ sent to reducer 2
Here, each mapper has data for all 3 reducers. Mapper 1 copies to 3 reducers, Mapper 2 copies to 3 reducers—total 2*3=6 copy operations, which is the absolute maximum.
Contrast: When We Don’t Hit m*r
To clarify, let’s look at a scenario where we fall short: if Mapper 1 only outputs keys that all hash to partition 0 (like apple, apricot, avocado), it only needs to copy data to 1 reducer. Even if Mapper 2 still copies to all 3, total operations would be 1+3=4—way less than 2*3=6.
The bottom line: maximum copy operations occur when every mapper’s intermediate data spans every reducer partition, with no partitions left empty for any mapper.
内容的提问来源于stack exchange,提问作者HungryBird

