spark rdd median 中位数求解
lookup(key)
Return the list of values in the RDD for key key. This operation is done efficiently if the RDD has a known partitioner by only searching the partition that the key maps to.
>>> l = range(1000)
>>> rdd = sc.parallelize(zip(l, l), 10)
>>> rdd.lookup(42) # slow
[42]
>>> sorted = rdd.sortByKey()
>>> sorted.lookup(42) # fast
[42]
>>> sorted.lookup(1024)
[]
>>> rdd2 = sc.parallelize([(('a', 'b'), 'c')]).groupByKey()
>>> list(rdd2.lookup(('a', 'b'))[0])
['c']
You need to sort RDD and take element in the middle or average of two elements. Here is example with RDD[Int]:
import org.apache.spark.SparkContext._
val rdd: RDD[Int] = ???
val sorted = rdd.sortBy(identity).zipWithIndex().map {
case (v, idx) => (idx, v)
}
val count = sorted.count()
val median: Double = if (count % 2 == 0) {
val l = count / 2 - 1
val r = l + 1
(sorted.lookup(l).head + sorted.lookup(r).head).toDouble / 2
} else sorted.lookup(count / 2).head.toDouble
实验:
all_data = sc.parallelize([25,1,2,3,4,5,6,7,8,100]) all_data.sortBy(lambda x:x).zipWithIndex().map(lambda x: (x[1],x[0])).collect [(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 25), (9, 100)]
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
· 全程不用写代码,我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15:你的「微服务管家」又秀新绝活了
· DeepSeek 开源周回顾「GitHub 热点速览」