Java 在烫伤中生成列表[字符串]的差异
我有一个Java 在烫伤中生成列表[字符串]的差异,java,scala,scalding,Java,Scala,Scalding,我有一个记录:TypedType[(String,util.List[String])]在我的工作中,第一个值是一个id,第二个值是一个内容列表。想象一下: ("1", ["a","b","c"]) ("1", ["a","b","c"]) ("1", ["a","b","c"]) ("2", ["a","b"]) ("2", ["a","b","c"]) ("3", ["a","b","c"]) 在records.groupBy(u._1)之后,我只想输出对于给定id彼此不同的记录。对于上面
记录:TypedType[(String,util.List[String])]
在我的工作中,第一个值是一个id,第二个值是一个内容列表。想象一下:
("1", ["a","b","c"])
("1", ["a","b","c"])
("1", ["a","b","c"])
("2", ["a","b"])
("2", ["a","b","c"])
("3", ["a","b","c"])
在records.groupBy(u._1)
之后,我只想输出对于给定id彼此不同的记录。对于上面的输入,输出应该是:
("2", ["a","b"])
("2", ["a","b","c"])
我是烫伤新手。实现这一目标的优雅方式是什么?我不知道这一点对你是否至关重要(你的收藏是否特别庞大?),但在普通的老Scala中,我会:
// Given:
val records = Seq( "1" -> List("a", "b", "c"), "1" -> List("a", "b", "c"), "1" -> List("a", "b", "c"), "2" -> List("a", "b"), "2" -> List("a", "b", "c"), "3" -> List("a", "b", "c"), "3" -> List("d")
val distinctValues = records.groupBy(_._1).map { case (k, v) => k -> v.toSet }
// => Map(2 -> Set((2,List(a, b)), (2,List(a, b, c))), 1 -> Set((1,List(a, b, c))), 3 -> Set((3,List(a, b, c)), (3,List(d))))
val havingMultipleDistinct = distinctValues.map { case (k, v) => v.size > 1 }
// => Map(2 -> Set((2,List(a, b)), (2,List(a, b, c))), 3 -> Set((3,List(a, b, c)), (3,List(d))))
val asRecords = havingMultipleDistinct.values.flatten
// => List((2,List(a, b)), (2,List(a, b, c)), (3,List(a, b, c)), (3,List(d)))
如果每个键的值的大小足够小,可以放入内存中,那么像这样的操作就可以了:
records
.group
.toSet
.filter(_.size > 1)
.flatten
如果它太大,则可以将管道本身连接起来:
val grouped = records.group
grouped
.join(grouped)
.collect { case(k, (a, b)) if a != b => k -> a }
是的,它必须在集群上运行。烫伤是根本