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I have
val a = List((1,2), (1,3), (3,4), (3,5), (4,5))
I am using A.groupBy(_._1) which is groupBy with the first element. But, it gives me output as
Map(1 -> List((1,2) , (1,3)) , 3 -> List((3,4), (3,5)), 4 -> List((4,5)))
But, I want answer as
Map(1 -> List(2, 3), 3 -> List(4,5) , 4 -> List(5))
So, how can I do this?
815
Scala groupBy is used for grouping of elements based on some criteria defined as a predicate inside the function. This function internally converts the collection into map object and this map object work on key value pair. This is basically used for storing and retrieving of the elements.
The GROUP BY clause is used to group the rows based on a set of specified grouping expressions and compute aggregations on the group of rows based on one or more specified aggregate functions.
groupBy as defined on TraversableLike produces an immutable. Map , so you can't make this method produce something else. The order of the elements in each entry is already preserved, but not the order of the keys. The keys are the result of the function supplied, so they don't really have an order.
You can do that by following up with mapValues (and a map over each value to extract the second element):
scala> a.groupBy(_._1).mapValues(_.map(_._2))
res2: scala.collection.immutable.Map[Int,List[Int]] = Map(4 -> List(5), 1 -> List(2, 3), 3 -> List(4, 5))
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