[英]group by with foldleft scala
I have the following list in input: 输入中有以下列表:
val listInput1 =
List(
"itemA,CATs,2,4",
"itemA,CATS,3,1",
"itemB,CATQ,4,5",
"itemB,CATQ,4,6",
"itemC,CARC,5,10")
and I want to write a function in scala using groupBy and foldleft ( just one function) in order to sum up third and fourth colum for lines having the same title(first column here), the wanted output is : 并且我想使用groupBy和foldleft在scala中编写一个函数(仅一个函数),以便对具有相同标题的行的第三和第四列(此处为第一列)求和,所需的输出为:
val listOutput1 =
List(
"itemA,CATS,5,5",
"itemB,CATQ,8,11",
"itemC,CARC,5,10"
)
def sumIndex (listIn:List[String]):List[String]={
listIn.map(_.split(",")).groupBy(_(0)).map{
case (title, label) =>
"%s,%s,%d,%d".format(
title,
label.head.apply(1),
label.map(_(2).toInt).sum,
label.map(_(3).toInt).sum)}.toList
} }
Kind regards 亲切的问候
You can solve it with a single foldLeft, iterating the input list only once. 您可以使用一个foldLeft来解决它,只对输入列表进行一次迭代。 Use a Map to aggregate the result. 使用地图汇总结果。
listInput1.map(_.split(",")).foldLeft(Map.empty[String, Int]) {
(acc: Map[String, Int], curr: Array[String]) =>
val label: String = curr(0)
val oldValue: Int = acc.getOrElse(label, 0)
val newValue: Int = oldValue + curr(2).toInt + curr(3).toInt
acc.updated(label, newValue)
}
result: Map(itemA -> 10, itemB -> 19, itemC -> 15) 结果:Map(itemA-> 10,itemB-> 19,itemC-> 15)
If you have a list as 如果您有一个清单
val listInput1 =
List(
"itemA,CATs,2,4",
"itemA,CATS,3,1",
"itemB,CATQ,4,5",
"itemB,CATQ,4,6",
"itemC,CARC,5,10")
Then you can write a general function that can be used with foldLeft
and reduceLeft
as 然后,您可以编写一个可以与foldLeft
和reduceLeft
一起使用的通用函数 , foldLeft
reduceLeft
:
def accumulateLeft(x: Map[String, Tuple3[String, Int, Int]], y: Map[String, Tuple3[String, Int, Int]]): Map[String, Tuple3[String, Int, Int]] ={
val key = y.keySet.toList(0)
if(x.keySet.contains(key)){
val oldTuple = x(key)
x.updated(key, (y(key)._1, oldTuple._2+y(key)._2, oldTuple._3+y(key)._3))
}
else{
x.updated(key, (y(key)._1, y(key)._2, y(key)._3))
}
}
and you can call them as 你可以称他们为
listInput1
.map(_.split(","))
.map(array => Map(array(0) -> (array(1), array(2).toInt, array(3).toInt)))
.foldLeft(Map.empty[String, Tuple3[String, Int, Int]])(accumulateLeft)
.map(x => x._1+","+x._2._1+","+x._2._2+","+x._2._3)
.toList
//res0: List[String] = List(itemA,CATS,5,5, itemB,CATQ,8,11, itemC,CARC,5,10)
listInput1
.map(_.split(","))
.map(array => Map(array(0) -> (array(1), array(2).toInt, array(3).toInt)))
.reduceLeft(accumulateLeft)
.map(x => x._1+","+x._2._1+","+x._2._2+","+x._2._3)
.toList
//res1: List[String] = List(itemA,CATS,5,5, itemB,CATQ,8,11, itemC,CARC,5,10)
Similarly you can just interchange the variables in the general function so that it can be used with foldRight
and reduceRight
as 同样,您可以在常规函数中互换变量,以便可以将foldRight
和reduceRight
与
def accumulateRight(y: Map[String, Tuple3[String, Int, Int]], x: Map[String, Tuple3[String, Int, Int]]): Map[String, Tuple3[String, Int, Int]] ={
val key = y.keySet.toList(0)
if(x.keySet.contains(key)){
val oldTuple = x(key)
x.updated(key, (y(key)._1, oldTuple._2+y(key)._2, oldTuple._3+y(key)._3))
}
else{
x.updated(key, (y(key)._1, y(key)._2, y(key)._3))
}
}
and calling the function would give you 并调用该函数会给你
listInput1
.map(_.split(","))
.map(array => Map(array(0) -> (array(1), array(2).toInt, array(3).toInt)))
.foldRight(Map.empty[String, Tuple3[String, Int, Int]])(accumulateRight)
.map(x => x._1+","+x._2._1+","+x._2._2+","+x._2._3)
.toList
//res2: List[String] = List(itemC,CARC,5,10, itemB,CATQ,8,11, itemA,CATs,5,5)
listInput1
.map(_.split(","))
.map(array => Map(array(0) -> (array(1), array(2).toInt, array(3).toInt)))
.reduceRight(accumulateRight)
.map(x => x._1+","+x._2._1+","+x._2._2+","+x._2._3)
.toList
//res3: List[String] = List(itemC,CARC,5,10, itemB,CATQ,8,11, itemA,CATs,5,5)
So you don't really need a groupBy
and can use any of the foldLeft
, foldRight
, reduceLeft
or reduceRight
functions to get your desired output. 因此,您实际上并不需要groupBy
,可以使用foldLeft
, foldRight
, reduceLeft
或reduceRight
函数中的任何一个来获取所需的输出。
The logic in your code looks sound, here it is with a case class
implemented as that handles edge cases more cleanly: 代码中的逻辑看起来很合理,这里是用一个case class
实现的,因为它可以更清晰地处理边缘案例:
// represents a 'row' in the original list
case class Item(
name: String,
category: String,
amount: Int,
price: Int
)
// safely converts the row of strings into case class, throws exception otherwise
def stringsToItem(strings: Array[String]): Item = {
if (strings.length != 4) {
throw new Exception(s"Invalid row: ${strings.foreach(print)}; must contain only 4 entries!")
} else {
val n = strings.headOption.getOrElse("N/A")
val cat = strings.lift(1).getOrElse("N/A")
val amt = strings.lift(2).filter(_.matches("^[0-9]*$")).map(_.toInt).getOrElse(0)
val p = strings.lastOption.filter(_.matches("^[0-9]*$")).map(_.toInt).getOrElse(0)
Item(n, cat, amt, p)
}
}
// original code with case class and method above used
listInput1.map(_.split(","))
.map(stringsToItem)
.groupBy(_.name)
.map { case (name, items) =>
Item(
name,
category = items.head.category,
amount = items.map(_.amount).sum,
price = items.map(_.price).sum
)
}.toList
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