[英]How do add values of selective rows from a list in an functional style?
我以命令式的方式解决了我的问题,但它看起来非常难看。 我怎样才能做得更好(更优雅,更简洁,更实用 - 最后是Scala)。 应跳过与前一行具有相同值但具有不同字母的行,应添加行的所有其他值。
val row1 = new Row(20, "A", true) // add value
val row2 = new Row(30, "A", true) // add value
val row3 = new Row(40, "A", true) // add value
val row4 = new Row(40, "B", true) // same value as the previous element & different letter -> skip row
val row5 = new Row(60, "B", true) // add value
val row6 = new Row(70, "B", true) // add value
val row7 = new Row(70, "B", true) // same value as the previous element, but the same letter -> add value
val rows = List(row1, row2, row3, row4, row5, row6, row7)
var previousLetter = " "
var previousValue = 0.00
var countSkip = 0
for (row <- rows) {
if (row.value == previousValue && row.letter != previousLetter) {
row.relevant = false
countSkip += 1
}
previousLetter = row.letter
previousValue = row.value
}
// get sum
val sumValue = rows.filter(_.relevant == true).map(_.value) reduceLeftOption(_ + _)
val sum = sumValue match {
case Some(d) => d
case None => 0.00
}
assert(sum == 290)
assert(countSkip == 1)
提前致谢
Twistleton
(rows.head :: rows).sliding(2).collect{
case List(Row(v1,c1), Row(v2,c2)) if ! (v1 == v2 && c1 != c2) => v2 }.sum
我认为当Row是一个案例类(删除布尔值)时,最短(防弹)的解决方案是
(for ((Row(v1,c1), Row(v2,c2)) <- (rows zip rows.take(1) ::: rows) if (v1 != v2 || c1 == c2)) yield v1).sum
其他一些解决方案不处理list-is-empty的情况,但这很大程度上是因为sliding
有一个错误,如果列表太短,它将返回部分列表。 对我来说更清楚(也是防弹)是:
(rows zip rows.take(1) ::: rows).collect{
case (Row(v1,c1), Row(v2,c2)) if (v1 != v2 || c1 == c2) => v1
}.sum
(如果将它保持在一行,则只有两个字符)。 如果你还需要跳过这个号码,
val indicated = (rows zip rows.take(1) ::: rows).collect {
case (Row(v1,c1), Row(v2,c2)) => (v1, v1 != v2 || c1 == c2)
}
val countSkip = indicated.filterNot(_._2).length
val sum = indicated.filter(_._2).map(_._1).sum
折起来:
scala> rows.foldLeft((row1, 0))((p:(Row,Int), r:Row) => (r, p._2 + (if (p._1.value == r.value && p._1.letter != r.letter) 0 else r.value)))._2
res2: Int = 290
(new Row(0, " ", true) +: rows).sliding(2).map { case List(r1, r2) =>
if (r1.value != r2.value || r1.letter == r2.letter) { r2.value }
else { 0 }
}.sum
当然,如果你不需要它,你可以删除Row
的布尔成员
减少它:
rows.reduceLeft { (prev, curr) =>
if (prev.value == curr.value && prev.letter != curr.letter) {
curr.relevant = false
countSkip += 1
}
curr
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.