如何在scala中按星期劃分一個列表

case class Test(dayOfWeek:Int,b:Int=Random.nextInt) 
val data=(3 to 100).map(_ % 7).map(Test(_))

如何將數據拆分成組，每個組有一個星期的數據，如果一個星期沒有完成，也有一個組。因此，組應該是如何在scala中按星期劃分一個列表

Group 1: (3,4,5,6) // the number here is the dayOfWeek 
Group 2: (0,1,2,3,4,5,6) 
Group 3: (0,1,2,3,4,5,6) 
... 
last Group:(0,1,2)

來源

2014-07-24 Daniel Wu

Scala的集合是真的很強大，這應該做幾行：

val (firstWeek, nextWeeks) = data.span(_.dayOfWeek != 0) 
val weeks = (firstWeek :: nextWeeks.grouped(7).toList).dropWhile(_.isEmpty)

看看這個文檔的span和groupedhere。

println(weeks.zipWithIndex.map { 
    case (week, i) => s"Group $i: (${week.map(_.dayOfWeek).mkString(",")})" 
}.mkString("\n"))

輸出：

Group 0: (3,4,5,6) 
Group 1: (0,1,2,3,4,5,6) 
Group 2: (0,1,2,3,4,5,6) 
[snip] 
Group 12: (0,1,2,3,4,5,6) 
Group 13: (0,1,2,3,4,5,6) 
Group 14: (0,1,2)

來源

2014-07-24 09:58:36 Utaal

是的，這是我尋找的優雅！ –

@如果其中一個答案滿足你，考慮接受它（http://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work）。 – Utaal

您可以摺疊做到這一點：

case class Test(dayOfWeek: Int, b: Int = scala.util.Random.nextInt) 
val data = (3 to 100).map(_ % 7).map(Test(_)) 

val spans = 
    data.foldLeft(Vector(Vector.empty[Test])) { 
    case (zs :+ z, e) => 
     if (e.dayOfWeek == 0) 
     if (z.nonEmpty) 
      (zs :+ z) :+ Vector(e) 
     else 
      zs :+ Vector(e) 
     else 
     zs :+ (z :+ e) 
    } 

for ((g, i) <- spans.zipWithIndex) { 
    println(f"Group $i: (${g.map(_.dayOfWeek).mkString(",")})") 
}

輸出：

Group 0: (3,4,5,6) 
Group 1: (0,1,2,3,4,5,6) 
Group 2: (0,1,2,3,4,5,6) 
Group 3: (0,1,2,3,4,5,6) 
Group 4: (0,1,2,3,4,5,6) 
Group 5: (0,1,2,3,4,5,6) 
Group 6: (0,1,2,3,4,5,6) 
Group 7: (0,1,2,3,4,5,6) 
Group 8: (0,1,2,3,4,5,6) 
Group 9: (0,1,2,3,4,5,6) 
Group 10: (0,1,2,3,4,5,6) 
Group 11: (0,1,2,3,4,5,6) 
Group 12: (0,1,2,3,4,5,6) 
Group 13: (0,1,2,3,4,5,6) 
Group 14: (0,1,2)

來源

2014-07-24 03:35:00 dhg

我認爲，如果第一個元素是0，你會在一開始就空組結束了。例如，'val data =（0到10）.map（_％7）.map（Test（_））' –

@將更新。 – dhg

這裏是一個遞歸版本的一般序列工作，現在需要weekDay功能。

def groupByWeek[T](s: Seq[T], maxDay: Int = 6, weekDay: T => Int) = { 
    @scala.annotation.tailrec 
    def recurse(r: Seq[T], results: Seq[Seq[T]]): Seq[Seq[T]] = 
    r.splitAt(r.indexWhere(item => weekDay(item) == maxDay)) match { 
     case (hd, tl) if (hd.isEmpty && tl.isEmpty) => results 
     case (hd, tl) if (hd.isEmpty) => results :+ tl 
     case (hd, tl) => recurse(tl.tail, results :+ (hd :+ tl.head)) 
    } 
    recurse(s,Seq.empty) 
    } 
}

這樣調用：

val weeks = groupByWeek(data, weekDay = (x:Test) => x.dayOfWeek)

而且你可以看到羣：

println(weeks.zipWithIndex.map { 
    case (week, i) => s"Group $i: (${week.map(_.dayOfWeek).mkString(",")})" 
}.mkString("\n"))

，輸出：

Group 0: (3,4,5,6) 
Group 1: (0,1,2,3,4,5,6) 
Group 2: (0,1,2,3,4,5,6) 
[snip] 
Group 12: (0,1,2,3,4,5,6) 
Group 13: (0,1,2,3,4,5,6) 
Group 14: (0,1,2)

來源

2014-07-24 03:35:07

如何在scala中按星期劃分一個列表

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