Scala:如何扁平一组文件的列表

Question

我有一个List[Set[Path]]：

更新：集合中的每条路径都是唯一的，并表示特定的目录位置。没有副本。所以，我要找的是路径元素/

val miceData = List(Set(C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test7.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test2.txt,     C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test6.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test5.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test8.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test3.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\aPowerPoint.pptx,     C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test1.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory\Test4.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test10.txt),    Set(C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test6.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test3.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test4.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test70.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test8.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test5.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\DeeplyNestedDirectory\FlatDirectory2\Test2.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test2.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test3.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test1.txt), Set(C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test80.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test7.txt, C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test40.txt, C:\Users\lulu\Documents\GitHub\data_mining_folder\FlatDirectory\Test6.txt,     C:\Users\lulu\Documents\mice_data\data_mining_folder\FlatDirectory\Test5.txt), Set(C:\Users\lulu\Documents\mice_data\data_mining_folder\zipfile.zip), Set(C:\Users\lulu\Documents\mice_data\data_mining_folder\micetest.txt,C:\Users\lulu\Documents\mice_data\data_mining_folder\riley.jpg))

在这个Sets中有5个List，每个Set持有Path(s)。如果我计算正确的话，这类Paths的总数是28。

现在，我想了解Path元素在这个List中的所有Sets中的总数。

我本可以在代码上游的某个区域完成此计算，但我现在很想这么做，并在此过程中了解Scala的更多信息。

类似于：

val totalPaths = <<iterate over this List and count all the paths>>

我想要最短的，最惯用的代码来完成这一点。

Answer 1

val totalPaths = miceData.map(_.size).sum

如果您有副本，您可以：

val totalPaths = miceData.flatten.distinct.size

Answer 2

val paths = for {              //gives you a list of all paths on all sets
    set <- miceData
    path <- set
   } yield path


val totalPaths = paths.toSet.size    // converting it to set will remove duplicates if any

Answer 3

我觉得扁平化就足够了

val toto = List(Set(1,2,3), Set(6,7,8))
println(toto.flatten.count)