我正在使用以下代码遍历目录并将文件添加mutable Map
到mutable ListBuffer
. 键是目录名,值是该目录下的文件列表。
def WalkDir(path:String): scala.collection.mutable.HashMap[String,scala.collection.mutable.ListBuffer[String]] = {
def collectFiles(p:Path, allFiles: scala.collection.mutable.HashMap[String,scala.collection.mutable.ListBuffer[String]]): = {
val fileIterator = ... //List all the files
while (fileIterator.hasNext()) {
val f = fileIterator.next()
if (f is not directory) { // pseduo code
val dir = f.getDir() // pseudo code
if (! allFiles.contains(f.getDir().toString())
{
val files = new scala.collection.mutable.ListBuffer[String]
files += (f.path().toString()) // psudeo-code file path
allFiles.put(dir.toString(), files)
}
else {
val t_list = allFiles.get(dir.toString()
allFiles.put(dir.toString(), t_list += f.path().toString())
}
}
if (f is directory) collectFiles(f.getPath(), allFiles)
}
collectFiles(path, new scala.collection.mutable.HashMap[String, scala.collection.mutable.ListBuffer[String]])
}
如何将返回值转换为Map[String, List[String]]
?是否有更“实用”的方式来实现相同的结果?