我看到Scala的collection.mutable.PriorityQueue
有些奇怪的行为。我正在执行外部排序并使用1M记录对其进行测试。每次运行测试并验证10-20条记录之间的结果都没有正确排序。我将scala PriorityQueue
实现替换为java.util.PriorityQueue
,它可以100%正常运行。有什么想法吗?
这是代码(对不起,它有点长......)。我使用http://sortbenchmark.org/
中的工具gensort -a 1000000
和valsort
对其进行测试
def externalSort(inFileName: String, outFileName: String)
(implicit ord: Ordering[String]): Int = {
val MaxTempFiles = 1024
val TempBufferSize = 4096
val inFile = new java.io.File(inFileName)
/** Partitions input file and sorts each partition */
def partitionAndSort()(implicit ord: Ordering[String]):
List[java.io.File] = {
/** Gets block size to use */
def getBlockSize: Long = {
var blockSize = inFile.length / MaxTempFiles
val freeMem = Runtime.getRuntime().freeMemory()
if (blockSize < freeMem / 2)
blockSize = freeMem / 2
else if (blockSize >= freeMem)
System.err.println("Not enough free memory to use external sort.")
blockSize
}
/** Sorts and writes data to temp files */
def writeSorted(buf: List[String]): java.io.File = {
// Create new temp buffer
val tmp = java.io.File.createTempFile("external", "sort")
tmp.deleteOnExit()
// Sort buffer and write it out to tmp file
val out = new java.io.PrintWriter(tmp)
try {
for (l <- buf.sorted) {
out.println(l)
}
} finally {
out.close()
}
tmp
}
val blockSize = getBlockSize
var tmpFiles = List[java.io.File]()
var buf = List[String]()
var currentSize = 0
// Read input and divide into blocks
for (line <- io.Source.fromFile(inFile).getLines()) {
if (currentSize > blockSize) {
tmpFiles ::= writeSorted(buf)
buf = List[String]()
currentSize = 0
}
buf ::= line
currentSize += line.length() * 2 // 2 bytes per char
}
if (currentSize > 0) tmpFiles ::= writeSorted(buf)
tmpFiles
}
/** Merges results of sorted partitions into one output file */
def mergeSortedFiles(fs: List[java.io.File])
(implicit ord: Ordering[String]): Int = {
/** Temp file buffer for reading lines */
class TempFileBuffer(val file: java.io.File) {
private val in = new java.io.BufferedReader(
new java.io.FileReader(file), TempBufferSize)
private var curLine: String = ""
readNextLine() // prep first value
def currentLine = curLine
def isEmpty = curLine == null
def readNextLine() {
if (curLine == null) return
try {
curLine = in.readLine()
} catch {
case _: java.io.EOFException => curLine = null
}
if (curLine == null) in.close()
}
override protected def finalize() {
try {
in.close()
} finally {
super.finalize()
}
}
}
val wrappedOrd = new Ordering[TempFileBuffer] {
def compare(o1: TempFileBuffer, o2: TempFileBuffer): Int = {
ord.compare(o1.currentLine, o2.currentLine)
}
}
val pq = new collection.mutable.PriorityQueue[TempFileBuffer](
)(wrappedOrd)
// Init queue with item from each file
for (tmp <- fs) {
val buf = new TempFileBuffer(tmp)
if (!buf.isEmpty) pq += buf
}
var count = 0
val out = new java.io.PrintWriter(new java.io.File(outFileName))
try {
// Read each value off of queue
while (pq.size > 0) {
val buf = pq.dequeue()
out.println(buf.currentLine)
count += 1
buf.readNextLine()
if (buf.isEmpty) {
buf.file.delete() // don't need anymore
} else {
// re-add to priority queue so we can process next line
pq += buf
}
}
} finally {
out.close()
}
count
}
mergeSortedFiles(partitionAndSort())
}
答案 0 :(得分:4)
我的测试没有显示PriorityQueue中的任何错误。
import org.scalacheck._
import Prop._
object PriorityQueueProperties extends Properties("PriorityQueue") {
def listToPQ(l: List[String]): PriorityQueue[String] = {
val pq = new PriorityQueue[String]
l foreach (pq +=)
pq
}
def pqToList(pq: PriorityQueue[String]): List[String] =
if (pq.isEmpty) Nil
else { val h = pq.dequeue; h :: pqToList(pq) }
property("Enqueued elements are dequeued in reverse order") =
forAll { (l: List[String]) => l.sorted == pqToList(listToPQ(l)).reverse }
property("Adding/removing elements doesn't break sorting") =
forAll { (l: List[String], s: String) =>
(l.size > 0) ==>
((s :: l.sorted.init).sorted == {
val pq = listToPQ(l)
pq.dequeue
pq += s
pqToList(pq).reverse
})
}
}
scala> PriorityQueueProperties.check
+ PriorityQueue.Enqueued elements are dequeued in reverse order: OK, passed
100 tests.
+ PriorityQueue.Adding/removing elements doesn't break sorting: OK, passed
100 tests.
如果你能以某种方式减少输入足以制作一个测试用例,那会有所帮助。
答案 1 :(得分:1)
我用五百万次输入多次运行它,输出匹配预期。我从查看你的代码的猜测是你的订购是问题(即它给出了不一致的答案。)