我已经实现了 k-means 算法，我想通过使用 Java 8 流和多核处理来加快我的过程。
我在 Java 7 中得到了这段代码:
//Step 2: For each point p:
//find nearest clusters c
//assign the point p to the closest cluster c
for (Point p : points) {
double minDst = Double.MAX_VALUE;
int minClusterNr = 1;
for (Cluster c : clusters) {
double tmpDst = determineDistance(p, c);
if (tmpDst < minDst) {
minDst = tmpDst;
minClusterNr = c.clusterNumber;
}
}
clusters.get(minClusterNr - 1).points.add(p);
}
//Step 3: For each cluster c
//find the central point of all points p in c
//set c to the center point
ArrayList<Cluster> newClusters = new ArrayList<Cluster>();
for (Cluster c : clusters) {
double newX = 0;
double newY = 0;
for (Point p : c.points) {
newX += p.x;
newY += p.y;
}
newX = newX / c.points.size();
newY = newY / c.points.size();
newClusters.add(new Cluster(newX, newY, c.clusterNumber));
}
我想使用带有并行流的 Java 8 来加速这个过程。 我尝试了一下并提出了这个解决方案:
points.stream().forEach(p -> {
minDst = Double.MAX_VALUE; //<- THESE ARE GLOBAL VARIABLES NOW
minClusterNr = 1; //<- THESE ARE GLOBAL VARIABLES NOW
clusters.stream().forEach(c -> {
double tmpDst = determineDistance(p, c);
if (tmpDst < minDst) {
minDst = tmpDst;
minClusterNr = c.clusterNumber;
}
});
clusters.get(minClusterNr - 1).points.add(p);
});
ArrayList<Cluster> newClusters = new ArrayList<Cluster>();
clusters.stream().forEach(c -> {
newX = 0; //<- THESE ARE GLOBAL VARIABLES NOW
newY = 0; //<- THESE ARE GLOBAL VARIABLES NOW
c.points.stream().forEach(p -> {
newX += p.x;
newY += p.y;
});
newX = newX / c.points.size();
newY = newY / c.points.size();
newClusters.add(new Cluster(newX, newY, c.clusterNumber));
});
这种带有流的解决方案比没有流的解决方案要快得多。我想知道这是否已经使用了多核处理？否则为什么它会突然快两倍？
without streams : Elapsed time: 202 msec & with streams : Elapsed time: 116 msec
另外，在任何这些方法中使用 parallelStream 来加速它们是否有用？当我将流更改为 stream().parallel().forEach(CODE) 时，它现在所做的就是导致 ArrayOutOfBounce 和 NullPointer 异常
---- 编辑(按要求添加源代码，以便您自己尝试)----
--- 聚类.java ---
package algo;
import java.awt.Color;
import java.awt.Graphics2D;
import java.awt.image.BufferedImage;
import java.util.ArrayList;
import java.util.Random;
import java.util.function.BiFunction;
import graphics.SimpleColorFun;
/**
* An implementation of the k-means-algorithm.
* <p>
* Step 0: Determine the max size of the canvas
* <p>
* Step 1: Place clusters at random
* <p>
* Step 2: For each point p:<br>
* find nearest clusters c<br>
* assign the point p to the closest cluster c
* <p>
* Step 3: For each cluster c<br>
* find the central point of all points p in c<br>
* set c to the center point
* <p>
* Stop when none of the cluster x,y values change
* @author makt
*
*/
public class Clustering {
private BiFunction<Integer, Integer, Color> colorFun = new SimpleColorFun();
// private BiFunction<Integer, Integer, Color> colorFun = new GrayScaleColorFun();
public Random rngGenerator = new Random();
public double max_x;
public double max_y;
public double max_xy;
//---------------------------------
//TODO: IS IT GOOD TO HAVE THOUSE VALUES UP HERE?
double minDst = Double.MAX_VALUE;
int minClusterNr = 1;
double newX = 0;
double newY = 0;
//----------------------------------
public boolean workWithStreams = false;
public ArrayList<ArrayList<Cluster>> allGeneratedClusterLists = new ArrayList<ArrayList<Cluster>>();
public ArrayList<BufferedImage> allGeneratedImages = new ArrayList<BufferedImage>();
public Clustering(int seed) {
rngGenerator.setSeed(seed);
}
public Clustering(Random rng) {
rngGenerator = rng;
}
public void setup(int centroidCount, ArrayList<Point> points, int maxIterations) {
//Step 0: Determine the max size of the canvas
determineSize(points);
ArrayList<Cluster> clusters = new ArrayList<Cluster>();
//Step 1: Place clusters at random
for (int i = 0; i < centroidCount; i++) {
clusters.add(new Cluster(rngGenerator.nextInt((int) max_x), rngGenerator.nextInt((int) max_y), i + 1));
}
int iterations = 0;
if (workWithStreams) {
allGeneratedClusterLists.add(doClusteringWithStreams(points, clusters));
} else {
allGeneratedClusterLists.add(doClustering(points, clusters));
}
iterations += 1;
//do until maxIterations is reached or until none of the cluster x and y values change anymore
while (iterations < maxIterations) {
//Step 2: happens inside doClustering
if (workWithStreams) {
allGeneratedClusterLists.add(doClusteringWithStreams(points, allGeneratedClusterLists.get(iterations - 1)));
} else {
allGeneratedClusterLists.add(doClustering(points, allGeneratedClusterLists.get(iterations - 1)));
}
if (!didPointsChangeClusters(allGeneratedClusterLists.get(iterations - 1), allGeneratedClusterLists.get(iterations))) {
break;
}
iterations += 1;
}
System.out.println("Finished with " + iterations + " out of " + maxIterations + " max iterations");
}
/**
* checks if the cluster x and y values changed compared to the previous x and y values
* @param previousCluster
* @param currentCluster
* @return true if any cluster x or y values changed, false if all of them they are the same
*/
private boolean didPointsChangeClusters(ArrayList<Cluster> previousCluster, ArrayList<Cluster> currentCluster) {
for (int i = 0; i < previousCluster.size(); i++) {
if (previousCluster.get(i).x != currentCluster.get(i).x || previousCluster.get(i).y != currentCluster.get(i).y) {
return true;
}
}
return false;
}
/**
*
* @param points - all given points
* @param clusters - its point list gets filled in this method
* @return a new Clusters Array which has an <b> empty </b> point list.
*/
private ArrayList<Cluster> doClustering(ArrayList<Point> points, ArrayList<Cluster> clusters) {
//Step 2: For each point p:
//find nearest clusters c
//assign the point p to the closest cluster c
for (Point p : points) {
double minDst = Double.MAX_VALUE;
int minClusterNr = 1;
for (Cluster c : clusters) {
double tmpDst = determineDistance(p, c);
if (tmpDst < minDst) {
minDst = tmpDst;
minClusterNr = c.clusterNumber;
}
}
clusters.get(minClusterNr - 1).points.add(p);
}
//Step 3: For each cluster c
//find the central point of all points p in c
//set c to the center point
ArrayList<Cluster> newClusters = new ArrayList<Cluster>();
for (Cluster c : clusters) {
double newX = 0;
double newY = 0;
for (Point p : c.points) {
newX += p.x;
newY += p.y;
}
newX = newX / c.points.size();
newY = newY / c.points.size();
newClusters.add(new Cluster(newX, newY, c.clusterNumber));
}
allGeneratedImages.add(createImage(clusters));
return newClusters;
}
/**
* Does the same as doClustering but about twice as fast!<br>
* Uses Java8 streams to achieve this
* @param points
* @param clusters
* @return
*/
private ArrayList<Cluster> doClusteringWithStreams(ArrayList<Point> points, ArrayList<Cluster> clusters) {
points.stream().forEach(p -> {
minDst = Double.MAX_VALUE;
minClusterNr = 1;
clusters.stream().forEach(c -> {
double tmpDst = determineDistance(p, c);
if (tmpDst < minDst) {
minDst = tmpDst;
minClusterNr = c.clusterNumber;
}
});
clusters.get(minClusterNr - 1).points.add(p);
});
ArrayList<Cluster> newClusters = new ArrayList<Cluster>();
clusters.stream().forEach(c -> {
newX = 0;
newY = 0;
c.points.stream().forEach(p -> {
newX += p.x;
newY += p.y;
});
newX = newX / c.points.size();
newY = newY / c.points.size();
newClusters.add(new Cluster(newX, newY, c.clusterNumber));
});
allGeneratedImages.add(createImage(clusters));
return newClusters;
}
//draw all centers from clusters
//draw all points
//color points according to cluster value
private BufferedImage createImage(ArrayList<Cluster> clusters) {
//add 10% of the max size left and right to the image bounds
//BufferedImage bi = new BufferedImage((int) (max_xy * 1.05), (int) (max_xy * 1.05), BufferedImage.TYPE_BYTE_INDEXED);
BufferedImage bi = new BufferedImage((int) (max_xy * 1.05), (int) (max_xy * 1.05), BufferedImage.TYPE_INT_ARGB); // support 32-bit RGBA values
Graphics2D g2d = bi.createGraphics();
int numClusters = clusters.size();
for (Cluster c : clusters) {
//color points according to cluster value
Color col = colorFun.apply(c.clusterNumber, numClusters);
//draw all points
g2d.setColor(col);
for (Point p : c.points) {
g2d.fillRect((int) p.x, (int) p.y, (int) (max_xy * 0.02), (int) (max_xy * 0.02));
}
//draw all centers from clusters
g2d.setColor(new Color(160, 80, 80, 200)); // use RGBA: transparency=200
g2d.fillOval((int) c.x, (int) c.y, (int) (max_xy * 0.03), (int) (max_xy * 0.03));
}
return bi;
}
/**
* Calculates the euclidean distance without square root
* @param p
* @param c
* @return
*/
private double determineDistance(Point p, Cluster c) {
//math.sqrt not needed because the relative distance does not change by applying the square root
// return Math.sqrt(Math.pow((p.x - c.x), 2)+Math.pow((p.y - c.y),2));
return Math.pow((p.x - c.x), 2) + Math.pow((p.y - c.y), 2);
}
//TODO: What if coordinates can also be negative?
private void determineSize(ArrayList<Point> points) {
for (Point p : points) {
if (p.x > max_x) {
max_x = p.x;
}
if (p.y > max_y) {
max_y = p.y;
}
}
if (max_x > max_y) {
max_xy = max_x;
} else {
max_xy = max_y;
}
}
}
--- 点.java ---
package algo;
public class Point {
public double x;
public double y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
public Point(double x, double y) {
this.x = x;
this.y = y;
}
}
--- 集群.java ---
package algo;
import java.util.ArrayList;
public class Cluster {
public double x;
public double y;
public int clusterNumber;
public ArrayList<Point> points = new ArrayList<Point>();
public Cluster(double x, double y, int clusterNumber) {
this.x = x;
this.y = y;
this.clusterNumber = clusterNumber;
}
}
--- SimpleColorFun.java ---
package graphics;
import java.awt.Color;
import java.util.function.BiFunction;
/**
* Simple function for selection a color for a specific cluster identified with an integer-ID.
*
* @author makl, hese
*/
public class SimpleColorFun implements BiFunction<Integer, Integer, Color> {
/**
* Selects a color value.
* @param n current index
* @param numCol number of color-values possible
*/
@Override
public Color apply(Integer n, Integer numCol) {
Color col = Color.BLACK;
//color points according to cluster value
switch (n) {
case 1:
col = Color.RED;
break;
case 2:
col = Color.GREEN;
break;
case 3:
col = Color.BLUE;
break;
case 4:
col = Color.ORANGE;
break;
case 5:
col = Color.MAGENTA;
break;
case 6:
col = Color.YELLOW;
break;
case 7:
col = Color.CYAN;
break;
case 8:
col = Color.PINK;
break;
case 9:
col = Color.LIGHT_GRAY;
break;
default:
break;
}
return col;
}
}
--- Main.java ---(用一些时间记录机制替换秒表 - 我从我们的工作环境中得到这个)
package main;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Random;
import java.util.concurrent.TimeUnit;
import javax.imageio.ImageIO;
import algo.Clustering;
import algo.Point;
import eu.lbase.common.util.Stopwatch;
// import persistence.DataHandler;
public class Main {
private static final String OUTPUT_DIR = (new File("./output/withoutStream")).getAbsolutePath() + File.separator;
private static final String OUTPUT_DIR_2 = (new File("./output/withStream")).getAbsolutePath() + File.separator;
public static void main(String[] args) {
Random rng = new Random();
int numPoints = 300;
int seed = 2;
ArrayList<Point> points = new ArrayList<Point>();
rng.setSeed(rng.nextInt());
for (int i = 0; i < numPoints; i++) {
points.add(new Point(rng.nextInt(1000), rng.nextInt(1000)));
}
Stopwatch stw = Stopwatch.create(TimeUnit.MILLISECONDS);
{
// Stopwatch start
System.out.println("--- Started without streams ---");
stw.start();
Clustering algo = new Clustering(seed);
algo.setup(8, points, 25);
// Stopwatch stop
stw.stop();
System.out.println("--- Finished without streams ---");
System.out.printf("Elapsed time: %d msec%n%n", stw.getElapsed());
System.out.printf("Writing images to '%s' ...%n", OUTPUT_DIR);
deleteOldFiles(new File(OUTPUT_DIR));
makeImages(OUTPUT_DIR, algo);
System.out.println("Finished writing.\n");
}
{
System.out.println("--- Started with streams ---");
stw.start();
Clustering algo = new Clustering(seed);
algo.workWithStreams = true;
algo.setup(8, points, 25);
// Stopwatch stop
stw.stop();
System.out.println("--- Finished with streams ---");
System.out.printf("Elapsed time: %d msec%n%n", stw.getElapsed());
System.out.printf("Writing images to '%s' ...%n", OUTPUT_DIR_2);
deleteOldFiles(new File(OUTPUT_DIR_2));
makeImages(OUTPUT_DIR_2, algo);
System.out.println("Finished writing.\n");
}
}
/**
* creates one image for each iteration in the given directory
* @param algo
*/
private static void makeImages(String dir, Clustering algo) {
int i = 1;
for (BufferedImage img : algo.allGeneratedImages) {
try {
String filename = String.format("%03d.png", i);
ImageIO.write(img, "png", new File(dir + filename));
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
i++;
}
}
/**
* deletes old files from the target directory<br>
* Does <b>not</b> delete directories!
* @param dir - directory to delete files from
* @return
*/
private static boolean deleteOldFiles(File file) {
File[] allContents = file.listFiles();
if (allContents != null) {
for (File f : allContents) {
deleteOldFiles(f);
}
}
if (!file.isDirectory()) {
return file.delete();
}
return false;
}
}
最佳答案
当你想高效地使用 Streams 时，你应该停止使用 forEach 来基本上编写与循环相同的代码，而是了解 aggregate operations .另见综合package documentation .
线程安全的解决方案可能看起来像
points.stream().forEach(p -> {
Cluster min = clusters.stream()
.min(Comparator.comparingDouble(c -> determineDistance(p, c))).get();
// your original code used the custerNumber to lookup the Cluster in
// the list, don't know whether this is this really necessary
min = clusters.get(min.clusterNumber - 1);
// didn't find a better way considering your current code structure
synchronized(min) {
min.points.add(p);
}
});
List<Cluster> newClusters = clusters.stream()
.map(c -> new Cluster(
c.points.stream().mapToDouble(p -> p.x).sum()/c.points.size(),
c.points.stream().mapToDouble(p -> p.y).sum()/c.points.size(),
c.clusterNumber))
.collect(Collectors.toList());
}
但是您没有提供足够的上下文来测试它。有一些悬而未决的问题，例如您使用 Cluster 实例的 clusterNumber 来查看 clusters 列表；我不知道 clusterNumber 是否代表我们已经拥有的 Cluster 实例的实际列表索引，也就是说，如果这是不必要的冗余，或者具有不同的含义。
我也不知道比同步特定 Cluster 以使其列表线程安全的操作更好的解决方案(给定您当前的代码结构)。仅当您决定使用并行流时才需要这样做，即 points.parallelStream().forEach(p -> …)，其他操作不受影响。
您现在有几个流可以并行和顺序尝试，以找出您从哪里获益。通常，只有其他流才能带来显着的好处，如果有的话……
关于java - for each loops as streams in Java8 - k-means，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/48704263/