《Java 8函数式编程》读书记录(1)

Chapter 6 DataParallelism

6.1 并行化流操作

Data parallelism is a way to split up work to be done on many cores at the same time .

计算一组专辑的曲目总长度，拿到每张album, 得到List< Track>信息， 通过flatMap组成新的Track Stream, 然后mapToInt计算每个Track的length：

public int serialArraySum() {
return albums.stream()
.flatMap(Album::getTracks)
.mapToInt(Track::getLength)
.sum();
}

/*
public Stream<Track> getTracks() {
return tracks.stream(); // List<Track> Stream
}

*/

并行化处理：

public int parallelArraySum() {
return albums.parallelStream()
.flatMap(Album::getTracks)
.mapToInt(Track::getLength)
.sum();
}

根据性能好坏，we can split up common data source from the core library into three main groups by performance characteristics(性能) :

The good(性能好)

An

ArrayList

, an

array

, or the

IntStream.range

constructor. These data sources all support random access, which means they can be split up arbitrarily with ease.

The Okay(性能一般)

The

HashSet

and

TreeSet

. You can’t easily decompose these with perfect amounts of balance, but most of the time it’s possible to do so.

The Bad(性能差)

Some data structures just don’t split well; for example, they may take O(N) time to decompose. Example here include a

LinkedList

, which is computationally hard to split in half.Also,

Streams.iterate

and

BufferedReader.lines

have unknown length at the beginning, so it’s pretty hard to estimate when to split these sources.

在讨论流中单独操作每一块的种类时，we can differentiate between two types of stream operation: stateless(无状态的) and stateful(有状态的)。Stateless operations need to maintain no concept of state over the whole operation; stateful operations have the overhead and constraint of maintaining state.

If you can get a way with using stateless operations, then you will get better parallel performance. Examples of stateless operations include

map

,

filter

, and

flatMap

;

sorted

,

distinct

, and

limit

are stateful.

6.2 Parallel Array Operations(并行化数组操作)

These operations are all located on the utility class Arrays.

Parallel operations on arrays:



Initializing an array using a for loop:

public static double[] imperativeInitilize(int size) {
double[] values = new double[size];
for(int i = 0; i < values.length;i++) {
values[i] = i;
}
return values;
}

Using

parallelSetAll

method in order to do this easily in parallel.

eg:

public static double[] parallelInitialize(int size) {
double[] values = new double[size];
Arrays.parallelSetAll(values, i -> i);
return values;
}

6.3 More Example

// 求平方和
public static int sumOfSquares(IntStream range) {
return range.parallel()
.map(x -> x * x)
.sum();
}

public static int sequentialSumOfSquares(IntStream range) {
return range.map(x -> x * x)
.sum();
}

// 并行化执行，注意初始值不能设为其他值，除了1
// 初值须为函数的恒等值，用恒等值和其他值做reduce操作时，
// 其他值保持不变
public static int multiplyThrough(List<Integer> numbers) {
return 5 * numbers.parallelStream()
.reduce(1, (acc, x) -> x * acc);
}

在来比较一个：

// fast
public int fastSumOfSquares() {
return arrayListOfNumbers.parallelStream()
.mapToInt(x -> x * x)
.sum();
}
// slowly
public int slowSumOfSquares() {
return linkedListOfNumbers.parallelStream()
.map(x -> x * x)
.reduce(0, (acc, x) -> acc + x);
}

Chapter 7. Testing, Debugging, and Refactoring

7.1 Refactor example

ThreadLocal<Album> thisAlbum = new ThreadLocal<Album>() {
@Override
protected Album initialValue() {
return database.lookupCurrentAlbum();
}
}

// other
ThreadLocal<Album> thisAlbum = ThreadLocal.withInitial(() -> database.lookupCurrentAlbum());

More examples:

1) refactor step1

/**
* Album  专辑
* Track  专辑里的歌曲
*/
// 计算所有专辑歌曲的时间
public long countRunningTime(List<Album> albums) {
long count = 0;
for (Album album : albums) {
for (Track track : album.getTrackList()) {
count += track.getLength();
}
}
return count;
}

// 计算所有专辑有多少音乐家
public long countMusicians(List<Album> albums) {
long count = 0;
for (Album album : albums) {
count += album.getMusicianList().size();
}
return count;
}

// 计算所有专辑有多少首歌曲
public long countTracks(List<Album> albums) {
long count = 0;
for (Album album : albums) {
count += album.getTrackList().size();
}
return count;
}

2) refactor step2

/**
* Album  专辑
* Track  专辑里的歌曲
*/
// 计算所有专辑歌曲的时间
public long countRunningTime(List<Album> albums) {
return albums.stream().mapToLong(
album -> album.getTracks().mapToLong(track -> track.getLength()).sum()
).sum();
}

// 计算所有专辑有多少音乐家
public long countMusicians(List<Album> albums) {
return albums.stream().mapToLong(
album -> album.getMusicianList().size()
).sum();
}

// 计算所有专辑有多少首歌曲
public long countTracks(List<Album> albums) {
return albums.stream().mapToLong(
album -> album.getTrackList().size()
).sum();
}

3) refactor step3

ToLongFunction

/**
* Album  专辑
* Track  专辑里的歌曲
*/
// 计算所有专辑歌曲的时间
public long countRunningTime(List<Album> albums) {
return contFeature(album -> album.getTracks().mapToLong(track -> track.getLength()).sum(), albums);
}

// 计算所有专辑有多少音乐家
public long countMusicians(List<Album> albums) {
return contFeature(album -> album.getMusicianList().size(), albums);
}

// 计算所有专辑有多少首歌曲
public long countTracks(List<Album> albums) {
return contFeature(album -> album.getTrackList().size(), albums);
}

public long contFeature(ToLongFunction<Album> function, List<Album> albums) {
return albums.stream().mapToLong(function).sum();
}

7.2 Unit Test

// Converting strings into their uppercase equivalents
public static List<String> allToUpperCase(List<String> words) {
return words.stream().map(str -> str.toUpperCase()).collect(Collectors.toList());
}

public static List<String> elementFirstToUpperCaseLambdas(List<String> words) {
return words.stream().map(
value -> {
char firstChar = Character.toUpperCase(value.charAt(0));
return firstChar + value.substring(1);
}
).collect(Collectors.toList());
}

Do use method references:

public class UnitTest {
public static List<String> elementFirstToUpperCaseLambdasMethodRef(List<String> words) {
return words.stream().map(
UnitTest::firstToUpperCase
).collect(Collectors.toList());
}

public static String firstToUpperCase(String value) {
char firstChar = Character.toUpperCase(value.charAt(0));
return firstChar + value.substring(1);
}
}

The Solution: peek

public static Set<String> forEachLoggingFailure(Album album) {
album.getMusicians()
.filter(artist -> artist.getName().startsWith("The"))
.map(artist -> artist.getNationality())
.forEach(nationality -> System.out.println("Found: " + nationality));

Set<String> nationalities
= album.getMusicians()
.filter(artist -> artist.getName().startsWith("The"))
.map(artist -> artist.getNationality())
.collect(Collectors.toSet());
return nationalities;
}

The streams library contains a method that lets you look at each value in turn and also lets you continue to operate on the same underlying stream.It’s called peek.

public static Set<String> nationalityReportUsingPeek(Album album) {
Set<String> nationalities
= album.getMusicians()
.filter(artist -> artist.getName().startsWith("The"))
.map(artist -> artist.getNationality())
.peek(nation -> System.out.println("Found nationality: " + nation))
.collect(Collectors.toSet());
return nationalities;
}

这样还可以使用peek进行日志记录。

Chapter 8. Design and Architectural Principles

The critical design tool for software development is a mind well educated in design principles. It is not…technology.