java8新特性（Stream API）
Stream API的操作步骤：
1、创建Stream
2、中间操作
3、终止操作（终端操作）
//1. 创建 Stream
@Test
public void test1(){
//1. Collection 提供了两个方法 stream() 与 parallelStream()
List<String> list = new ArrayList<>();
Stream<String> stream = list.stream(); //获取一个顺序流
Stream<String> parallelStream = list.parallelStream(); //获取一个并行流
//2. 通过 Arrays 中的 stream() 获取一个数组流
Integer[] nums = new Integer[10];
Stream<Integer> stream1 = Arrays.stream(nums);
//3. 通过 Stream 类中静态方法 of()
Stream<Integer> stream2 = Stream.of(1,2,3,4,5,6);
//4. 创建无限流
//迭代
Stream<Integer> stream3 = Stream.iterate(0, (x) -> x + 2).limit(10);
stream3.forEach(System.out::println);
//生成
Stream<Double> stream4 = Stream.generate(Math::random).limit(2);
stream4.forEach(System.out::println);
}
中间操作：
筛选与切片：
filter-接收Lambda表达式，从流中排出某些元素
limit--截断流，使元素个数不超过指定格式
skip--跳过元素，返回一个扔掉了前n个元素的流，若流中元素不足n个，则返回一个空流，与limit互补
distinct--筛选，通过所生成元素的hashCode()和equals()方法去除重复元素
filter：
//内部迭代：迭代操作 Stream API 内部完成
@Test
public void test2(){
//所有的中间操作不会做任何的处理
Stream<Employee> stream = emps.stream()
.filter((e) -> {
System.out.println("正在执行中间操作");
return e.getAge() <= 35;
});
//只有当做终止操作时，所有的中间操作会一次性的全部执行，称为“惰性求值”
stream.forEach(System.out::println);
}
//外部迭代
@Test
public void test3(){
Iterator<Employee> it = emps.iterator();
while(it.hasNext()){
System.out.println(it.next());
}
}
limit：
@Test
public void test4(){
emps.stream()
.filter((e) -> {
System.out.println("xxx"); // && ||
return e.getSalary() >= 5000;
}).limit(3)
.forEach(System.out::println);
}
skip：
@Test
public void test5(){
emps.parallelStream()
.filter((e) -> e.getSalary() >= 5000)
.skip(2)
.forEach(System.out::println);
}
distinct：
@Test
public void test6(){
emps.stream()
.distinct()
.forEach(System.out::println);
}
映射：
map——接收 Lambda ， 将元素转换成其他形式或提取信息。接收一个函数作为参数，该函数会被应用到每个元素上，并将其映射成一个新的元素。
flatMap——接收一个函数作为参数，将流中的每个值都换成另一个流，然后把所有流连接成一个流
public static Stream<Character> filterCharacter(String str){
List<Character> list = new ArrayList<>();
for (Character ch : str.toCharArray()) {
list.add(ch);
}
return list.stream();
}
@Test
public void test1(){
Stream<String> str = emps.stream()
.map((e) -> e.getName());
System.out.println("-------------------------------------------");
List<String> strList = Arrays.asList("aaa", "bbb", "ccc", "ddd", "eee");
Stream<String> stream = strList.stream()
.map(String::toUpperCase);
stream.forEach(System.out::println);
Stream<Stream<Character>> stream2 = strList.stream()
.map(TestStreamAPI1::filterCharacter);
stream2.forEach((sm) -> {
sm.forEach(System.out::println);
});
System.out.println("---------------------------------------------");
Stream<Character> stream3 = strList.stream()
.flatMap(TestStreamAPI1::filterCharacter);
stream3.forEach(System.out::println);
}
排序：
sorted()——自然排序
sorted(Comparator com)——定制排序
@Test
public void test2(){
emps.stream()
.map(Employee::getName)
.sorted()
.forEach(System.out::println);
System.out.println("------------------------------------");
emps.stream()
.sorted((x, y) -> {
if(x.getAge() == y.getAge()){
return x.getName().compareTo(y.getName());
}else{
return Integer.compare(x.getAge(), y.getAge());
}
}).forEach(System.out::println);
}
查找与匹配：
allMatch——检查是否匹配所有元素
anyMatch——检查是否至少匹配一个元素
noneMatch——检查是否没有匹配的元素
findFirst——返回第一个元素
findAny——返回当前流中的任意元素
count——返回流中元素的总个数
max——返回流中最大值
min——返回流中最小值
List<Employee> emps = Arrays.asList(
new Employee(102, "李四", 59, 6666.66, Status.BUSY),
new Employee(101, "张三", 18, 9999.99, Status.FREE),
new Employee(103, "王五", 28, 3333.33, Status.VOCATION),
new Employee(104, "赵六", 8, 7777.77, Status.BUSY),
new Employee(104, "赵六", 8, 7777.77, Status.FREE),
new Employee(104, "赵六", 8, 7777.77, Status.FREE),
new Employee(105, "田七", 38, 5555.55, Status.BUSY)
);
@Test
public void test1(){
boolean bl = emps.stream()
.allMatch((e) -> e.getStatus().equals(Status.BUSY));
System.out.println(bl);
boolean bl1 = emps.stream()
.anyMatch((e) -> e.getStatus().equals(Status.BUSY));
System.out.println(bl1);
boolean bl2 = emps.stream()
.noneMatch((e) -> e.getStatus().equals(Status.BUSY));
System.out.println(bl2);
}
@Test
public void test2(){
Optional<Employee> op = emps.stream()
.sorted((e1, e2) -> Double.compare(e1.getSalary(), e2.getSalary()))
.findFirst();
System.out.println(op.get());
System.out.println("--------------------------------");
Optional<Employee> op2 = emps.parallelStream()
.filter((e) -> e.getStatus().equals(Status.FREE))
.findAny();
System.out.println(op2.get());
}
@Test
public void test3(){
long count = emps.stream()
.filter((e) -> e.getStatus().equals(Status.FREE))
.count();
System.out.println(count);
Optional<Double> op = emps.stream()
.map(Employee::getSalary)
.max(Double::compare);
System.out.println(op.get());
Optional<Employee> op2 = emps.stream()
.min((e1, e2) -> Double.compare(e1.getSalary(), e2.getSalary()));
System.out.println(op2.get());
}
//注意：流进行了终止操作后，不能再次使用
@Test
public void test4(){
Stream<Employee> stream = emps.stream()
.filter((e) -> e.getStatus().equals(Status.FREE));
long count = stream.count();
stream.map(Employee::getSalary)
.max(Double::compare);
}
规约：
reduce(T identity, BinaryOperator) / reduce(BinaryOperator) ——可以将流中元素反复结合起来，得到一个值。
@Test
public void test1(){
List<Integer> list = Arrays.asList(1,2,3,4,5,6,7,8,9,10);
Integer sum = list.stream()
.reduce(0, (x, y) -> x + y);
System.out.println(sum);
System.out.println("----------------------------------------");
Optional<Double> op = emps.stream()
.map(Employee::getSalary)
.reduce(Double::sum);
System.out.println(op.get());
}
//需求：搜索名字中 “六” 出现的次数
@Test
public void test2(){
Optional<Integer> sum = emps.stream()
.map(Employee::getName)
.flatMap(TestStreamAPI1::filterCharacter)
.map((ch) -> {
if(ch.equals('六'))
return 1;
else
return 0;
}).reduce(Integer::sum);
System.out.println(sum.get());
}
collect——将流转换为其他形式。接收一个 Collector接口的实现，用于给Stream中元素做汇总的方法
@Test
public void test3(){
List<String> list = emps.stream()
.map(Employee::getName)
.collect(Collectors.toList());
list.forEach(System.out::println);
System.out.println("----------------------------------");
Set<String> set = emps.stream()
.map(Employee::getName)
.collect(Collectors.toSet());
set.forEach(System.out::println);
System.out.println("----------------------------------");
HashSet<String> hs = emps.stream()
.map(Employee::getName)
.collect(Collectors.toCollection(HashSet::new));
hs.forEach(System.out::println);
}
@Test
public void test4(){
Optional<Double> max = emps.stream()
.map(Employee::getSalary)
.collect(Collectors.maxBy(Double::compare));
System.out.println(max.get());
Optional<Employee> op = emps.stream()
.collect(Collectors.minBy((e1, e2) -> Double.compare(e1.getSalary(), e2.getSalary())));
System.out.println(op.get());
Double sum = emps.stream()
.collect(Collectors.summingDouble(Employee::getSalary));
System.out.println(sum);
Double avg = emps.stream()
.collect(Collectors.averagingDouble(Employee::getSalary));
System.out.println(avg);
Long count = emps.stream()
.collect(Collectors.counting());
System.out.println(count);
System.out.println("--------------------------------------------");
DoubleSummaryStatistics dss = emps.stream()
.collect(Collectors.summarizingDouble(Employee::getSalary));
System.out.println(dss.getMax());
}
//分组
@Test
public void test5(){
Map<Status, List<Employee>> map = emps.stream()
.collect(Collectors.groupingBy(Employee::getStatus));
System.out.println(map);
}
//多级分组
@Test
public void test6(){
Map<Status, Map<String, List<Employee>>> map = emps.stream()
.collect(Collectors.groupingBy(Employee::getStatus, Collectors.groupingBy((e) -> {
if(e.getAge() >= 60)
return "老年";
else if(e.getAge() >= 35)
return "中年";
else
return "成年";
})));
System.out.println(map);
}
//分区
@Test
public void test7(){
Map<Boolean, List<Employee>> map = emps.stream()
.collect(Collectors.partitioningBy((e) -> e.getSalary() >= 5000));
System.out.println(map);
}
@Test
public void test8(){
String str = emps.stream()
.map(Employee::getName)
.collect(Collectors.joining("," , "----", "----"));
System.out.println(str);
}
/* 1. 给定一个数字列表，如何返回一个由每个数的平方构成的列表呢？ ，给定【1，2，3，4，5】， 应该返回【1，4，9，16，25】。 */ @Test public void test1(){ Integer[] nums = new Integer[]{1,2,3,4,5}; Arrays.stream(nums) .map((x) -> x * x) .forEach(System.out::println); }
/* 2. 怎样用 map 和 reduce 方法数一数流中有多少个Employee呢？ */ List<Employee> emps = Arrays.asList( new Employee(102, "李四", 59, 6666.66, Status.BUSY), new Employee(101, "张三", 18, 9999.99, Status.FREE), new Employee(103, "王五", 28, 3333.33, Status.VOCATION), new Employee(104, "赵六", 8, 7777.77, Status.BUSY), new Employee(104, "赵六", 8, 7777.77, Status.FREE), new Employee(104, "赵六", 8, 7777.77, Status.FREE), new Employee(105, "田七", 38, 5555.55, Status.BUSY) ); @Test public void test2(){ Optional<Integer> count = emps.stream() .map((e) -> 1) .reduce(Integer::sum); System.out.println(count.get()); }
//交易员类 public class Trader { private String name; private String city; ｝
//交易类 public class Transaction { private Trader trader; private int year; private int value; ｝
List<Transaction> transactions = null; @Before public void before(){ Trader raoul = new Trader("Raoul", "Cambridge"); Trader mario = new Trader("Mario", "Milan"); Trader alan = new Trader("Alan", "Cambridge"); Trader brian = new Trader("Brian", "Cambridge"); transactions = Arrays.asList( new Transaction(brian, 2011, 300), new Transaction(raoul, 2012, 1000), new Transaction(raoul, 2011, 400), new Transaction(mario, 2012, 710), new Transaction(mario, 2012, 700), new Transaction(alan, 2012, 950) ); }
//1. 找出2011年发生的所有交易， 并按交易额排序（从低到高） @Test public void test1(){ transactions.stream() .filter((t) -> t.getYear() == 2011) .sorted((t1, t2) -> Integer.compare(t1.getValue(), t2.getValue())) .forEach(System.out::println); }
//2. 交易员都在哪些不同的城市工作过？ @Test public void test2(){ transactions.stream() .map((t) -> t.getTrader().getCity()) .distinct() .forEach(System.out::println); }
//3. 查找所有来自剑桥的交易员，并按姓名排序 @Test public void test3(){ transactions.stream() .filter((t) -> t.getTrader().getCity().equals("Cambridge")) .map(Transaction::getTrader) .sorted((t1, t2) -> t1.getName().compareTo(t2.getName())) .distinct() .forEach(System.out::println); }
public static Stream<String> filterCharacter(String str){ List<String> list = new ArrayList<>(); for (Character ch : str.toCharArray()) { list.add(ch.toString()); } return list.stream(); } //4. 返回所有交易员的姓名字符串，按字母顺序排序 @Test public void test4(){ transactions.stream() .map((t) -> t.getTrader().getName()) .sorted() .forEach(System.out::println); System.out.println("-----------------------------------"); String str = transactions.stream() .map((t) -> t.getTrader().getName()) .sorted() .reduce("", String::concat); System.out.println(str); System.out.println("------------------------------------"); transactions.stream() .map((t) -> t.getTrader().getName()) .flatMap(TestTransaction::filterCharacter) .sorted((s1, s2) -> s1.compareToIgnoreCase(s2)) .forEach(System.out::print); }
//5. 有没有交易员是在米兰工作的？ @Test public void test5(){ boolean bl = transactions.stream() .anyMatch((t) -> t.getTrader().getCity().equals("Milan")); System.out.println(bl); }
//6. 打印生活在剑桥的交易员的所有交易额 @Test public void test6(){ Optional<Integer> sum = transactions.stream() .filter((e) -> e.getTrader().getCity().equals("Cambridge")) .map(Transaction::getValue) .reduce(Integer::sum); System.out.println(sum.get()); }
//7. 所有交易中，最高的交易额是多少 @Test public void test7(){ Optional<Integer> max = transactions.stream() .map((t) -> t.getValue()) .max(Integer::compare); System.out.println(max.get()); }
//8. 找到交易额最小的交易 @Test public void test8(){ Optional<Transaction> op = transactions.stream() .min((t1, t2) -> Integer.compare(t1.getValue(), t2.getValue())); System.out.println(op.get()); }
