Stream API练习实例

别团等shy哥发育

发布于 2023-02-25 16:42:05

3840

文章目录

1、实体类

交易员类

//交易员类
public class Trader {

	private String name;
	private String city;

	public Trader() {
	}

	public Trader(String name, String city) {
		this.name = name;
		this.city = city;
	}

	public String getName() {
		return name;
	}

	public void setName(String name) {
		this.name = name;
	}

	public String getCity() {
		return city;
	}

	public void setCity(String city) {
		this.city = city;
	}

	@Override
	public String toString() {
		return "Trader [name=" + name + ", city=" + city + "]";
	}

}

交易类

//交易类
public class Transaction {

	private Trader trader;
	private int year;
	private int value;

	public Transaction() {
	}

	public Transaction(Trader trader, int year, int value) {
		this.trader = trader;
		this.year = year;
		this.value = value;
	}

	public Trader getTrader() {
		return trader;
	}

	public void setTrader(Trader trader) {
		this.trader = trader;
	}

	public int getYear() {
		return year;
	}

	public void setYear(int year) {
		this.year = year;
	}

	public int getValue() {
		return value;
	}

	public void setValue(int value) {
		this.value = value;
	}

	@Override
	public String toString() {
		return "Transaction [trader=" + trader + ", year=" + year + ", value="
				+ value + "]";
	}

}

2、实例

测试数据

		List<Transaction> transactions = null;
	
		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、给定一个数字列表，返回一个由每个数的平方构成的列表

@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、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);
	}

3、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);
	}

4、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);
	}

5、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);
	}

6、是否有在米兰工作的交易员

@Test
	public void test5(){
		boolean bl = transactions.stream()
					.anyMatch((t) -> t.getTrader().getCity().equals("Milan"));
		
		System.out.println(bl);
	}

7、打印生活在剑桥的交易员的所有交易额

@Test
	public void test6(){
		//1
		Optional<Integer> sum = transactions.stream()
					.filter((e) -> e.getTrader().getCity().equals("Cambridge"))
					.map(Transaction::getValue)
					.reduce(Integer::sum);
		
		System.out.println(sum.get());
		//2
		Integer test = transactions.stream()
				.filter((e) -> e.getTrader().getCity().equals("Cambridge"))
				.collect(Collectors.summingInt(Transaction::getValue));
		System.out.println(test);
	}

8、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());
	}

9、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());
	}