Java8 Stream流常用方法及例子

交易員和交易的實體類的定義如下：

Trader.class

public class Trader {

    private String name;
    private String city;

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

    public String getName() {
        return this.name;
    }

    public String getCity() {
        return this.city;
    }

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

    public String toString() {
        return "Trader:" + this.name + " in " + this.city;
    }
}

Transaction.class

public class Transaction {

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

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

    public Trader getTrader() {
        return this.trader;
    }

    public int getYear() {
        return this.year;
    }

    public int getValue() {
        return this.value;
    }

    public String toString() {
        return "{" + this.trader + ", " +
                "year: " + this.year + ", " +
                "value:" + this.value + "}";
    }
}

問題：

(1) 找出2011年發生的所有交易，並按交易額排序（從低到高）。
(2) 交易員都在哪些不同的城市工作過？
(3) 查找所有來自於劍橋的交易員，並按姓名排序。
(4) 返回所有交易員的姓名字符串，按字母順序排序。
(5) 有沒有交易員是在米蘭工作的？
(6) 打印生活在劍橋的交易員的所有交易額。
(7) 所有交易中，最高的交易額是多少？
(8) 找到交易額最小的交易。

解答：

import java.util.Arrays;
import java.util.List;
import java.util.Optional;

import static java.util.Comparator.comparing;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

public class PuttingIntoPractice {

    public static void main(String[] args) {
        Trader raoul = new Trader("Raoul", "Cambridge");
        Trader mario = new Trader("Mario", "Milan");
        Trader alan = new Trader("Alan", "Cambridge");
        Trader brian = new Trader("Brian", "Cambridge");

        List<Transaction> 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)
        );

        // Query 1: Find all transactions from year 2011 and sort them by value (small to high).
        List<Transaction> tr2011 = transactions.stream()
                .filter(transaction -> transaction.getYear() == 2011)
                .sorted(comparing(Transaction::getValue))
                .collect(toList());
        System.out.println(tr2011);

        // Query 2: What are all the unique cities where the traders work?
        List<String> cities =
                transactions.stream()
                        .map(transaction -> transaction.getTrader().getCity())
                        .distinct()
                        .collect(toList());
        System.out.println(cities);

        // Query 3: Find all traders from Cambridge and sort them by name.

        List<Trader> traders =
                transactions.stream()
                        .map(Transaction::getTrader)
                        .filter(trader -> trader.getCity().equals("Cambridge"))
                        .distinct()
                        .sorted(comparing(Trader::getName))
                        .collect(toList());
        System.out.println(traders);


        // Query 4: Return a string of all traders’ names sorted alphabetically.

        String traderStr =
                transactions.stream()
                        .map(transaction -> transaction.getTrader().getName())
                        .distinct()
                        .sorted()
                        .reduce("", (n1, n2) -> n1 + n2);
        System.out.println(traderStr);

        // 請注意，此解決方案效率不高（所有字符串都被反復連接，每次迭代的時候都要建立一個新的String對象）。
        // 這里有一個更為高效的解決方案，它像下面這樣使用joining（其內部會用到StringBuilder）：
        String traderStr_2 =
                transactions.stream()
                        .map(transaction -> transaction.getTrader().getName())
                        .distinct()
                        .sorted()
                        .collect(joining());
        // Query 5: Are there any trader based in Milan?

        boolean milanBased =
                transactions.stream()
                        .anyMatch(transaction -> transaction.getTrader()
                                .getCity()
                                .equals("Milan")
                        );
        System.out.println(milanBased);


        // Query 6: Update all transactions so that the traders from Milan are set to Cambridge.
        transactions.stream()
                .map(Transaction::getTrader)
                .filter(trader -> trader.getCity().equals("Milan"))
                .forEach(trader -> trader.setCity("Cambridge"));
        System.out.println(transactions);


        // Query 7: What's the highest value in all the transactions?
        int highestValue =
                transactions.stream()
                        .map(Transaction::getValue)
                        .reduce(0, Integer::max);
        System.out.println(highestValue);

        // tips:流支持min和max方法，它們可以接受一個Comparator作為參數，指定計算最小或最大值時要比較哪個鍵值：
        Optional<Transaction> smallestTransaction =
                transactions.stream()
                        .min(comparing(Transaction::getValue));
    }


}