Pre
都学了这俩，是不是该出来练一练了？
练习
基础数据
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) );
(1) 找出2011年发生的所有交易，并按交易额排序（从低到高）
List<Transaction> collect = transactions.stream() .filter(t -> t.getYear() == 2011) .sorted(Comparator.comparing(Transaction::getValue)) .collect(toList());
(2) 交易员都在哪些不同的城市工作过？
List<String> collect1 = transactions.stream() .map(t->t.getTrader().getCity()) .distinct() .collect(toList());
(3) 查找所有来自于剑桥的交易员，并按姓名排序。
transactions.stream().map(t->t.getTrader()) .filter(x -> x.getCity().equals("Cambridge")) .distinct() .sorted(Comparator.comparing(Trader::getName)) .forEach(System.out::println);
(4) 返回所有交易员的姓名字符串，按字母顺序排序。
String value = transactions.stream().map(t -> t.getTrader().getName()) .distinct() .sorted() .reduce("", (name1, name2) -> name1 + name2); System.out.println(value);
请注意，此解决方案效率不高（所有字符串都被反复连接，每次迭代的时候都要建立一个新的 String 对象）。
更为高效的解决方案，它像下面这样使用 joining
String traderStr = transactions.stream() .map(transaction -> transaction.getTrader().getName()) .distinct() .sorted() .collect(joining());
(5) 有没有交易员是在米兰工作的？
boolean liveInMilan1 = transactions.stream().anyMatch(t -> t.getTrader().getCity().equals("Milan")); boolean liveInMilan2 = transactions.stream().map(Transaction::getTrader).anyMatch(t -> t.getCity().equals("Milan")); System.out.println(liveInMilan1); System.out.println(liveInMilan2);
(6) 打印生活在剑桥的交易员的所有交易额。
transactions.stream().filter(t -> t.getTrader().getCity().equals("Cambridge")) .map(Transaction::getValue) .forEach(System.out::println);
(7) 所有交易中，最高的交易额是多少？
Optional<Integer> maxValue = transactions.stream().map(Transaction::getValue).reduce((i, j) -> i > j ? i : j); System.out.println(maxValue.get());
(8) 找到交易额最小的交易
Optional<Integer> minValue = transactions.stream().map(Transaction::getValue).reduce(Integer::min); System.out.println(minValue.get());
还可以做得更好。流支持 min 和 max 方法，它们可以接受一个 Comparator 作为参数，指定计算最小或最大值时要比较哪个键值：
Optional<Transaction> smallestTransaction = transactions.stream() .min(comparing(Transaction::getValue));
附 Trader & Transaction
package com.artisan.java8.stream.excise; public class Trader{ private final String name; private final 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 String toString(){ return "Trader:"+this.name + " in " + this.city; } }
package com.artisan.java8.stream.excise; public class Transaction { private final Trader trader; private final int year; private final 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 + "}"; } }
