Java: 如何使用函数式编程正确操作 BigDecimal 数组?

Java: How to properly manipulate BigDecimal array using functional programming?

那么,如何使用函数式编程获得此代码的结果:

public static final List<BigDecimal> numbers = Arrays.asList(
    new BigDecimal("15"), new BigDecimal("10"), new BigDecimal("17"),
    new BigDecimal("30"), new BigDecimal("18"), new BigDecimal("23"),
    new BigDecimal("5"), new BigDecimal("12") );

BigDecimal totalOfReducedNumbers = BigDecimal.ZERO;
   for(BigDecimal number : numbers) {
       if(number.compareTo(BigDecimal.valueOf(20)) > 0) 
           totalOfReducedNumbers = 
         totalOfReducedNumbers.add(number.multiply(BigDecimal.valueOf(0.9)));
   }
System.out.println("Total of reduced numbers: " + totalOfReducedNumbers);

抛出“总减少数:47.7

如何使用 map()、reduce() 等函数式编程工具 获得相同的结果?

通过执行相同的操作,首先过滤值(您只需要大于 20 的值)。然后将这些值乘以 0.9,最后通过执行加法来减少项。喜欢,

BigDecimal TWENTY = BigDecimal.valueOf(20);
BigDecimal POINT9 = BigDecimal.valueOf(0.9);
System.out.println("Total of reduced numbers: " + numbers.stream()
        .filter(x -> x.compareTo(TWENTY) > 0)
        .map(x -> x.multiply(POINT9)).reduce((a, b) -> a.add(b)).get());

输出(按要求)

Total of reduced numbers: 47.7

而且,正如评论中所建议的,我们可以通过方法参考进一步改进,使用 orElse 比原始 get() 更安全。喜欢,

System.out.println("Total of reduced numbers: " + numbers.stream() 
        .filter(x -> x.compareTo(TWENTY) > 0) 
        .map(x -> x.multiply(POINT9)) 
        .reduce(BigDecimal::add) 
        .orElse(BigDecimal.ZERO));

for 循环实现是典型的 "reduce" 模式(在某些其他语言中也称为 "fold"、"Aggregate" 模式)

您正在寻找 filter --> reduce 方法:

BigDecimal reduce = 
         numbers.stream()
                .filter(n -> n.compareTo(BigDecimal.valueOf(20)) > 0)
                .reduce(BigDecimal.ZERO, 
                       (a, b) -> a.add(b.multiply(BigDecimal.valueOf(0.9))));

您可以进一步减少通过缓存 BigDecimal.valueOf(20)BigDecimal.valueOf(0.9) 构造的对象数量,如@Elliot 的回答所示,即:

BigDecimal TWENTY = BigDecimal.valueOf(20);
BigDecimal POINT9 = BigDecimal.valueOf(0.9);
BigDecimal reduce = 
             numbers.stream()
                    .filter(n -> n.compareTo(TWENTY) > 0)
                    .reduce(BigDecimal.ZERO, 
                           (a, b) -> a.add(b.multiply(POINT9)));