Java流。对对象流中的两个字段求和

Question

我有这样的想法：

Integer totalIncome = carDealer.getBrands().stream().mapToInt(brand -> brand.getManufacturer().getIncome()).sum();
Integer totalOutcome = carDealer.getBrands().stream().mapToInt(brand -> brand.getManufacturer().getOutcome()).sum();

我怎么能把它写在一个流中呢？收集f.e.对

编辑：

谢谢你们的评论、回答和参与。我会对使用流解决这个问题的不同方法提出问题。你对此有什么看法：

final IncomeAndOutcome incomeAndOutcome = carDealer.getBrands()
                    .stream()
                    .map(Brand::getManufacturer)
                    .map(IncomeAndOutcome::of)
                    .reduce(IncomeAndOutcome.ZERO, IncomeAndOutcome::sum);

static class IncomeAndOutcome {

    private static final IncomeAndOutcome ZERO = of(0, 0);

    @Getter
    private final int income;

    @Getter
    private final int outcome;

    public static IncomeAndOutcome of(final int income, final int outcome) {
        return new IncomeAndOutcome(income, outcome);
    }

    public static IncomeAndOutcome of(final Manufacturer manufacturer) {
        return new IncomeAndOutcome(manufacturer.getIncome(), manufacturer.getOutcome());
    }

    IncomeAndOutcome(final int income, final int outcome) {
        this.income = income;
        this.outcome = outcome;
    }

    IncomeAndOutcome sum(final IncomeAndOutcome incomeAndOutcome) {
        return of(this.income + incomeAndOutcome.getIncome(), this.outcome + incomeAndOutcome.getOutcome());
    }
}




Answer 1

没有正确的测量——一切都是猜测。我唯一同意的论点是可读性——这里几乎没有这种情况；但是如果你出于学术目的想知道这个，你可以这样做：

int[] result = carDealer.getBrands()
         .stream()
         .map(brand -> new int[]{brand.getManufacturer().getIncome(),
                                 brand.getManufacturer().getOutcome()})
         .collect(Collector.of(
                    () -> new int[2],
                    (left, right) -> {
                        left[0] += right[0];
                        left[1] += right[1];
                    },
                    (left, right) -> {
                        left[0] += right[0];
                        left[1] += right[1];
                        return left;
                    }));