Java流.在对象流中求和两个字段

Question

我有这样的事情:

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

我怎么能在一个流中写出来？Pair<Integer, Integer>用totalIncome和收集fe totalOutcome？

编辑:

谢谢你们的评论,答案和参与.我会对使用流来解决该问题的不同方法提出疑问.你怎么看待这个:

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

如果没有正确测量——一切都是猜测。我唯一同意的论点是关于可读性——这里的情况几乎不是这样；但如果您出于学术目的想了解这一点，您可以这样做：

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;
                    }));