如何使用流平均大小数？

Question

我想采取以下方法：

public BigDecimal mean(List<BigDecimal> bigDecimals, RoundingMode roundingMode) {
    BigDecimal sum = BigDecimal.ZERO;
    int count=0;
    for(BigDecimal bigDecimal : bigDecimals) {
        if(null != bigDecimal) {
            sum = sum.add(bigDecimal);
            count++;
        }
    }
    return sum.divide(new BigDecimal(count), roundingMode);
}

并使用Streams api更新它。这是我到目前为止得到的：

public BigDecimal average(List<BigDecimal> bigDecimals, RoundingMode roundingMode) {
    BigDecimal sum = bigDecimals.stream()
        .map(Objects::requireNonNull)
        .reduce(BigDecimal.ZERO, BigDecimal::add);
    long count = bigDecimals.stream().filter(Objects::nonNull).count();
    return sum.divide(new BigDecimal(count), roundingMode);
}

有没有办法在不流式传输两次（第二次获得计数）的情况下做到这一点？

Answer 1

或者，您可以使用此Collector实现：

class BigDecimalAverageCollector implements Collector<BigDecimal, BigDecimalAccumulator, BigDecimal> {

    @Override
    public Supplier<BigDecimalAccumulator> supplier() {
        return BigDecimalAccumulator::new;
    }

    @Override
    public BiConsumer<BigDecimalAccumulator, BigDecimal> accumulator() {
        return BigDecimalAccumulator::add;
    }

    @Override
    public BinaryOperator<BigDecimalAccumulator> combiner() {
        return BigDecimalAccumulator::combine;
    }

    @Override
    public Function<BigDecimalAccumulator, BigDecimal> finisher() {
        return BigDecimalAccumulator::getAverage;
    }

    @Override
    public Set<Characteristics> characteristics() {
        return Collections.emptySet();
    }

    @NoArgsConstructor
    @AllArgsConstructor
    static class BigDecimalAccumulator {
        @Getter private BigDecimal sum = BigDecimal.ZERO;
        @Getter private BigDecimal count = BigDecimal.ZERO;

        BigDecimal getAverage() {
           return BigDecimal.ZERO.compareTo(count) == 0 ?
                  BigDecimal.ZERO :
                  sum.divide(count, 2, BigDecimal.ROUND_HALF_UP);
        }

        BigDecimalAccumulator combine(BigDecimalAccumulator another) {
            return new BigDecimalAccumulator(
                    sum.add(another.getSum()),
                    count.add(another.getCount())
            );
        }

        void add(BigDecimal successRate) {
            count = count.add(BigDecimal.ONE);
            sum = sum.add(successRate);
        }
    }

}

这样使用：

BigDecimal mean = bigDecimals.stream().collect(new BigDecimalAverageCollector());

注意：示例使用Project Lombok注释来缩短粘合代码。

Answer 2

你不需要流两次。只需调用List。size（）用于计数：

public BigDecimal average(List<BigDecimal> bigDecimals, RoundingMode roundingMode) {
    BigDecimal sum = bigDecimals.stream()
        .map(Objects::requireNonNull)
        .reduce(BigDecimal.ZERO, BigDecimal::add);
    return sum.divide(new BigDecimal(bigDecimals.size()), roundingMode);
}

Answer 3

BigDecimal[] totalWithCount
                = bigDecimals.stream()
                .filter(bd -> bd != null)
                .map(bd -> new BigDecimal[]{bd, BigDecimal.ONE})
                .reduce((a, b) -> new BigDecimal[]{a[0].add(b[0]), a[1].add(BigDecimal.ONE)})
                .get();
BigDecimal mean = totalWithCount[0].divide(totalWithCount[1], roundingMode);

对于那些认为有帮助的人，代码的可选文本描述（如果您发现代码足够自解释，请忽略）：

• 将大小数列表转换为流