使用Stream和java8在一个字符串中返回最小，最大，平均，总和，薪金计数

Question

我有一份以薪水为特征的雇员List 。 为什么此代码不起作用？

String joined = employees.stream().collect(
    Collectors.summingInt(Employee::getSalary),
    Collectors.maxBy(Comparator.comparing(Employee::getSalary)),
    Collectors.minBy(Comparator.comparing(Employee::getSalary)),
    Collectors.averagingLong((Employee e) ->e.getSalary() * 2),
    Collectors.counting(),
    Collectors.joining(", "));

我正在使用一套收集器。

Answer 1

请注意，当前您尝试获取的不是最高/最低薪水，而是拥有该薪水的雇员。 如果您实际上想获得最高/最低工资本身（数字），则可以使用Collectors.summarizingInt()一次计算这些特征：

IntSummaryStatistics stats = employees.stream()
    .collect(Collectors.summarizingInt(Employee::getSalary));

如果要将它们连接为字符串，则可以使用：

String statsString = Stream.of(stats.getSum(), stats.getMax(), stats.getMin(), 
                               stats.getAverage()*2, stats.getCount())
                           .map(Object::toString)
                           .collect(Collectors.joining(", "));

---

如果您确实想获得薪水最高/最低的Employee，则此处IntSummaryStatistics不会为您提供帮助。 但是，您可以改为创建收集器流：

String result = Stream.<Collector<Employee,?,?>>of(
            Collectors.summingInt(Employee::getSalary),
            Collectors.maxBy(Comparator.comparing(Employee::getSalary)),
            Collectors.minBy(Comparator.comparing(Employee::getSalary)),
            Collectors.averagingLong((Employee e) ->e.getSalary() * 2),
            Collectors.counting())
        .map(collector -> employees.stream().collect(collector))
        .map(Object::toString)
        .collect(Collectors.joining(", "));

请注意，通过这种方式，您将获得类似（取决于Employee.toString()实现的输出：

1121, Optional[Employee [salary=1000]], Optional[Employee [salary=1]], 560.5, 4

不要忘记maxBy / minBy返回Optional 。

---

如果您对第一个解决方案不满意，并且由于某种原因不想多次迭代输入，则可以使用以下方法创建组合的收集器：

/**
 * Returns a collector which joins the results of supplied collectors
 * into the single string using the supplied delimiter.
 */
@SafeVarargs
public static <T> Collector<T, ?, String> joining(CharSequence delimiter, 
        Collector<T, ?, ?>... collectors) {
    @SuppressWarnings("unchecked")
    Collector<T, Object, Object>[] cs = (Collector<T, Object, Object>[]) collectors;
    return Collector.<T, Object[], String>of(
        () -> Stream.of(cs).map(c -> c.supplier().get()).toArray(), 
        (acc, t) -> IntStream.range(0, acc.length)
            .forEach(idx -> cs[idx].accumulator().accept(acc[idx], t)), 
        (acc1, acc2) -> IntStream.range(0, acc1.length)
            .mapToObj(idx -> cs[idx].combiner().apply(acc1[idx], acc2[idx]))
            .toArray(), 
        acc -> IntStream.range(0, acc.length)
            .mapToObj(idx -> cs[idx].finisher().apply(acc[idx]).toString())
            .collect(Collectors.joining(delimiter)));
}

有这样的方法，你可以写

String stats = employees.stream().collect(joining(", ",
        Collectors.summingInt(Employee::getSalary),
        Collectors.maxBy(Comparator.comparing(Employee::getSalary)),
        Collectors.minBy(Comparator.comparing(Employee::getSalary)),
        Collectors.averagingLong((Employee e) ->e.getSalary() * 2),
        Collectors.counting()));

Answer 2

我终于找到了解决方案。。谢谢尝试的家伙

String s = employees.stream().mapToDouble(a>a.getSalary()).summaryStatistics().toString();

这是输出：

 DoubleSummaryStatistics{count=21, sum=17200,000000, min=100,000000,
 average=819,047619, max=2100,000000}