使用 Stream 和 java8 以一个字符串返回工资的最小值、最大值、平均值、总和、计数

问题描述 投票:0回答:3

我有

List
的员工,他们的特点是工资。 为什么这段代码不起作用?

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。

java java-8 java-stream collectors
3个回答
5
投票

请注意,当前您试图获取的不是最高/最低工资,而是具有此类工资的员工。如果您实际上想要拥有最高/最低工资本身(数字),那么可以使用

Collectors.summarizingInt()
:

立即计算这些特征
IntSummaryStatistics stats = employees.stream()
    .collect(Collectors.summarizingInt(Employee::getSalary));

如果你想将它们连接成字符串,你可以使用:

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

如果您确实想获得最高/最低工资的员工,这里

IntSummaryStatistics
不会帮助您。但是,您可以创建收集器流:

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,您将得到类似的输出(取决于

Employee.toString()
实现:

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,您可以使用如下方法创建组合收集器:

/**
 * 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)));
}

有了这样的方法就可以写了

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()));

1
投票

我终于找到了解决方案..感谢你们的尝试

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}

0
投票

如果我们只想查找所有工资的总和,可以使用以下解决方案。

Integer a= emp.stream().collect(Collectors.summingInt(Employe::getSalary));
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