在 Java 8 中查找列表的最大值、最小值、总和和平均值

Question

如何在Java 8中求下列列表中数字的最大值、最小值、总和和平均值？

List<Integer> primes = Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19, 23, 29);

Answer 1

有一个类名，IntSummaryStatistics

例如：

List<Integer> primes = Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19, 23, 29);
IntSummaryStatistics stats = primes.stream()
                                     .mapToInt((x) -> x)
                                     .summaryStatistics();
System.out.println(stats);

输出：

IntSummaryStatistics{count=10, sum=129, min=2, average=12.900000, max=29}

希望对你有帮助

Read about IntSummaryStatistics

Answer 2

我认为了解一个问题的多个解决方案以选择最适合该问题的解决方案总是很好的。以下是一些其他解决方案：

final List<Integer> primes = Arrays.asList(2, 3, 5, 7, 11, 13, 17, 19, 23, 29);

找到最大值

// MAX -- Solution 1
primes.stream() //
    .max(Comparator.comparing(i -> i)) //
    .ifPresent(max -> System.out.println("Maximum found is " + max));

// MAX -- Solution 2
primes.stream() //
    .max((i1, i2) -> Integer.compare(i1, i2)) //
    .ifPresent(max -> System.out.println("Maximum found is " + max));

// MAX -- Solution 3
int max = Integer.MIN_VALUE;
for (int i : primes) {
    max = (i > max) ? i : max;
}
if (max == Integer.MIN_VALUE) {
    System.out.println("No result found");
} else {
    System.out.println("Maximum found is " + max);
}

// MAX -- Solution 4 
max = Collections.max(primes);
System.out.println("Maximum found is " + max);

求最小值

// MIN -- Solution 1
primes.stream() //
    .min(Comparator.comparing(i -> i)) //
    .ifPresent(min -> System.out.println("Minimum found is " + min));

// MIN -- Solution 2
primes.stream() //
    .max(Comparator.comparing(i -> -i)) //
    .ifPresent(min -> System.out.println("Minimum found is " + min));

// MIN -- Solution 3
int min = Integer.MAX_VALUE;
for (int i : primes) {
    min = (i < min) ? i : min;
}
if (min == Integer.MAX_VALUE) {
    System.out.println("No result found");
} else {
    System.out.println("Minimum found is " + min);
}

// MIN -- Solution 4
min = Collections.min(primes);
System.out.println("Minimum found is " + min);

求平均值

// AVERAGE -- Solution 1
primes.stream() //
    .mapToInt(i -> i) //
    .average() //
    .ifPresent(avg -> System.out.println("Average found is " + avg));

// AVERAGE -- Solution 2
int sum = 0;
for (int i : primes) {
    sum+=i;
}
if(primes.isEmpty()){
    System.out.println("List is empty");
} else {
    System.out.println("Average found is " + sum/(float)primes.size());         
}

求和

// SUM -- Solution 1
int sum1 = primes.stream() //
    .mapToInt(i -> i) //
    .sum(); //
System.out.println("Sum found is " + sum1);

// SUM -- Solution 2
int sum2 = 0;
for (int i : primes) {
    sum2+=i;
}
System.out.println("Sum found is " + sum2);

但要尽可能简洁，所以我最喜欢的是：

// Find a maximum with java.Collections
Collections.max(primes);

// Find a minimum with java.Collections 
Collections.min(primes);

顺便说一下，Oracle 教程是个金矿：https://docs.oracle.com/javase/tutorial/collections/streams/reduction.html

Answer 3

    //By using lambda
    int sum = primes.stream().mapToInt(a->a).sum();
    System.out.println(sum);
    int min = primes.stream().mapToInt(a->a).min().orElse(0);
    System.out.println(min);
    int max = primes.stream().mapToInt(a->a).max().orElse(0);
    System.out.println(max);
    double average = primes.stream().mapToInt(a->a).average().orElse(0);
    System.out.println(average);
    
    //By using Collections
    System.out.println(Collections.min(primes));
    System.out.println(Collections.max(primes));