Question

如何在Java 8中找到以下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

有一个班级名称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}

希望有所帮助

Read about IntSummaryStatistics

Answer 2

我认为知道一个以上的问题解决方案来挑选最适合问题的权利总是好的。以下是其他一些解决方案：

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);

找到最低

// 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);

查找平均值

// 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());         
}

查找总和

// 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);

但请尽量做到最好，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 Tutorial是一个金矿：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));

在Java 8中查找列表的最大值，最小值，总和和平均值

3 个答案: