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中查找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}

希望能帮助到你

阅读有关IntSummaryStatistics的信息

Answer 2

我认为知道一个以上的问题解决方案来选择最适合该问题的解决方案总是很高兴的。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

// 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);

但是要尽可能地谨慎，所以我的最爱是：

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

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

顺便说一下，Oracle教程是一个金矿：https : //docs.oracle.com/javase/tutorial/collections/streams/reduction.html