在 Java 8 中查找列表的最大值、最小值、總和和平均值

Question

如何在Java 8中找到以下列表中數字的最大值，最小值，總和和平均值？

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

Answer 1

有一個類名， 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);

輸出：

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

希望能幫助到你

閱讀 IntSummaryStatistics

Answer 2

我認為知道一個問題的多個解決方案來選擇最適合問題的解決方案總是好的。 以下是一些其他解決方案：

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);

但要盡可能簡潔，所以我最喜歡的是：

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

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

順便說一句，Oracle 教程是一個金礦： 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));