问如何将2D字符串数组的结果除以子数组的长度？
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Stack Overflow用户

提问于 2021-08-15 22:57:03

回答 1查看 33关注 0票数 0

如何将二维数组的结果除以子数组的长度。在第三个问题上我需要帮助。Where要求输入缺少的代码行，以查找每个跑步者的平均时间。使用averageVal。

下面是问题“我们给了你一个名为averageVal的变量，它当前存储了0。编辑这行代码来找出每个跑步者的平均时间。”

public class RowMajor {
    public static void main(String[] args) {
        // Given runner lap data
        double[][] times = {{64.791, 75.972, 68.950, 79.039, 73.006, 74.157}, {67.768, 69.334, 70.450, 67.667, 75.686, 76.298}, {72.653, 77.649, 74.245, 62.121, 63.379, 79.354}};
        
        // Replace the incorrect for loop headers, use the iterators 'outer' and 'inner' for the outer and inner loops
        double runnerTime = 0.0;
        for(int outer = 0; outer < times.length; outer++) {
            runnerTime = 0.0;
            for(int inner = 0; inner < times[outer].length; inner++) {
                System.out.println("Runner index: " + outer + ", Time index: " + inner);
                // Enter the missing line of code to sum up the values in each row. Use the variable runnerTime
                runnerTime += times[outer][inner];
        
            }

            // I need help with this question below.
            // Enter the missing line of code to find the average time of each runner. Use the variable averageVal
            double averageVal = 0;
            // Below is what I tried but it keep showing an error "The correct average time for runner 0 was not printed to the console."
            double averageTime = runnerTime / times[outer].length;

            System.out.println("Sum of runner " + outer + " times: " + runnerTime + averageTime);
            System.out.println("Average of runner " + outer + ": " + averageVal + averageTime);
        }
    }
}

java

multidimensional-array

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回答 1

Stack Overflow用户

发布于 2021-08-16 01:22:15

我想这一行已经完成了任务

double averageTime = runnerTime / times[outer].length;

问题是当你用

System.out.println("Average of runner " + outer + ": " + averageVal + averageTime);

averageVal和averageTime没有加在一起(我不明白你为什么要加它们)。它们是附加的。

例如：

    int one = 1;
    int three = 3;
    System.out.println("" + one + three);

输出是13，而不是4。

原因是，当您一起打印一个基元时，当字符串，基元将是整数的auto boxed和语句变成"".toString() + one.toString() + three.toString()。因为toString()返回字符串，所以运算符+会将它们链接在一起。

我整理了您的代码，并发布了下面的输出

public class RowMajor {
    public static void main(String[] args) {
        // Given runner lap data
        double[][] times = { { 64.791, 75.972, 68.950, 79.039, 73.006, 74.157 },
                { 67.768, 69.334, 70.450, 67.667, 75.686, 76.298 },
                { 72.653, 77.649, 74.245, 62.121, 63.379, 79.354 } };

        // Replace the incorrect for loop headers, use the iterators 'outer' and 'inner'
        // for the outer and inner loops
        double runnerTime = 0.0;
        for (int outer = 0; outer < times.length; outer++) {
            runnerTime = 0.0;
            for (int inner = 0; inner < times[outer].length; inner++) {
                System.out.println("Runner index: " + outer + ", Time index: " + inner);
                // Enter the missing line of code to sum up the values in each row. Use the
                // variable runnerTime
                runnerTime += times[outer][inner];

            }

            // I need help with this question below.
            // Enter the missing line of code to find the average time of each runner. Use
            // the variable averageVal
            double averageVal = runnerTime / times[outer].length;

            System.out.println("Sum of runner " + outer + " times: " + runnerTime);
            System.out.println("Average of runner " + outer + ": " + averageVal);

        }
    }
}

控制台输出

Runner index: 0, Time index: 0
Runner index: 0, Time index: 1
Runner index: 0, Time index: 2
Runner index: 0, Time index: 3
Runner index: 0, Time index: 4
Runner index: 0, Time index: 5
Sum of runner 0 times: 435.9149999999999
Average of runner 0: 72.65249999999999
Runner index: 1, Time index: 0
Runner index: 1, Time index: 1
Runner index: 1, Time index: 2
Runner index: 1, Time index: 3
Runner index: 1, Time index: 4
Runner index: 1, Time index: 5
Sum of runner 1 times: 427.2030000000001
Average of runner 1: 71.20050000000002
Runner index: 2, Time index: 0
Runner index: 2, Time index: 1
Runner index: 2, Time index: 2
Runner index: 2, Time index: 3
Runner index: 2, Time index: 4
Runner index: 2, Time index: 5
Sum of runner 2 times: 429.401
Average of runner 2: 71.56683333333334