Java 计算平均执行时间

Question

我想计算 x 次运行的平均执行时间(即 10)...我可以使用 main 方法中的循环轻松执行 10 次，但如何存储执行时间并计算平均值？我认为这非常简单，但我现在一片空白......提前致谢!

import java.util.Arrays;
import java.util.Random;

public class OptQSort1 {

static boolean insertionSortCalled = false;
private static final Random random = new Random();
private static final int RANDOM_INT_RANGE = 9999;

private static int[] randomArray(int size) {

    // Randomize data (array)
    final int[] arr = new int[size];
    for (int i = 0; i < arr.length; i++) {
        arr[i] = random.nextInt(RANDOM_INT_RANGE);
    }
    return arr;
}

// Sort
private static void sort(int[] arr) {
    if (arr.length > 0)
        sortInPlace(arr, 0, arr.length - 1);
}

private static void sortInPlace(int[] arr, int left, int right) {



    // OptQSort1:
    int size = right - left + 1;
    if (size < 10 && !insertionSortCalled) {
        insertionSortCalled = true;
        insertionSort(arr, 0, arr.length - 1);
    }

    if (left >= right)
        return; // sorted

    final int range = right - left + 1;
    int pivot = random.nextInt(range) + left;

    int newPivot = partition(arr, left, right, pivot);

    sortInPlace(arr, left, newPivot - 1);
    sortInPlace(arr, newPivot + 1, right);

}

private static int partition(int[] arr, int left, int right, int pivot) {

    int pivotVal = arr[pivot];
    swapArrayVals(arr, pivot, right);

    int storeIndex = left;
    for (int i = left; i <= (right - 1); i++) {
        if (arr[i] < pivotVal) {
            swapArrayVals(arr, i, storeIndex);
            storeIndex++;
        }
    }

    swapArrayVals(arr, storeIndex, right);

    return storeIndex;
}

private static void swapArrayVals(int[] arr, int from, int to) {
    int fromVal = arr[from];
    int toVal = arr[to];
    arr[from] = toVal;
    arr[to] = fromVal;
}

public static void insertionSort(int[] arr, int left, int right) {
    int in, out;

    for (out = left + 1; out <= right; out++) {
        int temp = arr[out];
        in = out;

        while (in > left && arr[in - 1] >= temp) {
            arr[in] = arr[in - 1];
            --in;
        }
        arr[in] = temp;
    }

}

public static void main(String[] args) {

    long StartTime = System.nanoTime();

    int runCount = 0;

    // Array size
    int[] arr = randomArray(1000);
    int[] copy = Arrays.copyOf(arr, arr.length);

    // Print original data (array)
    System.out.println("The starting/unsorted array: \n"
            + Arrays.toString(arr));

    sort(arr);

    do {
    // check the result
    Arrays.sort(copy);
    if (Arrays.equals(arr, copy)) {
        System.out.println("The ending/sorted array: \n"
                + Arrays.toString(arr));


        // print time
        long TotalTime = System.nanoTime() - StartTime;
        System.out.println("Total elapsed time (milliseconds) " + "is: "
                + TotalTime + "\n");

        runCount++;
    } 
    }   while (runCount < 10);
}

} 

Answer 1

您可以通过测量代码 10 次迭代的总时间，然后将其除以 10 来计算平均值。

例如:

public static void main(String[] args) {
    long start = System.currentTimeMillis();
    for (int i = 0; i < 10; ++i) {
        doSort();
    }
    long elapsed = System.currentTimeMillis() - start;
    long average = elapsed / 10;
}

作为一个有用的提示，请使用命名常量而不是文字值来表示迭代次数:

private final static int ITERATIONS = 10;
public static void main(String[] args) {
    long start = System.currentTimeMillis();
    for (int i = 0; i < ITERATIONS; ++i) {
        doSort();
    }
    long elapsed = System.currentTimeMillis() - start;
    long average = elapsed / ITERATIONS;
}

这意味着如果您想运行 50 或 100 次迭代，则只需更改一处的数字。

您还应该意识到，从此类实验中获得准确的计时结果是非常困难的。最好包含一个“预热”阶段，以允许 JIT 评估和优化代码，并进行更多的迭代:

private static final int WARMUP_ITERATIONS = 10000;
private static final int RUN_ITERATIONS = 100000;
public static void main(String[] args) {
    // Warmup with no timing
    for (int i = 0; i < WARMUP_ITERATIONS; ++i) {
        doSort();
    }

    // Now the real test
    long start = System.currentTimeMillis();
    for (int i = 0; i < RUN_ITERATIONS; ++i) {
        doSort();
    }
    long elapsed = System.currentTimeMillis() - start;
    long average = elapsed / RUN_ITERATIONS;
}