为了确保线程具有“相似”的工作负载,找到均匀的分布很重要。当线程数量与元素数量相比“高”时,这一点尤为重要。对于这种情况,应该确保线程负责的元素数相差至多1。
为了达到这个目的,你可以计算除以元素数量(在你的情况下数组长度)除以线程数量的余数,并在任务中逐个分配这个余数。
前段时间我有同样的问题。实际上,我试图以稍微更一般的形式解决它,对于某些类需要计算开始和结束任意范围的间隔的指数(其不需要以索引0
)。下面从这个类是“提取”:
import java.util.Arrays;
public class EvenTaskDistribution
{
public static void main(String[] args)
{
test(22, 4);
test(21, 4);
test(100, 3);
test( 3, 4);
}
private static void test(int numElements, int parallelism)
{
int taskSizes[] = computeTaskSizes(parallelism, 0, numElements);
System.out.printf("Distributing %4d elements among %4d threads: %s\n",
numElements, parallelism, Arrays.toString(taskSizes));
}
public static int[] computeTaskSizes(
int parallelism, int globalMin, int globalMax)
{
if (parallelism <= 0)
{
throw new IllegalArgumentException(
"Parallelism must be positive, but is " + parallelism);
}
if (globalMin > globalMax)
{
throw new IllegalArgumentException(
"The global minimum may not be larger than the global " +
"maximum. Global minimum is "+globalMin+", " +
"global maximum is "+globalMax);
}
int range = globalMax - globalMin;
if (range == 0)
{
return new int[0];
}
int numTasks = Math.min(range, parallelism);
int localRange = (range - 1)/numTasks + 1;
int spare = localRange * numTasks - range;
int currentIndex = globalMin;
int taskSizes[] = new int[numTasks];
for (int i = 0; i < numTasks; i++)
{
final int min = currentIndex;
final int max = min + localRange - (i < spare ? 1 : 0);
taskSizes[i] = max - min;
currentIndex = max;
}
return taskSizes;
}
}
输出是
Distributing 22 elements among 4 threads: [5, 5, 6, 6]
Distributing 21 elements among 4 threads: [5, 5, 5, 6]
Distributing 100 elements among 3 threads: [33, 33, 34]
Distributing 3 elements among 4 threads: [1, 1, 1]
(最后一个显示的极端案例一个一个可能要考虑到例如,一个可能。期望[1,1,1,0]
,但这可以根据应用情况轻松调整)。
事实上,你这样做是为了在线程之间分割工作,这在很大程度上是不相关的 - 你似乎在问如何将一个数组分割成N个大致相同大小的块。 –