我正在尝试在浮点值的输入上完成我的顺序代码的1D中值滤波器的并行版本的编码。我一直在使用一些较小长度的测试值来编写并行版本，最后似乎让代码工作，但是现在在+ -36000元素的大型数据集上，代码似乎只是增加CPU资源而不是完成而我的顺序版本在相同的给定输入上完成。有谁知道我哪里出错了？
测试输入有效：2,6,80,3,1并正确生成：2,6,6,3,1
// median filter code
public class MedianFilter extends RecursiveAction {
// filter class variables
float[] numbers;
int filter;
int window;
int length;
int lo;
int hi;
// sequential cutoff set to predetermined value
static final int SEQUENTIAL_CUTOFF = 500;
// array used for recursive calls in parallel code
float[] filtered;
float[] result;
public MedianFilter(float[] numbers, int filter, int lo, int hi) {
this.numbers = numbers;
this.filter = filter;
this.lo = lo;
this.hi = hi;
length = numbers.length;
// the section of floats to be filtered (median section)
window = (filter - 1) / 2;
}
protected void compute() {
filtered = new float[length];
result = new float[length];
if ((hi - lo) < SEQUENTIAL_CUTOFF) {
for (int a = lo; a < hi; a++) {
// iterate window through all elements of the array
for (int i = 0; i < length; i++) {
// fetch boundary elements
if (i < window || i >= length - window) {
result[i] = numbers[i];
}
// fetch elements within filter window
else {
for (int j = 0; j < filter; j++) {
filtered[j] = numbers[i - window + j];
}
// order elements
for (int j = 0; j < filtered.length / 2; j++) {
// get the position of the smallest float
int min = j;
for (int k = j + 1; k < filter; k++) {
if (filtered[k] < filtered[min]) {
min = k;
}
// reorder array for minimum element
float temp = filtered[j];
filtered[j] = filtered[min];
filtered[min] = temp;
}
// result
result[i] = filtered[window];
}
}
}
}
} else {
MedianFilter left = new MedianFilter(filtered, filter, lo, (hi + lo) / 2);
MedianFilter right = new MedianFilter(filtered, filter, (hi + lo) / 2, hi);
left.fork();
right.compute();
left.join();
}
}