我需要找到一个双精度数组的中值(在 Java 中),而不需要修改它(所以选择不可用)或分配大量新内存。我也不关心找到确切的中位数,但在 10% 以内是可以的(所以如果中位数将排序数组拆分为 40%-60% 就可以了)。
我怎样才能有效地实现这一目标?
考虑到 rfreak、ILMTitan 和 Peter 的建议,我编写了以下代码:
public static double median(double[] array) {
final int smallArraySize = 5000;
final int bigArraySize = 100000;
if (array.length < smallArraySize + 2) { // small size, so can just sort
double[] arr = array.clone();
Arrays.sort(arr);
return arr[arr.length / 2];
} else if (array.length > bigArraySize) { // large size, don't want to make passes
double[] arr = new double[smallArraySize + 1];
int factor = array.length / arr.length;
for (int i = 0; i < arr.length; i++)
arr[i] = array[i * factor];
return median(arr);
} else { // average size, can sacrifice time for accuracy
final int buckets = 1000;
final double desiredPrecision = .005; // in percent
final int maxNumberOfPasses = 10;
int[] histogram = new int[buckets + 1];
int acceptableMin, acceptableMax;
double min, max, range, scale,
medianMin = -Double.MAX_VALUE, medianMax = Double.MAX_VALUE;
int sum, numbers, bin, neighborhood = (int) (array.length * 2 * desiredPrecision);
for (int r = 0; r < maxNumberOfPasses; r ++) { // enter search for number around median
max = -Double.MAX_VALUE; min = Double.MAX_VALUE;
numbers = 0;
for (int i = 0; i < array.length; i ++)
if (array[i] > medianMin && array[i] < medianMax) {
if (array[i] > max) max = array[i];
if (array[i] < min) min = array[i];
numbers ++;
}
if (min == max) return min;
if (numbers <= neighborhood) return (medianMin + medianMax) / 2;
acceptableMin = (int) (numbers * (50d - desiredPrecision) / 100);
acceptableMax = (int) (numbers * (50d + desiredPrecision) / 100);
range = max - min;
scale = range / buckets;
for (int i = 0; i < array.length; i ++)
histogram[(int) ((array[i] - min) / scale)] ++;
sum = 0;
for (bin = 0; bin <= buckets; bin ++) {
sum += histogram[bin];
if (sum > acceptableMin && sum < acceptableMax)
return ((.5d + bin) * scale) + min;
if (sum > acceptableMax) break; // one bin has too many values
}
medianMin = ((bin - 1) * scale) + min;
medianMax = (bin * scale) + min;
for (int i = 0; i < histogram.length; i ++)
histogram[i] = 0;
}
return .5d * medianMin + .5d * medianMax;
}
}
这里我考虑到数组的大小。如果它很小,那么只需排序并获得真正的中位数。如果它非常大,则对其进行采样并获取样本的中值,否则迭代地对值进行分箱并查看中值是否可以缩小到可接受的范围。
我对这段代码没有任何问题。如果有人看到它有问题,请告诉我。
谢谢你。