我有一种方法可以计算时间序列的移动中值。与移动平均线一样，它使用固定的窗口或周期（有时称为回溯期）。如果周期为 10，它将创建一个包含前 10 个值 (0-9) 的数组，然后找到它们的中值。它将重复此操作，将窗口增加 1 步（现在的值为 1-10）等等......因此是移动的部分。这个过程与移动平均线完全相同。
中值通过以下方式找到：
- 对数组的值进行排序
- 如果数组中有奇数个值，则取中间值。5 个值的排序数组的中位数将是第三个值。
- 如果数组中有偶数个值，则取中间值每一侧的两个值并将它们平均。6 个值的排序数组的中位数是 (2nd + 3rd) / 2。
我创建了一个函数，它通过填充 a List<double>、调用List<>.Sort()，然后找到适当的值来计算它。
计算上是正确的，但我想知道是否有办法提高这种计算的性能。也许通过手动滚动排序double[]而不是使用列表。
我的实现如下：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Moving_Median_TimeSeries
{
class Program
{
static void Main(string[] args)
{
// created a a sample test time series of 10 days
DateTime Today = DateTime.Now;
var TimeSeries = new SortedList<DateTime, double>();
for (int i = 0; i < 10; i++)
TimeSeries.Add(Today.AddDays(i), i * 10);
// write out the time series
Console.WriteLine("Our time series contains...");
foreach (var item in TimeSeries)
Console.WriteLine(" {0}, {1}", item.Key.ToShortDateString(), item.Value);
// calculate an even period moving median
int period = 6;
var TimeSeries_MovingMedian = MovingMedian(TimeSeries, period);
// write out the result of the calculation
Console.WriteLine("\nThe moving median time series of {0} periods contains...", period);
foreach (var item in TimeSeries_MovingMedian)
Console.WriteLine(" {0}, {1}", item.Key.ToShortDateString(), item.Value);
// calculate an odd period moving median
int period2 = 5;
var TimeSeries_MovingMedian2 = MovingMedian(TimeSeries, period);
// write out the result of the calculation
Console.WriteLine("\nThe moving median time series of {0} periods contains...", period2);
foreach (var item in TimeSeries_MovingMedian2)
Console.WriteLine(" {0}, {1}", item.Key.ToShortDateString(), item.Value);
}
public static SortedList<DateTime, double> MovingMedian(SortedList<DateTime, double> TimeSeries, int period)
{
var result = new SortedList<DateTime, double>();
for (int i = 0; i < TimeSeries.Count(); i++)
{
if (i >= period - 1)
{
// add all of the values used in the calc to a list...
var values = new List<double>();
for (int x = i; x > i - period; x--)
values.Add(TimeSeries.Values[x]);
// ... and then sort the list <- there might be a better way than this
values.Sort();
// If there is an even number of values in the array (example 10 values), take the two mid values
// and average them. i.e. 10 values = (5th value + 6th value) / 2.
double median;
if (period % 2 == 0) // is any even number
median = (values[(int)(period / 2)] + values[(int)(period / 2 - 1)]) / 2;
else // is an odd period
// Median equals the middle value of the sorted array, if there is an odd number of values in the array
median = values[(int)(period / 2 + 0.5)];
result.Add(TimeSeries.Keys[i], median);
}
}
return result;
}
}
}