TimeSpan 和双舍入误差

发布于 2024-09-11 10:49:10 字数 1253 浏览 11 评论 0原文

我正在处理物理实体，例如时间、速度和距离，并执行简单的数学运算，例如距离 = 时间 * 速度等。速度和距离是四舍五入到第 8 位数字的双精度值，对于时间值，是 .NET TimeSpan被使用。由于 TimeSpan 四舍五入到最接近的毫秒，我遇到了舍入错误，因此我需要编写自定义舍入方法，将所有计算四舍五入到最接近的毫秒。例如（为简单起见，舍入到第 8 位数字被省略）：

  static void Main(string[] args) {
     var dist = 1.123451;
     var speed = 1.123452;

     var timeA = TimeSpan.FromHours(dist / speed);
     var timeB = timeA + TimeSpan.FromMilliseconds(1);

     var distA = _round(timeA.TotalHours * speed);
     var distB = _round(timeB.TotalHours * speed);

     var timeA1 = TimeSpan.FromHours(distA / speed);
     var timeB1 = TimeSpan.FromHours(distB / speed);

     // a correct implementation should give both the following vars true
     var isDistributive = distA == dist;
     var isPrecise = (timeB1 - timeA1) == TimeSpan.FromMilliseconds(1);
  }

  public static double _round(double d) {
     // Q: what should be here?
  }

使用 Math.Round(d, 6) 是分配性的，但会损失精度（精确到约 4 毫秒）
使用 Math.Round(d, 7) 精确到单个毫秒，但不是分配性的（上面的 distA 将是 1.1234511 ！= 1.123451）
使用以下内容（舍入到最接近的毫秒）似乎是正确的，但舍入代码本身会引入其自己的双精度错误：
```
 public static double _round(double d) {
    var pre = 3600000.0;
    返回 Math.Round(d * pre) / pre;
  }
```

谢谢，鲍里斯。

原文

I'm dealing with physical entities, such as time, speed and distance, and performing simple math, such as distance = time * speed, etc. Speed and Distance are double values rounded to 8th digit, and for Time values a .NET TimeSpan is used. Since TimeSpan rounds to a closest millisecond, I'm getting rounding errors, so I need to write custom rounding method that rounds all calculations to the closest millisecond. For example (rounding to 8th digit is omitted for simplicity):

  static void Main(string[] args) {
     var dist = 1.123451;
     var speed = 1.123452;

     var timeA = TimeSpan.FromHours(dist / speed);
     var timeB = timeA + TimeSpan.FromMilliseconds(1);

     var distA = _round(timeA.TotalHours * speed);
     var distB = _round(timeB.TotalHours * speed);

     var timeA1 = TimeSpan.FromHours(distA / speed);
     var timeB1 = TimeSpan.FromHours(distB / speed);

     // a correct implementation should give both the following vars true
     var isDistributive = distA == dist;
     var isPrecise = (timeB1 - timeA1) == TimeSpan.FromMilliseconds(1);
  }

  public static double _round(double d) {
     // Q: what should be here?
  }

Using Math.Round(d, 6) is distributive, but loses precision (precise up to ~4 msec)
Using Math.Round(d, 7) is precise to a single msec, but not distributive (distA above will be 1.1234511 != 1.123451)
Using the following (round to closest msec) seems to be correct, but the rounding code itself introduces its own double precision errors:
```
  public static double _round(double d) {
    var pre = 3600000.0;
    return Math.Round(d * pre) / pre;
  }
```

Thanks,
Boris.

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