如何处理多个 Enumerable.Zip 调用?

How to deal with multiple Enumerable.Zip calls?

我有几个带有传感器数据的枚举

time_elapsed、speed_x、speed_y、speed_z、海拔、纬度、经度……

每个列表都有相同数量的元素。

我想将所有列表的数据合并成一个状态项序列。

class Status 
{
    public int TimeElapsed {get; set; }
    public double SpeedX {get; set; }
    public double SpeedY {get; set; }
    public double SpeedZ {get; set; }
   ...
}

想过用Enumerable.Zip的方法,但是看起来真的很麻烦:

var statuses = time_elapsed
    .Zip(speed_x, (a, b) => new { a,  b})
    .Zip(speed_y, (c, d) => new { c,  d})
    .Zip(speed_z, (e, f) => new { e , f})
    .Select(x => new Status
    {
        Time = x.e.c.a,
        SpeedX = x.e.c.b,
        SpeedY = x.e.d,
        SpeedZ = x.f
        // ...
    });

如您所见,它来自所有这些匿名类型的可读性。

有没有更好的方法来做到这一点而不至于失去理智?

您在这里可以做的不多,但您可以提高可读性并删除一种匿名类型:

var statuses = time_elapsed
    .Zip(speed_x, (time, speedX) => new {time, speedX})
    .Zip(speed_y, (x, speedY) => new {x.time, x.speedX, speedY})
    .Zip(speed_z, (x, speedZ) => new Status
    {
        TimeElapsed = x.time,
        SpeedX = x.speedX,
        SpeedY = x.speedY,
        SpeedZ = speedZ
    });

您也可以使用这种方法:

int minSize = new IList[] {time_elapsed, speed_x, speed_y, speed_z}.Min(c => c.Count);
IEnumerable<Status> statuses = Enumerable.Range(0, minSize)
    .Select(i => new Status
    {
        TimeElapsed = time_elapsed[i],
        SpeedX = speed_x[i],
        SpeedY = speed_y[i],
        SpeedZ = speed_z[i],
    });

@Michael Liu 的 link 指出了 Jon Skeet 的一些代码,它为您的问题提供了一个干净的解决方案,它只是创建一个处理 3 个序列的自定义 zip:

    static IEnumerable<TResult> Zip<TFirst, TSecond, TThird, TResult>(
    IEnumerable<TFirst> first,
    IEnumerable<TSecond> second,
    IEnumerable<TThird> third,
    Func<TFirst, TSecond, TThird, TResult> resultSelector)
    {
        using (IEnumerator<TFirst> iterator1 = first.GetEnumerator())
        using (IEnumerator<TSecond> iterator2 = second.GetEnumerator())
        using (IEnumerator<TThird> iterator3 = third.GetEnumerator())
        {
            while (iterator1.MoveNext() && iterator2.MoveNext() && iterator3.MoveNext())
            {
                yield return resultSelector(iterator1.Current, iterator2.Current, iterator3.Current);
            }
        }
    }