使用LINQ将多个列表合并到一个列表中

Question

有没有一种灵活的方法可以使用LINQ将多个列表合并到一个列表中以有效地复制它？

public class RGB
{
    public int Red { get; set; }
    public int Green { get; set; }
    public int Blue { get; set; }
    public RGB(int red, int green, int blue) { Red = red; Green = green; Blue = blue; }
}

public void myFunction()
{
    List<int> red = new List<int> { 0x00, 0x03, 0x06, 0x08, 0x09 };
    List<int> green = new List<int> { 0x00, 0x05, 0x06, 0x07, 0x0a };
    List<int> blue = new List<int> { 0x00, 0x02, 0x03, 0x05, 0x09 };

    List<RGB> colors = new List<RGB>();

    colors.Add(new RGB(red[0], green[0], blue[0]));
    colors.Add(new RGB(red[1], green[1], blue[1]));
    colors.Add(new RGB(red[2], green[2], blue[2]));
    colors.Add(new RGB(red[3], green[3], blue[3]));
    colors.Add(new RGB(red[4], green[4], blue[4]));
}

或者,由于列表单独到达,因此按顺序合并它们更有效,如下所示.

public class RGB
{
    public int Red { get; set; }
    public int Green { get; set; }
    public int Blue { get; set; }

    public RGB(int red, int green, int blue) { Red = red; Green = green; Blue = blue; }
}

public void myFunction()
{
    List<int> red = new List<int> { 0x00, 0x03, 0x06, 0x08, 0x09 };

    List<RGB> colors = new List<RGB>();

    colors.Add(new RGB(red[0], 0, 0));
    colors.Add(new RGB(red[1], 0, 0));
    colors.Add(new RGB(red[2], 0, 0));
    colors.Add(new RGB(red[3], 0, 0));
    colors.Add(new RGB(red[4], 0, 0));

    List<int> green = new List<int> { 0x00, 0x05, 0x06, 0x07, 0x0a };

    colors[0].Green = green[0];
    colors[1].Green = green[1];
    colors[2].Green = green[2];
    colors[3].Green = green[3];
    colors[4].Green = green[4];

    List<int> blue = new List<int> { 0x00, 0x02, 0x03, 0x05, 0x09 };

    colors[0].Blue = blue[0];
    colors[1].Blue = blue[1];
    colors[2].Blue = blue[2];
    colors[3].Blue = blue[3];
    colors[4].Blue = blue[4];
}

Answer 1

你基本上试图压缩三个集合.如果只Zip()支持LINQ 方法同时支持两个以上.但唉,它一次只支持两个.但我们可以使它工作:

var reds = new List<int> { 0x00, 0x03, 0x06, 0x08, 0x09 };
var greens = new List<int> { 0x00, 0x05, 0x06, 0x07, 0x0a };
var blues = new List<int> { 0x00, 0x02, 0x03, 0x05, 0x09 };

var colors =
    reds.Zip(greens.Zip(blues, Tuple.Create),
        (red, tuple) => new RGB(red, tuple.Item1, tuple.Item2)
    )
    .ToList();

当然,编写扩展方法来做三个(或更多)并不是非常痛苦.

public static IEnumerable<TResult> Zip<TFirst, TSecond, TThird, TResult>(
    this IEnumerable<TFirst> first,
    IEnumerable<TSecond> second,
    IEnumerable<TThird> third,
    Func<TFirst, TSecond, TThird, TResult> resultSelector)
{
    using (var enum1 = first.GetEnumerator())
    using (var enum2 = second.GetEnumerator())
    using (var enum3 = third.GetEnumerator())
    {
        while (enum1.MoveNext() && enum2.MoveNext() && enum3.MoveNext())
        {
            yield return resultSelector(
                enum1.Current,
                enum2.Current,
                enum3.Current);
        }
    }
}

这使事情变得更好:

var colors =
    reds.Zip(greens, blues,
        (red, green, blue) => new RGB(red, green, blue)
    )
    .ToList();

Answer 2

是的 - 你可以这样做:

List<int> red = new List<int> { 0x00, 0x03, 0x06, 0x08, 0x09 };
List<int> green = new List<int> { 0x00, 0x05, 0x06, 0x07, 0x0a };
List<int> blue = new List<int> { 0x00, 0x02, 0x03, 0x05, 0x09 };

List<RGB> colors = Enumerable
    .Range(0, red.Count)
    .Select(i => new RGB(red[i], green[i], blue[i]))
    .ToList();

Answer 3

像这样使用 SelectMany：

List_A.Select(a => a.List_B).SelectMany(s => s).ToList();