将两个double 数组合并为double

我经常有两个数组，我需要将它们组合成一个矩阵（相同的长度和类型）。 我想知道是否有一种更优雅的linq方式：

var result = new double[dt.Count, 2]; for (int i = 0; i < dt.Count; i++) { result[i, 0] = dts[i]; result[i, 1] = dt[i]; } 

我试过了

 var result = dts.zip(dt, (a,b) => new{a,b}) 

和：

 var result = dts.Concat(dt).ToArray() 

但是我想做的也不行……

Answer 1

框架中没有任何内容，但如果您需要做很多事情，可以轻松地为此定义自己的类似LINq的扩展方法：

 public static class Ex { public static T[,] To2DArray(this T[] left, T[] right) { if (left == null) throw new ArgumentNullException("left"); if (right == null) throw new ArgumentNullException("right"); if (left.Length != right.Length) throw new ArgumentException("input arrays should have the same length"); var result = new T[left.Length, 2]; for (int i = 0; i < left.Length; i++) { result[i, 0] = right[i]; result[i, 1] = left[i]; } return result; } } 

使用如下：

 var combined = left.To2DArray(right); 

编辑：重新审视这个答案，我意识到有一个更通用的解决方案。 像这样的东西：

 public static class ArrayConvert { public static T[,] To2DArray(params T[][] arrays) { if (arrays == null) throw new ArgumentNullException("arrays"); foreach (var a in arrays) { if (a == null) throw new ArgumentException("can not contain null arrays"); if (a.Length != arrays[0].Length) throw new ArgumentException("input arrays should have the same length"); } var height = arrays.Length; var width = arrays[0].Length; var result = new T[width, height]; for (int i = 0; i < height; i++) for (int j = 0; j < width; j++) { result[i, j] = arrays[i][j]; } return result; } } 

然后可以使用以下内容：

 var convertedArray = ArrayConvert.To2DArray(new[]{1,2,3}, new[]{4,5,6}); 

Answer 2

好吧然后使用它

 class Program { static void Main(string[] args) { double[,] x = { { 1, 2, 3 }, { 4, 5, 6 } }; double[,] y = { { 7, 8, 9 }, { 10, 11, 12 } }; var xy = new StitchMatrix(x, y); Console.WriteLine("0,0=" + xy[0, 0]); // 1 Console.WriteLine("1,1=" + xy[1, 1]); // 5 Console.WriteLine("1,2=" + xy[1, 2]); // 6 Console.WriteLine("2,2=" + xy[2, 2]); // 9 Console.WriteLine("3,2=" + xy[3, 2]); // 12 } } class StitchMatrix { private T[][,] _matrices; private double[] _lengths; public StitchMatrix(params T[][,] matrices) { // TODO: check they're all same size _matrices = matrices; // call uperbound once for speed _lengths = _matrices.Select(m => m.GetUpperBound(0)).ToArray(); } public T this[double x, double y] { get { // find the right matrix double iMatrix = 0; while (_lengths[iMatrix] < x) { x -= (_lengths[iMatrix] + 1); iMatrix++; } // return value at cell return _matrices[iMatrix][x, y]; } } } 

Answer 3

这是另一种解决方案。 我“准备”LINQ处理的输入。 不确定这是优雅的，但它是LINQ：

  // the input double[] dts = { 1, 2, 3, 4, 5 }; double[] dt = { 10, 20, 30, 40, 50 }; // list of lists, for iterating the input with LINQ double[][] combined = { dts, dt }; var indexes = Enumerable.Range(0, dt.Length); var subIndexes = Enumerable.Range(0, 2); // the output var result = new double[dt.Length, 2]; var sss = from i in indexes from j in subIndexes select result[i, j] = combined[j][i]; // just to activate the LINQ iterator sss.ToList(); 

Answer 4

我建议不要直接在LINQ中这样做。 您可以编写一个通用方法来为您执行此操作，例如：

  public static T[,] To2DArray(this T[][] arr) { if (arr.Length == 0) { return new T[,]{}; } int standardLength = arr[0].Length; foreach (var x in arr) { if (x.Length != standardLength) { throw new ArgumentException("Arrays must have all the same length"); } } T[,] solution = new T[arr.Length, standardLength]; for (int i = 0; i < arr.Length; i++) { for (int j = 0; j < standardLength; j++) { solution[i, j] = arr[i][j]; } } return solution; } 

Answer 5

我知道这不是问题，但最优雅的答案是使用f＃：

 let combinearrays (arr1:array<'a>) (arr2:array<'a>) = let rws = arr1|> Array.length Array2D.init rws 2 (fun ij -> match j with |0 -> arr1.[i] |1 -> arr2.[i]) 

从约翰看到这里

Answer 6

这是解决方案在一个2Darrays中转换两个1darrays，我为自己开发。

 (from a1 in array1.Select((n,index)=>new{Index=index,c1=n}).ToList() join a2 in array2.Select((n,index)=>new {Index=index,c2=n}).ToList() on a1.Index equals a2.Index select new {c1,c2} ).ToArray()