如何在 C# 中按行对锯齿状数组进行排序? [英] How do I sort jagged array by row in C#?

问题描述

我有二维锯齿状数组.我想按任何行对其进行排序.

I have 2D jagged array. And I want to sort it by any rows.

我搜索并找到了按列排序的代码

I've searched and found code for sorting by columns

private static void Sort<T>(T[][] data, int col) 
{ 
    Comparer<T> comparer = Comparer<T>.Default;
    Array.Sort<T[]>(data, (x,y) => comparer.Compare(x[col],y[col])); 
}

我可以调整它以按任何行排序吗?

Can I adapt it for sort by any rows ?

感谢任何帮助.

我的锯齿状数组示例(已添加)

Sample of my jagged array (Added)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication3
{
    class Program
    {
        static void Main(string[] args)
        {
            int n = 10;
            int[][] capm = new int[3][];
            for (int i = 0; i <= 2; i++)
            {
                capm[i] = new int[n + 1];
            }
            Random rand = new Random();            
            for (int i = 1; i <= n; i++)
            {
                capm[1][i] = i;
            }

            for (int i = 1; i <= n; i++)
            {
                capm[2][i] = rand.Next(1, 6);
            }

            Sort(capm, 2);

            Console.ReadLine();
        }
            private static void Sort<T>(T[][] data, int col)    
            {  
                data = data.OrderBy(i => i[col]).ToArray();
            }
        }

    }

@Dani &@Martin 我希望我的锯齿状数组按 capm[2][] 排序.

@Dani & @Martin I want my jagged array to sort by capm[2][].

推荐答案

我能想到的唯一方法是按索引数组排序:

The only way I can think of doing this is sorting by an array of indices:

private static void Sort<T>(T[][] data, int row) 
{
    int[] Indices = new int[data[0].Length];
    for(int i = 0; i < Indices.Length; i++)
        Indices[i] = i;

    Comparer<T> comparer = Comparer<T>.Default;
    Array.Sort(Indices, (x, y) => comparer.Compare(data[row][x], data[row][y]);

    for(int i = 0; i < data.Length; i++)
    {
        T[] OldRow = (T[])data[i].Clone();
        for(int j = 0; j < OldRow.Length; j++)
            data[i][j] = OldRow[i][Indices[j]];
    }
}

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