如何找到總在二維陣列的每一列的

Question

注意：列數組和總應出現如下圖所示（數字不應 相同如下圖所示，數應該是隨機的）

5 1 1 1 1 1 1 1 1 1 
2 2 4 5 6 7 8 8 9 9 
1 1 1 1 1 1 1 1 1 1 
2 2 2 2 2 2 2 2 2 2 
3 3 3 3 3 3 3 3 3 3 
5 5 5 5 5 5 5 5 5 5 
4 4 4 4 4 4 4 4 4 4 
7 7 7 7 7 7 7 7 7 7 
3 3 3 3 3 3 3 3 3 3 
1 1 1 1 1 1 1 11 1 1 

每列的總

33 29 31 32 33 34 35 45 36 36 

`進入coddouble [，] A =新的雙[10，10]; 隨機x =新隨機（）; string s =「」;

 for (int r = 0; r < 10; r++) 
     { 
      for (int c = 0; c < 10; c++) 
      { 
       a[r, c] = x.Next(1, 21); 
       s = s + a[r, c].ToString() + "\t"; 


      } 
     } 
     textBox1.Text = s;e here` 

這是如何打印ARRY現在如何找到每一列

Answer 1

這是數組實現非常基本的總和。

有很多種方法，通過使用簡單的循環像這樣在陣列結構迭代實現this.One我的首選方式。

static int array[10][10]; 
//or like 
int[,] array = new int[,] 
{ 
    <declare array structure> 
}; 
int i, j, rowlength=10, columnlength=10, sum = 0; //assuming your order 10 as given example .To make it dynamic use array.Length 

sum = 0; 
for (j = 0; j < columnlength; ++j) 
{ 
    for (i = 0; i < rowlength; ++i) 
    { 
     sum = sum + array[i][j]; 
    } 
    Console.WriteLine(String.Format("Summation of {0}th column is {1}", j,sum)); 
    sum=0;  
} 

Answer 2

var arr2d = new int[5]; 
arr2d[0] = {1,2,3,4,5}; 
// etc... 

var arr = new int[arr2d.length]; 

for(int i=0; i<arr2d.length; i++) { 
    var item = arr2d[i]; 
    for(int j=0; j<item.length; j++) { 
     arr[i] += item[j]; 
    } 
} 

Answer 3

你可以試試下面的代碼。它的通用代碼，可以在任意數量的列上工作，不需要硬編碼列號。 columnSum列表包含個別列的總和

var array2D = Get2DArray(); //some function to populate, you can have your own way to populate your 2D array 

List<int> columnSum = new List<int>(); 

for (int j = 0; j < array2D.length ; ++j) 
{ 
    int sum = 0; 
    for (int i = 0; i < array2D[j].length; ++i) 
    { 
     sum = sum + array2D[i][j]; 
    } 

    columnSum.Add(sum); // columnSum List contain the sum of individual column 
} 

Answer 4

嘗試使用您的代碼。

我所做的就是和你添加到數組列。這是假設c是列，r是行。

double[,] a = new double[10, 10]; 
Random x = new Random(); 
string s = ""; 
var colsum = new List<int>(); 
for (int r = 0; r < 10; r++) 
{ 
    var col = 0; 
    for (int c = 0; c < 10; c++) 
    { 
     col += x.Next(1, 21); 
     a[r, c] = col; 
     s = s + a[r, c].ToString() + "\t"; 
    } 
    colsum.Add(col); 
} 
foreach(var col in colsum){ 
    Console.WriteLine(colsum.ToString()); 
} 

Answer 5

試試這個

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

namespace ConsoleApplication1 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 

      int[,] data = new int[,] { 
       {5, 1, 1, 1, 1, 1, 1, 1, 1, 1 }, 
       {2 ,2, 4, 5, 6, 7, 8, 8, 9, 9}, 
       {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, 
       {2, 2, 2, 2, 2, 2, 2, 2, 2, 2}, 
       {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}, 
       {5, 5, 5, 5, 5, 5, 5, 5, 5, 5}, 
       {4, 4, 4, 4, 4, 4, 4, 4, 4, 4}, 
       {7, 7, 7, 7, 7, 7, 7, 7, 7, 7}, 
       {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}, 
       {1, 1, 1, 1, 1, 1, 1, 11, 1, 1} 
      }; 
      int[] sum = new int[10]; 

      for (int row = 0; row < 10; row++) 
      { 
       for (int col = 0; col < 10; col++) 
       { 
        if (row == 0) 
        { 
         sum[col] = data[row, col]; 
        } 
        else 
        { 
         sum[col] += data[row, col]; 
        } 
       } 
      } 

      List<List<int>> data2 = new List<List<int>>() { 
       new List<int>() {5, 1, 1, 1, 1, 1, 1, 1, 1, 1 }, 
       new List<int>() {2 ,2, 4, 5, 6, 7, 8, 8, 9, 9}, 
       new List<int>() {1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, 
       new List<int>() {2, 2, 2, 2, 2, 2, 2, 2, 2, 2}, 
       new List<int>() {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}, 
       new List<int>() {5, 5, 5, 5, 5, 5, 5, 5, 5, 5}, 
       new List<int>() {4, 4, 4, 4, 4, 4, 4, 4, 4, 4}, 
       new List<int>() {7, 7, 7, 7, 7, 7, 7, 7, 7, 7}, 
       new List<int>() {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}, 
       new List<int>() {1, 1, 1, 1, 1, 1, 1, 11, 1, 1} 
      }; 

      //using linq 
      int[] sum2 = data2.Select(s => s.Select((t, i) => new { num = t, col = i })).SelectMany(u => u).GroupBy(v => v.col).ToList().Select(w => w.Select(x => x.num).Sum()).ToArray(); 

     } 
    } 
} 
​ 

Answer 6

這很簡單 - 只需2套for環路喜歡與你這段代碼

static double[] SumColumns(double[,] source) 
{ 
    int rowCount = source.GetLength(0); 
    int colCount = source.GetLength(1); 
    var colSums = new double[colCount]; 
    for (int row = 0; row < rowCount; row++) 
     for (int col = 0; col < colCount; col++) 
      colSums[col] += source[row, col]; 
    return colSums; 
} 

使用範例：

int rows = 10, cols = 10; 
var a = new double[rows, cols]; 
var x = new Random(); 
var sb = new StringBuilder(); 
for (int r = 0; r < rows; r++) 
{ 
    for (int c = 0; c < cols; c++) 
    { 
     a[r, c] = x.Next(1, 21); 
     if (c > 0) sb.Append("\t"); 
     sb.Append(a[r, c]); 
    } 
    sb.AppendLine(); 
} 
var colsSums = SumColumns(a); 
for (int c = 0; c < cols; c++) 
{ 
    if (c > 0) sb.Append("\t"); 
    sb.Append(colsSums[c]); 
} 
var s = sb.ToString(); 
textBox1.Text = s;