C#相互添加4个不同字符串数组的字符串元素

Question

我需要关于如何从 4 个不同的字符串数组创建新字符串数组的建议：

我们有 4 个字符串数组：

string[] arr1 = new string []  {"a1","a2","a3"..., "a30"};  
string[] arr2 = new string []  {"d10","d11","d12","d13","d14","d15"};  
string[] arr3 = new string []  {"f1","f2","f3"...,"f20"};  
string[] arr4 = new string []  {"s10","s11","s12","s13","s14"};  

我们需要像这样将所有 4 个数组的所有字符串元素相互添加：

a1+d10+f1+s10  
a2+d10+f1+s10

...       
a1+d11+f1+s10  
a2+d11+f1+s10  
...  
a30+d15+f20+s14

我的意思是按顺序排列的所有组合：arr1_element、arr2_element、arr3_element、arr4_element
所以结果数组会是这样的：
字符串[] arr5 = 新字符串 [] {"a1d10f1s10","a2d10f1s10 "....};

Answer 1

创建一个返回数组所有组合的递归方法：

static IEnumerable<string> GetCombinations(string[][] arrays, int len, string start) {
  foreach (string s in arrays[len - 1]) {
    if (len == 1) {
      yield return s + start;
    } else {
      foreach (string r in GetCombinations(arrays, len - 1, s + start)) {
        yield return r;
      }
    }
  }
}

用法：

string[][] arrays = { arr1, arr2, arr3, arr4 };
foreach (string s in GetCombinations(arrays, arrays.Length, string.Empty)) {
  Console.WriteLine(s);
}

Answer 2

曾几何时，我在一次编程比赛中使用了这个。它工作得很好。看看它是否适合你。

:)

 private static IEnumerable<string> 
    GetCombinations(string[] arr1, string[] arr2, string[] arr3, string[] arr4)
        {
            int i, j, k, l;
            i = j = k = l = 0;

            for (i = 0; i < arr4.Length; i++)
                for (j = 0; j < arr3.Length; j++)
                    for (k = 0; k < arr2.Length; k++)
                        for (l = 0; l < arr1.Length; l++)
                            yield return (arr1[l] + arr2[k] + arr3[j] + arr4[i]);

        }

示例用法：

public static void Main(string[] args)
{
    string[] arr1 = new string[] { "a1", "a2", "a3", "a30" };
    string[] arr2 = new string[] { "d10", "d11", "d12", "d13" };
    string[] arr3 = new string[] { "f1", "f2", "f3", "f20" };
    string[] arr4 = new string[] { "s10", "s11", "s13", "s14" };


    var list = GetCombinations(arr1, arr2, arr3, arr4);

    foreach (var item in list.Take<string>(10))
    {
        Console.WriteLine(item);
    }
}

Answer 3

LINQ 在这种情况下非常方便：

var arr1 = new [] { "a1", "a2", "a3", "a4", "a5" };
var arr2 = new [] { "b1", "b2", "b3", "b4", "b5" };
var arr3 = new [] { "c1", "c2", "c3", "c4", "c5" };
var arr4 = new [] { "d1", "d2", "d3", "d4", "d5" };

var arr5 = (from a in arr1
            from b in arr2
            from c in arr3
            from d in arr4
            select a + b + c + d).ToList();

这 5 行代码将返回：

a1b1c1
a1b1c2
a1b1c3
a1b1c4
a1b1c5
a1b2c1
a1b2c2
[...]   
a2b1c1
a2b1c2
a2b1c3
a2b1c4
a2b1c5
a2b2c1
a2b2c2
a2b2c3
a2b2c4
a2b2c5
a2b3c1
[...]
a5b5c4
a5b5c5

Answer 4

类似的东西可能会起作用...

List<string> AddEntries(List<List<string>> entries) {
   List<string> finalEntries = new List<string>();

   if (entries != null && entries.Length > 0) {
      if (entries.Length > 1) {
         foreach(string entry in entries) {
            foreach(string subentry in AddEntries(entries.Skip(1)) {
               finalEntries.Add(entry + subEntry);
            }
         }
      } else {
         foreach(string entry in entries[0]) { finalEntries.add(entry); }
      }
   }
   return finalEntries;
}

我为代码道歉，我这里没有编译器来测试。假设您正在使用带有扩展方法的 C# 3.5。我冒昧地将您的 string[] 更改为 List。如果您不使用 3.5，则需要编写自己的函数来修剪嵌套数组。

Answer 5

您需要对 IEnumerable 集合使用 C# 4.0 .Zip() 扩展方法。为此，该方法将委托作为参数。 示例：

   List<string> a = new List<string> { "code" };
   List<int> b = new List<int>() { 7 };
   var res =  a.Zip(b, (p1, p2) => p1.ToString() + p2.ToString());