c# - 如何使用 C# 将二维数组 string[,] 转换为字符串

Question

我有二维数组

var temp = new string[,] { { "1", "2", "3" }, { "4", "5", "6" }, { "7", "8", "9" } };

提醒：

string[,] != string[][]

我想转换成...

123

456

789

在这种情况下如何快速转换？

Answer 1

对于单个代码行，您可以使用：

var temp = new string[,] { { "1", "2", "3" }, { "4", "5", "6" }, { "7", "8", "9" } };
var result = string.Join("\r\n\r\n", 
    temp.OfType<string>()
    .Select((str, idx) => new {index = idx, value = str})
    .GroupBy(a => a.index/(temp.GetUpperBound(0) + 1))
    .Select(gr => gr.Select(n => n.value).ToArray())
    .Select(a => string.Join("", a.SelectMany(x => x)))
    .ToArray());

如果您不将数组定义为多维数组，则单行代码看起来会更好：

string[][] array2d = { new[] { "1", "2", "3" }, new[] { "4", "5", "6" }, new[] { "7", "8", "9" } };
string[][] jagged2d = { new[] { "1", "2", "3" }, new[] { "4", "5" }, new[] { "6" } };

string array2dConcatenate = string.Join("\r\n\r\n", array2d.Select(a => string.Join("", a.SelectMany(x => x))));
string jagged2dConcatenate = string.Join("\r\n\r\n", jagged2d.Select(a => string.Join("", a.SelectMany(x => x))));

只需连接一个多维数组，您可以使用：

string[,] multidimensional2d = { { "1", "2", "3" }, { "4", "5", "6" } };
string[,,] multidimensional3d = { { { "1", "2" }, { "3", "4" } }, { { "5", "6" }, { null, null } } };
string multidimensional2dConcatenate = string.Join(", ", multidimensional2d.OfType<string>());
string multidimensional3dConcatenate = string.Join(", ", multidimensional3d.OfType<string>());

要充分利用 linq，请参阅：何时在 Linq 中使用 Cast() 和 Oftype()

如果您想使用 3d 数组或保护 null，您可以执行以下操作：

string[][][] array3d = { new[] { new[] { "1", "2" } }, new[] { new[] { "3", "4" } }, new[] { new[] { "5", "6" } }, null };
string[][][] jagged3d = { new[] { new[] { "1", "2" }, new[] { "3" } }, new[] { new[] { "4" }, new[] { "5" } }, new[] { new[] { "6" }, null }, null };

string array3dConcatenate = string.Join("\r\n\r\n", array3d.Where(x => x != null).SelectMany(x => x).Where(x => x != null).Select(a => string.Join("", a.SelectMany(x => x))));
string jagged3dConcatenate = string.Join("\r\n\r\n", jagged3d.Where(x => x != null).SelectMany(x => x).Where(x => x != null).Select(a => string.Join("", a.SelectMany(x => x))));

Answer 2

private static string ObjectToString(IList<object> messages)
    {
        StringBuilder builder = new StringBuilder();
        foreach (var item in messages)
        {

            if (builder.Length > 0)
                builder.Append(" ");
            if (item is IList<object>)
                builder.Append(ObjectToString((IList<object>)item));
            else
                builder.Append(item);

        }

        return builder.ToString();
    }

Answer 3

这是一种带有 2 个嵌套循环的方法：

var temp = new string[,]{{"1","2","3"},{"4","5","6"}};
var output = new string[temp.GetUpperBound(0)+1];
for (int i = 0; i<=temp.GetUpperBound(0); i++)
{
    var sb = new StringBuilder(temp.GetUpperBound(1)+1);
    for (int j = 0; j<=temp.GetUpperBound(1); j++)
        sb.Append(temp[i,j]);
    output[i] = sb.ToString();
}

如果您认为可以将 2-d 数组视为 1-d 并绕过 2 个循环，这里有一些技巧：如何将一行值从 2D 数组复制到 1D 数组中？，但是为了能够使用这些技巧，您需要 char 数组，而不是您所要求的字符串数组。