c# - 在 LINQ 中，我可以选择多个项目吗？

Question

假设我有一个像这样的 string 数组:

string [] foos = {
    "abc",
    "def",
    "ghi"
};

我想要一个包含每个字符串及其反转的新集合。所以结果应该是:

    "abc",
    "cba",
    "def",
    "fed",
    "ghi",
    "ihg"

我可以遍历数组，但这很笨拙:

string Reverse (string str)
{
    return new string (str.Reverse ().ToArray ());
}

List<string> results = new List<string> ();
foreach (string foo in foos) {
    results.Add (foo);
    results.Add (Reverse(str));
}

有没有办法在 LINQ 中执行此操作？有点像

var results = from foo in foos
    select foo /* and also */ select Reverse(foo)

Answer 1

var result = from foo in foos
             from x in new string[] { foo, Reverse(foo) }
             select x;

或

var result = foos.SelectMany(foo => new string[] { foo, Reverse(foo) });

这会将 foos 中的每个 foo 映射到一个由两个元素组成的数组:foo 和 Reverse(foo)。然后它将所有这些双元素数组连接成一个大型可枚举数组。

{                           {                             {"abc",                          {                                "abc",                    "abc",                                "cba",                    "cba",                                },"def",           =>             {              =>                                "def",                    "def",                                "fed",                    "fed",                                }"ghi",                          {                                "ghi",                    "ghi",                                "ihg",                    "ihg",                                }}                           }                             }

---

If the order of your output is not so important, you can also concatenate the original array with the result of mapping each element in the original array using the Reverse method:

var result = foos.Concat(from foo in foos select Reverse(foo));

或

var result = foos.Concat(foos.Select(Reverse));

{                               {                               {"abc",                          "abc",                          "abc","def",           =>             "def",           =>             "def","ghi",                          "ghi",                          "ghi",}                               }                                concat                                {                                "cba",                          "cba",                                "fed",                          "fed",                                "ihg",                          "ihg",                                }                               }