Question

我有7个布尔值。基于它们的值，我需要构建一个字符串。最有效的方法是什么？

为了举例，我的布尔值是：

bool green = false;
bool red = true;
bool purple = true;
bool orange = false;
bool black = true;
bool blue = true;
bool brown = false;

每个布尔变量（理论上）都有一个与之关联的字符串。我想要生成的最后一个字符串取决于＆＃39; true＆＃39;值。

在这里，我想要生成的字符串是"redskins - ravens - raiders - texans"。

与＆＃39; false＆＃39;相关联的字符串变量不会包含在最终字符串中。

我正在处理的实际问题是规模较大，大约有50个布尔值，我的要求有50个与之关联的字符串。

使用if-else？

是否比使用大型StringBuilder方法更有效

我的变量名称不代表我需要输出的字符串。

Answer 1

bool green = false;
bool red = true;
bool purple = true;
bool orange = false;
bool black = true;
bool blue = true;
bool brown = false;

var bools = new Dictionary<string, bool>
{
     { $"{nameof(green)}", green}
    ,{ $"{nameof(red)}", red}
    ,{ $"{nameof(purple)}", purple}
    ,{ $"{nameof(orange)}", orange}
    ,{ $"{nameof(black)}", black}
    ,{ $"{nameof(blue)}", blue}
    ,{ $"{nameof(brown)}", brown}
};

Console.Write(string.Join("-", bools.Where(b => b.Value).Select(b => b.Key)));

Answer 2

我正在考虑使用字典进行配对。

    static void Main(string[] args)
    {
        Dictionary<string, string> pair = new Dictionary<string, string>();

        pair.Add("green", "green");
        pair.Add("red", "redskiins");
        pair.Add("purple", "ravens");
        pair.Add("orange", "orange");
        pair.Add("black", "raiders");
        pair.Add("blue", "texans");
        pair.Add("brown", "brown");

        List<string> display = new List<string>();

        bool green = false;
        bool red = true;
        bool purple = true;
        bool orange = false;
        bool black = true;
        bool blue = true;
        bool brown = false;

        if (green)
            display.Add(pair["green"]);
        if (red)
            display.Add(pair["red"]);
        if (purple)
            display.Add(pair["purple"]);
        if (orange)
            display.Add(pair["orange"]);
        if (black)
            display.Add(pair["black"]);
        if (blue)
            display.Add(pair["blue"]);
        if (brown)
            display.Add(pair["brown"]);

        string finalStr = string.Join(" - ", display);
        Console.WriteLine(finalStr);

        Console.Read();
    }

输出：

Answer 3

var boolArray = GetLotsOfBools();
var strArray = GetLotsOfStrings();

return string.Join(" - ",
   boolArray
   .Select((@bool, x) => new {
      Bool = @bool,
      Str = strArray[x]
   }).Where(pair => pair.Bool)
   .Select(pair => pair.Str)
   .ToArray()
);

Answer 4

我会做这样的事情，使用索引到输入字符串（颜色）的字典，并使用元组作为值：

        // Create your dictionary, indexed with the color, with a tuple as a value to describe
        // boolean state and matching string fragment
        var dictTeams = new Dictionary<string, Tuple<bool, string>>()
        {
            { "green", new Tuple<bool, string>(false, "packers") },
            { "red", new Tuple<bool, string>(true, "redskins") },
            { "purple", new Tuple<bool, string>(true, "ravens") },
            { "orange", new Tuple<bool, string>(false, "bears") },
            { "black", new Tuple<bool, string>(true, "raiders") },
            { "blue", new Tuple<bool, string>(true, "texans") },
            { "brown", new Tuple<bool, string>(false, "browns") }
        };

        // You can either initialize your dictionary all at once or fill it in later on.
        dictTeams.Add("white", new Tuple<bool, string>(true, "patriots"));

        // Select the keys of the entries marked true and aggregate them into a string.
        // Here you could also add a filter for only colors selected in another list by adding
        // a constraint in the Where clause using the key in addition to the boolean value.
        var outString = dictTeams.Where(c => c.Value.Item1)
                                 .Select(c => c.Value.Item2)
                                 .Aggregate((a, b) => a + " - " + b);

请注意，元组是不可变的，因此如果布尔状态需要在运行时更改，请为字典值创建两个字段类，这样您就可以执行以下操作：

dictTeams["green"].BoolState = false;

和

var outString = dictTeams.Where(c => c.Value.BoolState && lstUserColors.Contains(c.Key))
                         .Select(c => c.Value.StringFragment)
                         .Aggregate((a, b) => a + " - " + b);

C＃ - 基于一组布尔值构建字符串的最有效方法是什么

4 个答案: