一个数字在数组中出现的次数列表[重复]

Question

我必须创建一个由 n 个元素组成的数组，并找出 每个 数字出现的次数，如下所示：

Array: (-1.7 ; 3.0 ; 0.0 ; 1.5 ; 0.0 ; -1.7 ; 2.3 ; -1,7)
-1.7 appears 3 times
 3.0 appears 1 time
 0.0 appears 2 times
 1.5 appears 1 time
 2.3 appears 1 time

我试过了，我的代码是这样的：

        int n = 0;
        Console.WriteLine("Type in the size of your array...");
        n = int.Parse(Console.ReadLine());

        float[] vet = new float[n];
        int[] freq = new int[n];

        Console.WriteLine("Now, type in each element...");
        for(int i = 0; i < vet.Length; i++)
        {
            Console.Write("Position {0}: ", i);
            vet[i] = float.Parse(Console.ReadLine());
        }
        for(int j = 0; j< vet.Length; j++)
        {
            for(int k = 0; k< freq.Length; k++)
            {
                if(vet[j] == vet[k])
                {
                    freq[j]++;
                }
            }
        }
        Console.WriteLine("The number of times each element appears in the array is:");

        for(int l = 0; l< freq.Length; l++)
        {
            Console.WriteLine("{0} appears {1} time(s)", vet[l], freq[l]);
        }
        Console.ReadKey();

但是输出还是这样的：

The number of times each element appears in the array is:
-1.7 appears 3 time(s)
 3.0 appears 1 time(s)
 0.0 appears 2 time(s)
 1.5 appears 1 time(s)
 0.0 appears 2 time(s)
-1.7 appears 3 time(s)
 2.3 appears 1 time(s)
-1.7 appears 3 time(s)

我的问题是：我怎样才能让我的代码像第一个示例一样以打印一次重复数字的方式工作？

Answer 1

尝试改用Dictionary。代码如下：

    var arr = new double [] { -1.7, 3.0, 0.0, 1.5, 0.0, -1.7, 2.3, -1.7 };
    var dic = new Dictionary<double, int>();

    foreach (var element in arr)
    {
        if (dic.ContainsKey(element))
            dic[element]++;
        else
            dic[element] = 1;
    }

    foreach (var element in dic)
    {
        Console.WriteLine(element.Key + " appears " + element.Value + " time(s)");
    }

Answer 2

我个人会像这样使用 linq：

    decimal[] values = {-1.7m , 3.0m , 0.0m , 1.5m , 0.0m , -1.7m , 2.3m , -1,7m};
    
    var result = values
        .GroupBy(v => v)
        .ToDictionary(g => g.Key, g => g.Count());
    
    foreach(var kvp in result)
    {
        Console.WriteLine("{0} appears {1} time(s)", kvp.Key, kvp.Value);
    }

DotNetFiddle.Net

结果：

-1.7 出现 2 次​​p>

3.0 出现 1 次

0.0 出现 2 次​​p>

1.5 出现 1 次

2.3 出现 1 次

-1 出现 1 次

7 出现 1 次

Answer 3

非非非非非。您没有使用 C# 的强大功能。这也可能是一个 c++ 程序。更好的方法是使用类似的方法将数组转换为列表

List<float> list=vet.ToList()

那么你可以关注这里的信息：Get a list of distinct items and their count

我会写在这里：

var query = list.SelectMany(x => x.Names)
            .GroupBy(s => s)
            .Select(g => new { Name = g.Key, Count = g.Count() });

foreach(var result in query) {
    Console.WriteLine("Name: {0}, Count: {1}", result.Name, result.Count);
}

Answer 4

典型的解决方案是基于Linq：

using System.Linq;

...

var freqs = vet
  .GroupBy(item => item)
  .OrderBy(chunk => chunk.Key)
  .Select(chunk => 
     $"{chunk.Key,4:f1} appears {chunk.Count()} {(chunk.Count() > 1 ? "times" : "time")}");

Console.WriteLine("The number of times each element appears in the array is:");

Console.Write(string.Join(Environment.NewLine, freqs));

Console.ReadKey();

结果：

The number of times each element appears in the array is:
-1.7 appears 3 times
 0.0 appears 2 times
 1.5 appears 1 time
 2.3 appears 1 time
 3.0 appears 1 time