如何使用二维数组来计算和存储来自其他数组的值的频率

How to use a bidimensional array to calculate and store frequencies of values from other array?

本文关键字：其他 数组 频率 存储 何使用 二维数组 计算      更新时间：2023-10-16

我开始学习C编程，所以我不知道如何做基本的事情（可能是我所做的没有意义）。我要做的就是从用户那里获取 1-9 的未定义数量的整数并将其加载到数组中，当用户输入 0 或 10 时，循环结束并且数组完成，所以我有： int quals[SIZEQ] 和一个加载它的函数，它可以工作。现在我必须计算这个数组上引入的每个数字的频率。我想有很多方法可以做到这一点（我不知道），我正在尝试使用另一个二维数组来做到这一点，该数组有 2 行和 10 列，使用 1 行包含从 0 到 9 的值，将每个数字与数组资格的数量进行比较[]。另一行用于存储每个数字在资格上出现的次数。我不知道该怎么做，我尝试：

// 1st row of the freq array contains the 1-9 possible qualifications
// 2nd row to count the times that every value appears in qualifications array
// example: 1st row {0,1,2,3...} if 1 repeated 3 times:
//          2nd row {0,2,0,0...}  
int freq[2][10]={{0,1,2,3,4,5,6,7,8,9},{0,0,0,0,0,0,0,0,0,0}};
    int  j=0;   //var to load 0-9 positions of freq[1][j]
    for (int i = 0; i < SIZEQ; i ++)
    {
        if(qualifications[i]==freq[0][j]){
//what i want to do is to compare from 0 each cell of
//qual with freq with 1-9 numbers
        freq[1][j] = freq[1][j] + freq[1][j];
        //if detected a coincidence, increment the field of the row to
        //count repetitions
        //so, if qualifications[i] it's now '2' when compared to 
        //freq[0][2] (what it's 2) freq[1][2] it's now 1.  
        }
        j=i; 
        if(j>=9){ //to avoid that j be bigger than freq column size
        j=0;
        }
    }
    j=1; //skip 0
        while(j<=9){
            //j print 1-9 numbers and freq[1][j] print the number of
            //times it is repeated 
                cout << "Qualification " << j << " repeated " << freq[1][j] << " times." << endl;
            j++;
        }
}

在原始代码中，您不会根据"freq"中的每个可能值检查每个"限定"。

我认为这样的东西就是你要找的。 

#include <iostream>
using namespace std;
int main()
{
  //make an example 'qualifications'
  int SIZEQ = 5;
  const int SIZEQ = 5;
  int qualifications[SIZEQ] = {1,1,2,3,4};


  int freq[2][10]={{0,1,2,3,4,5,6,7,8,9},{0,0,0,0,0,0,0,0,0,0}};
  for (int i = 0; i < SIZEQ; i ++) // loop through each qualification
  {
    for (int j = 0; j < 10; j++)   // loop through each frequency to see if it matches
    {
      if(qualifications[i]==freq[0][j])
      {
        freq[1][j] += 1;
      }
    }
  }
  int k;
  k=1; //skip 0
  while(k<=9){
    //j print 1-9 numbers and freq[1][j] print the number of
    //times it is repeated
    cout << "Qualification " << k << " repeated " << freq[1][k] << " times." << endl;
    k++;
  }

  return 0;
}

输出应为： 

Qualification 1 repeated 2 times.
Qualification 2 repeated 1 times.
Qualification 3 repeated 1 times.
Qualification 4 repeated 1 times.
Qualification 5 repeated 0 times.
Qualification 6 repeated 0 times.
Qualification 7 repeated 0 times.
Qualification 8 repeated 0 times.
Qualification 9 repeated 0 times.
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