c++ - 如何使用二维数组计算和存储来自其他数组的值的频率?
问题描述
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)。我要做的是从用户那里获取一个未定义的 1-9 整数并将其加载到一个数组中,当用户输入 0 或 10 时,循环结束并且数组完成,所以我有: int 资格 [ 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中引入的每个数字的频率。我想有很多方法可以简单地做到这一点(我不知道),我正在尝试使用另一个二维数组来完成它,它有 2 行和 10 个列,使用 1 行来包含来自的值0 到 9 将每个数字与数组限定 [] 的数字进行比较。另一行存储每个数字在资格中出现的次数。我不知道该怎么做
// 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++;
}
}
解决方案
在您的原始代码中,您没有针对“频率”中的每个可能值检查每个“资格”。
我认为这样的东西是你正在寻找的。
#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;
}
输出应该是:
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.
推荐阅读
- python - Windows:已安装 numpy(conda):ModuleNotFoundError
- c++ - 将 C++ 随机数分布传递给函数
- java - Apache Beam 不将无限数据保存到文本文件
- python-2.7 - Python Orion 上下文代理令牌问题
- typescript - 编译器抱怨嵌套对象中可能存在未定义
- c# - C# 异步 UDP 服务器 - 如何将套接字传递给回调
- angular - Angular Materials - 输入框中的下拉菜单
- php - 如何仅从共享键/值对的数组中提取项目?
- python - 在 UWSGI 中创建的进程未在线程模式下完成
- database - 如何在 NoSQL 环境中创建以下数据结构