C++ 一维数组中数字的频率

Frequency of Numbers in a 1D Array

提问by faizann24

its been 6 hours since I have been writing the code but to no avail, I don't no where I am making the mistake but I am making some. Its a frequency output program and output should be like this:

自从我编写代码以来已经 6 个小时了，但无济于事，我不知道我在哪里犯了错误，但我正在犯一些错误。它的一个频率输出程序和输出应该是这样的：

array[8] = {6,1,7,8,6,6,1,9}

Output:

输出：

6:3
1:2
7:1
8:1
9:1

But its repeating the same numbers in my code. Any help would be much appreciable.

但它在我的代码中重复了相同的数字。任何帮助将是非常可观的。

int array[8] = {6,1,7,8,6,6,1,9};
int store[8];
int a =0;
int b =0;
int c=0;

int d = 0; 
store[d] = array[b];
for (d = 0; d < 8; d++){
  int count=0;
  c = d; 
  b = d; 
  for (int e = 0; e < d; e++){
    if (array[b] == store[e]){
      store[d] = array[b];
      b++;              
      e = 0;            

    }                   
    else   
    {      
      store[d] = array[b]; 
      break;    
    }       
  }        

  for ( int z = 0; z < 7; z++){ 
    if (store[d] == array[z])
    {          
      count++;              

    }                       
  }            
  cout << store[d] << ":" << count << endl;
}

回答by SlashDot

You may use a map first to store num->frequency and then a multimap to store freqeuncy => num.

您可以先使用映射来存储 num->frequency，然后使用多映射来存储 freqeuncy => num。

Hereis the working solution.

这是工作解决方案。

#include <map>
#include <algorithm>
#include <iostream>

int main()
{
    int array[8] = {6,1,7,8,6,6,1,9};

    // A map to store num => freq 
    std::map <int, int> freq;

    // A map to store freq(can be duplicate) => num 
    std::multimap <int, int> freqCounts;

    // Store num => frequency
    for (int i = 0 ; i < 8; i++)
    {
        freq[array[i]] += 1;
    }

    // Now Store freq => num
    for(auto const & iter : freq)
    {
        freqCounts.insert (std::pair<int,int>(iter.second, iter.first)); 
    }

    // Print in reverse order i.e. highest frequency first
    for (std::multimap<int,int>::reverse_iterator rit=freqCounts.rbegin(); rit!=freqCounts.rend(); ++rit)
    {
        std::cout << rit->second << " : " << rit->first << '\n';
    }
    return 0;
}

回答by Kerrek SB

You never seem to update the counters. Try this:

您似乎从未更新过计数器。尝试这个：

int array[8] = {6,1,7,8,6,6,1,9};

unsigned int store[10] = {};    // large enough to hold the largest array value,
                                // initialized to zero

for (int n : array) ++store[n]; // update counts

for (int i = 0; i != 10; ++i)
{
    std::cout << "Frequency of int " << i << " is " << store[i] << "\n";
}

If the set of values that occur is sparse, or includes negatives, or simply does not fit into a dense range of integers nicely, you can replace unsigned int[10]with an associative container, e.g.:

如果出现的值集是稀疏的，或者包含负数，或者根本不能很好地适应密集的整数范围，您可以unsigned int[10]用关联容器替换，例如：

std::map<int, unsigned int> store;

// algorithm as before

for (auto const & p : store)
{
    std::cout << "Frequency of " << p.first << " is " << p.second << "\n";
}

回答by Guffa

I'm not sure what you are trying to do with the arrays. I have tried to follow the logic, but it's hard to see it with all the anonymous variable names. It seems like you are trying to look for duplicates earlier in the array, but the variable enever gets any other value than 0, so you will only be comparing with the first item in the array.

我不确定你想用数组做什么。我试图遵循逻辑，但很难看到所有匿名变量名称。似乎您正在尝试在数组中较早地查找重复项，但该变量e从未获得除 之外的任何其他值0，因此您只会与数组中的第一项进行比较。

You can just look in the array itself for previous occurances, and once you know that the number is the first occurance, you only need to look for more occurances after it in the array:

您可以只在数组本身中查找先前出现的次数，一旦您知道该数字是第一次出现，您只需要在数组中查找它之后的更多出现次数：

int array[8] = {6,1,7,8,6,6,1,9};

for (int i = 0; i < 8; i++) {

  // look to the left in the array if the number was used before
  int found = 0;
  for (int j = 0; j < i; j++) {
    if (array[i] == array[j]) found++;
  }

  // go on if it's the first occurance
  if (found == 0) {

    // we know of one occurance
    int count = 1;

    // look to the right in the array for other occurances
    for (int j = i + 1; j < 8; j++) {
      if (array[i] == array[j]) count++;
    }

    cout << array[i] << ":" << count << endl;
  }
}

回答by Fidelius

I wanted to submit my solution which I think it′s a more easy one:

我想提交我的解决方案，我认为它更简单：

#include <iostream>
using namespace std;

int main() {
    int n; //number of Elements in the vector
    cin>>n;
    int vec[n]; //vector with all elements
    int v[n];  //vector with Elements without repetition
    int c[n];  // vector which stores the frequency of each element
    for(int i=0; i<n; i++)
        cin>>vec[i];
    int k=0; // number of Elements of the vector without Repetition, in the begining, it has 0 Elements
    int j=0; //logic Counter, could be replaced with bool
    for(int i=0; i<n; i++) {
        for(int h=0; h<=k; h++) {
            if(vec[i]==v[h]) {
                c[h]++;
                j=1;
                break;
            }
        }
//if element i of the original vector is equal to element h of the second vector, then increment the frequency of this element
        if(j==0) { //else if the element is not equal to any of the second vector, the Position of the 2nd vector is filled with the element, which in this case is the first of ist Kind.
            v[k]=vec[i];
            c[k]=1;
            k++;
        } //the number of Elements is increased by one to store another element;
        else {
            j=0;
        }
    }
    cout<<endl<<endl;
    for(int i=0; i<k; i++)
        cout<<v[i]<<":"<<c[i]<<endl;
    return 0;
}

回答by Sajjad Ali Vayani

/**
 * The methods counts the frequency of each element in an array.
 * 
 * Approach: The method checks if the element is already present in the <strong>Map of frequency</strong>.
 * If it is not present, add it to the map with the frequency 1 else put it in the map with 
 * an increment by one of it's existing frequency.
 * 
 * @param arr list of elements
 * @return frequency of each elements
 */
public static Map<Integer, Integer> countFrequency(int[] arr) {
    Map<Integer, Integer> frequency= new HashMap<Integer, Integer>();
    for(int i = 0; i < arr.length; i++) {
        if(frequency.get(arr[i])==null) {
            frequency.put(arr[i], 1);
        }
        else {
            frequency.put(arr[i],frequency.get(arr[i])+1);
        }
    }
    System.out.println("\nMap: "+frequency);
    return frequency;
}

回答by Bhaskar Chouhan

#include<iostream>
#include<conio.h>
using namespace std;
main()
{   int count[10],key[10],n=10,m;
    int i,j,k,temp;
    cout<<"Enter The Size Of Array:-\n";
    cin>>n;
    int a[n];
    cout<<"Enter The Elements in Array:-\n";
    for(i=0; i<n; i++)
        cin>>a[i];
    for(i=0; i<n; i++)
        for(j=0; j<n-1; j++)
        {   if(a[j]>a[j+1])
            {   temp=a[j];
                a[j]=a[j+1];
                a[j+1]=temp;
            }
        }
    for(i=0; i<n; i++)
        cout<<a[i]<<"\t";
    for(i=0,k=0; i<n; k++)
    {   count[k]=0;
        key[k]=a[i];
        for(j=i; j<n; j++)
        {   if(a[i]==a[j])
                count[k]++;
        }
        i=i+count[k];
    }
    for(i=0; i<k; i++)
        cout<<endl<<key[i]<<" Occurred "<<count[i]<<" Times\n";
    getch();
}