C++ Calculating the Mode of a Sorted Array

问题:

I have to write a C++ code that finds the median and mode of an array. I'm told that it's much easier to find the mode of an array AFTER the numbers have been sorted. I sorted the function but still cannot find the mode.

 int counter = 0;
    for (int pass = 0; pass < size - 1; pass++)
        for (int count = pass + 1; count < size; count++) {
            if (array [count] == array [pass])
                counter++;
            cout << "The mode is: " << counter << endl; 

回答1:

If the array has been sorted already, you can count the occurrences of a number at once. Then just save the number that has biggest occurrences. And you can find out the mode in only one for-loop. Otherwise, you'll have to do more than one for-loops. See a details example at the link below Find-the-Mode-of-a-Set-of-Numbers

Here is the code,

int number = array[0];
int mode = number;
int count = 1;
int countMode = 1;

for (int i=1; i<size; i++)
{
      if (array[i] == number) 
      { // count occurrences of the current number
         ++count;
      }
      else
      { // now this is a different number
            if (count > countMode) 
            {
                  countMode = count; // mode is the biggest ocurrences
                  mode = number;
            }
           count = 1; // reset count for the new number
           number = array[i];
  }
}

cout << "mode : " << mode << endl;

回答2:

One way is that you can use Run Length encoding. In Run Length encoding, representation would be like; (Item, Its frequency).

While doing so, keep track of the maximum frequency and Item. This will give you the mode once you complete the Run Length.

for example:

 1 1  2 2 2 3 3 4 5

It run length encoding would be

 {1, 2}, {2, 3}, {3, 2}, {4, 1}, {5, 1}

It needs O(n) space.

回答3:

This is how I did it, my solution will take a sorted vector as input. It has O(n) time complexity and can work with the case where there are more than 1 "mode" number in the vector.

void findMode(vector<double> data) {

double biggestMode = 1;
vector<double> mode, numbers;
numbers.push_back(data.at(0));
mode.push_back(1);
int count = 0;
for (int i = 1; i < data.size(); i++) {
    if (data.at(i) == numbers.at(count)) {
        mode.at(count)++;
    }
    else {
        if (biggestMode < mode.at(count)) {
            biggestMode = mode.at(count);
        }
        count++;
        mode.push_back(1);
        numbers.push_back(data.at(i));
    }
}

for (int i = 0; i < mode.size(); i++) {
    if (mode.at(i) == biggestMode)
        cout << numbers.at(i) << " ";
}
cout << endl;

}

回答4:

Here is the code snippet:

int number = array[0];
int mode = number;
int count = 1;
int countMode = 1;

for (int i=1; i<size; i++)
{
    if (array[i] == number) 
    {
        count++;
    }
    else
    {
        if (count > countMode) 
        {
            countMode = count;
            mode = number;
        }
        count = 1;
        number = array[i];
    }
}

cout << "mode : " << mode << endl;

回答5:

The "mode" is the value that occurs most often. If no number is repeated, then there is no mode for the list. So there would be no benefit to sorting if you needed to know the "mode".

Are you sure you are not referring to the median? The median is the middle number in a set. If you have 1,2,3,4,5 the Median (middle number) is the (total_number)/2) rounded up if it is odd, 2.5 -> 3 and our median would be 3. you can only really calculate the median if your numbers are sorted. If you have an even number in a set 1,2,3,4,5,6 your mode is slots 3,4 (coincidentally also, 3,4) (total_number)/2 slot and (total_number)/2 + 1 slot, for any even array of numbers.

http://www.purplemath.com/modules/meanmode.htm

回答6:

This code should give you the mode. If there are equal number of two different numbers, it will output the first of such.

int count = 1, mode = 0, m = 0, i = 1;
size_t sz = sizeof(array)/sizeof(*array);
while(i != sz+1) {
    if(array[i-1] != array[i]) {
        if(count > m) {
            mode = array[i-1];
            m = count;
            count = 1;
        }
    }
    else
        ++count;
    ++i;
}
std::cout << "mode: " << mode << std::endl;

回答7:

This code finds the mode in C++:

#include <iostream>
using namespace std;

int main(int argc, char** argv)
{
    int i,j,k=0,n,repeat_max=0,cn=0;
    int array1[50],mode[50],count[50]={0},c[50];

    cout<<"\n inter count:\t";
    cin>>n; 


    cout<<"\n";

    for(i=0;i<n;i++)
    cin>>array1[i];

    for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {

            if(array1[i]==array1[j])
            {   
                count[i]++;
                if(count[i]>=repeat_max)
                {
                    repeat_max=count[i];
                    mode[k++]=array1[i];        
                }
            }
        }
    }
    cout<<"\n================\n";
    for(i=1;i<k;i++)
    cout<<"\t mode[i]="<<mode[i]<<"\n";
    cout<<"\t\n\nrepeat array:"<<repeat_max;

    return 0;
}

回答8:

I did it this way:

    int main()
{ 
    int mode,modecount2,modecount1;
    bool is_nomode=false;
    vector<int> numbers = { 15,43,25,25,25,25,16,14,93,93,58,14,55,55,55,64,14,43,14,25,15,56,78,13,15,29,14,14,16 };
    sort(numbers);

    //If you uncomment the following part, you can see the sorted list of above numbers
    //for (int i = 0; i < numbers.size(); ++i) std::cout << numbers[i] << '\n';
    //keep_window_open();

    mode = numbers[0];
    modecount1 = 0;
    modecount2 = 1; //Obviously any number exists at least once!
    for (int i = 1; i < numbers.size(); ++i) {
        if(numbers[i]==numbers[i-1]) ++modecount2;
        else {
            if (modecount2 > modecount1) {
                mode = numbers[i - 1];
                modecount1 = modecount2;
            }
            else if (i != 1 && modecount2 == modecount1) { std::cout << "No mode!\n"; is_nomode = true; break; }
            modecount2 = 1;
        }
    }
    if(!is_nomode) std::cout << "Mode of these numbers is: " << mode << std::endl;
    keep_window_open();

Also you can add another 25 to the list of numbers and see what happens if two numbers have the same occurrence! I hope it helps.

回答9:

This code uses "map" to find out the MODE from the given array. It assumes the array is already sorted.

int findMode(int * arr, int arraySize)
{
    map<int, int> modeMap;
    for (int i = 0; i < arraySize; ++i) {
        ++modeMap[arr[i]];
    }

    auto x = std::max_element(modeMap.begin(), modeMap.end(),
        [](const pair<int, int>& a, const pair<int, int>& b) {
        return a.second < b.second; });

    return x->first;
}

回答10:

This had worked.

int vals[9];                
sort(vals, vals + 9);
int key = vals[0], value = 1,max_key=0,max_value=0;

for (int l = 1; l < 9; l++){
    if (key == vals[l]){
        value++;
    }
    else{
        if (value>max_value){
            max_key = vals[l-1];
            max_value = value;
        }
        key = vals[l];
        value = 1;
    }
}
cout<< "Mode: "<< max_key << endl;

回答11:

There is an old adage that states "If you put 10 programmers in a room and give them the same program to code you will get 12 different results", hence my version of answering your question. It may not be as fast (I'm planning on testing it's speed versus some of the other suggestions) but I feel it is easy to understand.

#include <iostream>

using namespace std;

int main ()
{
    short z[10];
    short maxCount = 0, curCount = 0, cur = 0, most = 0;

    for (int i = 0; i < 10; i++)
        {
         cout << "Enter a number: " << endl;
         cin >> z[i];
        }

    for (int i = 0; i < 10; i++)
        {
         cur = z[i];
            for (int a = i; a < 10; a++)
                {
                 if (cur == z[a])
                    {
                     curCount++;
                     cur = z[a];
                    }
                if (curCount > maxCount)
                   {
                    maxCount = curCount;
                    most = z[a];
                   }
            }
            curCount = 0;
        }

    cout << "the mode is : " << maxCount << ", the number is: " << most << endl;
}