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How to find multiple modes?

j rod (5)
I am writing a program that finds the mode of a vector. I need to figure out how to detect if there are multiple modes. Right now it is kind of a flop. How do I detect multiple modes?

Here's what I have so far. 

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  #include<iostream>
#include<vector>
using namespace std;

int main() {

vector<int>mode; 


int num;
int max = 0;
int result;


cout << "Give me numbers and I will find a mode. Enter -1 when you are ready.\n ";
while(num !=-1) {
num = 0;
cin >> num;
if(num!=-1){
mode.push_back(num);
}
} 
int freq = 0; 

for(int x=0; x<mode.size(); x++){///bubble sort technique

		for(int y=0; y<mode.size()-1; y++){

			if(mode[y]>mode[y+1]){
				int xd = mode[y+1];
				mode[y+1] = mode[y];
				mode[y] = xd;
			}
		}
	}
int mult[mode.size()] = {}; 
for (int i = 0;i < mode.size();i++) {
    
if(mode[i] == mode[i + 1]) {
           
freq++; 
} 
else {
  freq = 0;// if number[i] is equal to the next number,frequency increases.
}
if (freq > max) {
     
max = freq; 
result = mode[i]; // if the frequency is bigger than the max frequeny,the result should be the number[i]
mult[i] = mode[i];
result = 1;
}
else if (max == 0) {
     
result = 0;

}
}
if(result == 0) {
cout << "there is no mode." << endl;
}
else if(result == 1){
cout << "the mode is: " << max << endl;

}
}
coder777 (8444)
What exactly does not work?
Thomas1965 (4571)
I am not really sure what the "mode of the vector" is. Could you explain?
Imagine you have a following numbers {1, 2, 2, 3, 4, 5, 5, 5, 6}
What should the result be?
lastchance (6980)
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#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;


//======================================================================


template <class T> vector<T> mode( const vector<T> &v, int &maxCount )        // *** MUST SORT v BEFORE USE ***
{
   vector<T> m;                                  // to hold (and ultimately return) the most common values

   if ( v.size() == 0 )                          // trivial case; empty vector
   {
      maxCount = 0;
      return m;                
   }

   T current = v[0];                             // current is the "last known value"; initialise with start of the array
   int counter = 1;                              // counter holds the number of occurrences of current
   maxCount = counter;                           // maxCount holds the maximum number of occurrences of any value
   m.push_back( current );                       // m holds the modes (i.e. the most frequent values)

   for ( int i = 1; i < v.size(); i++ )          // loop through the remaining elements of v
   {
      if ( v[i] == current )                     // if the same as the current value ...
      {
         counter++;
         if ( counter > maxCount )                      // if a new modal value ...
         {
            m.clear();                                     // empty the old modal vector ...
            m.push_back( current );                        // ... and put the new mode in
            maxCount = counter;
         }
         else if ( counter == maxCount )                // if the same as the modal value ...
         {
            m.push_back( current );                        // add to the list of modes
         }
      }
      else                                       // start counting on a new value
      {
         current = v[i];
         counter = 1;
         if ( counter == maxCount ) m.push_back( current );        // add to modes only if modal count is 1
      }
   }

   return m;
}


//======================================================================


int main()
{
   vector<int> v = { 9, 2, 2, 7, 5, 2, 6, 5, 5, 1 };
   int maxNumber;

   sort( v.begin(), v.end() );

   vector<int> m = mode( v, maxNumber );

   cout << "Mode(s): ";   for ( auto e : m ) cout << e << " ";
   cout << "\nMaximum count: " << maxNumber;
}


@j rod - if you are put off by the templated function and want to restrict yourself to integers only then you can simplify lines 10-20 as
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vector<int> mode( vector<int> v, int &maxCount )        // *** MUST SORT v BEFORE USE ***
{
   vector<int> m;                                // to hold (and ultimately return) the most common values

   if ( v.size() == 0 )                          // trivial case; empty vector
   {
      maxCount = 0;
      return m;                
   }

   int current = v[0];                           // current is the "last known value"; initialise with start of the array
Last edited on
j rod (5)
@coder777
It does not output the correct mode.
@Thomas1965
The 'mode' is the value that occurs the most - with your value set {1, 2, 2, 3, 4, 5, 5, 5, 6}, output should be '5'
Thomas1965 (4571)
Just an idea:
Use a struct to store the count of each number entered and sort the vector accordingly. The mode(s) will be at the front of the vector.
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struct NumberInfo
{
  int num;
  int count;
};


int main()
{
  vector<NumberInfo> numbers;

  for (int num; cin >> num;)
  {
    auto pos = find_if(numbers.begin(), numbers.end(),
               [num](NumberInfo& info)
               {
                return info.num == num;
               });
    if (pos != numbers.end())
    {
      (*pos).count++;
    }
    else
    {
      numbers.push_back(NumberInfo{num, 1});
    }
  }
  // sort the vector according to the NumberInfo.count
  sort(numbers.begin(), numbers.end(), [](const NumberInfo& ni1, const NumberInfo& ni2)
  {
    return ni1.count > ni2.count;
  });
  
  // display the all the numbers and their count - just for test
  for (auto ni&: numbers)
    cout << ni.num << "\t" << ni.count << "\n";

  // TODO display the numbers with the same highest count 

1 2 3 4 4 5 6 6 7^Z
4       2
6       2
1       1
2       1
3       1
5       1
7       1
Press any key to continue . . .
coder777 (8444)
The 'mode' is the value that occurs the most
This should do the job:
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for (int i = 0;i < mode.size();i++) {
    
if(mode[i] == mode[i + 1]) {
           
freq++; 
} 
else {
  freq = 0;// if number[i] is equal to the next number,frequency increases.
}
if (freq > max) {
     
max = freq; 
result = mode[i]; // if the frequency is bigger than the max frequeny,the result should be the number[i]
mult[i] = mode[i];
result = 1;
}
else if (max == 0) {
     
result = 0;

}
}
if(mode.empty() result == 0) {
cout << "there is no mode." << endl;
}
else if(result == 1){
cout << "the mode is: " << result max << endl;

}
}
j rod (5)
Okay. Here is what I have so far. Still doesn't display if there are multiple modes.

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#include<iostream>
#include<vector>
using namespace std;

int main() {

vector<int>mode;


int num;
int maxi = 0;
vector<int>result;


cout << "Give me numbers and I will find a mode. Enter -1 when you are ready.\n ";
while(num !=-1) {
num = 0;
cin >> num;
if(num!=-1){
mode.push_back(num);
}
}
int freq = 0;

for(int x=0; x<mode.size(); x++){///bubble sort technique

		for(int y=0; y<mode.size()-1; y++){

			if(mode[y]>mode[y+1]){
				int xd = mode[y+1];
				mode[y+1] = mode[y];
				mode[y] = xd;
			}
		}
	}
int mult[mode.size()] = {};
for (int i = 0;i < mode.size();i++) {

if(mode[i] == mode[i + 1]) {

freq++;
}
else {
  freq = 0;// if number[i] is equal to the next number,frequency increases.
}
for(int i = 0; i < mode.size(); i++){
if (freq > maxi) {

maxi = freq;
result.push_back(mode[i]); // if the frequency is bigger than the max frequeny,the result should be the number[i]
mult[i] = mode[i];
}
}
}
if(result.size() == 1){
cout << "the mode is: " << result[0] << endl;
}
else{
    cout << "The modes are: ";
    for(int i = 0; i < result.size(); i++){
        cout << result[i] << " ";
    }
}
}


Thanks for your input :)
lastchance (6980)
@j rod
Is there any chance of making some cosmetic changes to your code to make it easier to read:
- what is in mode[] is NOT a mode: could you just call it numbers[] or something?
- you don't need mult[] at all (unless you are also going to do a histogram) - remove it?
- please INDENT CONSISTENTLY (preferably with spaces, not tabs)

Other things:
- when you go on to a new value freq needs to be set to 1, not 0 (line 44)
- you are going to use values beyond the end of your array in line 39
- when you get a single new mode ... you need to empty all previous ones, or they'll just accumulate.

Have a look at my earlier code sample - it should translate reasonably easily to what you are doing.
Last edited on
gunnerfunner (2127)
Have a look at my earlier code sample - it should translate reasonably easily to what you are doing

lastchance - you're being your usual modest self, in fact that program was just great and did everything OP wanted. I've saved it myself for future reference
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