在具有1000个元素的整数数组中查找多个模式

Question

So I need a way to find the mode(s) in an array of 1000 elements, with each element generated randomly using math.Random() from 0-300.

int[] nums = new int[1000];

    for(int counter = 0; counter < nums.length; counter++)
        nums[counter] = (int)(Math.random()*300);

int maxKey = 0;
    int maxCounts = 0;

    sortData(array);
    int[] counts = new int[301];

    for (int i = 0; i < array.length; i++)
    {
        counts[array[i]]++;
        if (maxCounts < counts[array[i]]) 
        {
            maxCounts = counts[array[i]];
            maxKey = array[i];
        }
    }

This is my current method, and it gives me the most occurring number, but if it turns out that something else occurred the same amount of times, it only outputs one number and ignore the rest.

WE ARE NOT ALLOWED TO USE ARRAYLIST or HASHMAP (teacher forbade it)

Please help me on how I can modify this code to generate an output of array that contains all the modes in the random array.

Thank you guys!

EDIT:

Thanks to you guys, I got it:

private static String calcMode(int[] array)
{
    int[] counts = new int[array.length];
    for (int i = 0; i < array.length; i++) {
        counts[array[i]]++;
    }
    int max = counts[0];
    for (int counter = 1; counter < counts.length; counter++) {
        if (counts[counter] > max) {
            max = counts[counter];
        }
    }

    int[] modes = new int[array.length];

    int j = 0;
    for (int i = 0; i < counts.length; i++) {
        if (counts[i] == max)
            modes[j++] = array[i];
    }

    toString(modes);
    return "";
}

public static void toString(int[] array)
{
    System.out.print("{");
    for(int element: array)
    {
        if(element > 0)
            System.out.print(element + " ");
    }
    System.out.print("}");
}

Answer 1

Look at this, not full tested. But I think it implements what @ajb said:

private static int[] computeModes(int[] array)
{
    int[] counts = new int[array.length];
    for (int i = 0; i < array.length; i++) {
        counts[array[i]]++;
    }
    int max = counts[0];
    for (int counter = 1; counter < counts.length; counter++) {
        if (counts[counter] > max) {
            max = counts[counter];
        }
    }

    int[] modes = new int[array.length];

    int j = 0;
    for (int i = 0; i < counts.length; i++) {
        if (counts[i] == max)
            modes[j++] = array[i];
    }

    return modes;
}

This will return an array int[] with the modes. It will contain a lot of 0 s, because the result array ( modes[] ) has to be initialized with the same length of the array passed. Since it is possible that every element appears just one time.

When calling it at the main method:

public static void main(String args[])
{
    int[] nums = new int[300];

    for (int counter = 0; counter < nums.length; counter++)
        nums[counter] = (int) (Math.random() * 300);

    int[] modes = computeModes(nums);
    for (int i : modes)
        if (i != 0) // Discard 0's
            System.out.println(i);
}

Answer 2

Your first approach is promising, you can expand it as follows:

for (int i = 0; i < array.length; i++)
{
    counts[array[i]]++;
    if (maxCounts < counts[array[i]]) 
    {
        maxCounts = counts[array[i]];
        maxKey = array[i];
    }
}

// Now counts holds the number of occurrences of any number x in counts[x]
// We want to find all modes: all x such that counts[x] == maxCounts

// First, we have to determine how many modes there are
int nModes = 0;

for (int i = 0; i < counts.length; i++)
{
    // increase nModes if counts[i] == maxCounts
}

// Now we can create an array that has an entry for every mode:
int[] result = new int[nModes];

// And then fill it with all modes, e.g:
int modeCounter = 0;
for (int i = 0; i < counts.length; i++)
{
    // if this is a mode, set result[modeCounter] = i and increase modeCounter  
}

return result;

Answer 3

THIS USES AN ARRAYLIST but I thought I should answer this question anyways so that maybe you can use my thought process and remove the ArrayList usage yourself. That, and this could help another viewer.

Here's something that I came up with. I don't really have an explanation for it, but I might as well share my progress:

Method to take in an int array, and return that array with no duplicates ints:

public static int[] noDups(int[] myArray) 
{ 
    // create an Integer list for adding the unique numbers to
    List<Integer> list = new ArrayList<Integer>();

    list.add(myArray[0]); // first number in array will always be first
    // number in list (loop starts at second number)

    for (int i = 1; i < myArray.length; i++)
    {
        // if number in array after current number in array is different
        if (myArray[i] != myArray[i - 1]) 
            list.add(myArray[i]); // add it to the list
    }

    int[] returnArr = new int[list.size()]; // create the final return array
    int count = 0;

    for (int x : list) // for every Integer in the list of unique numbers
    {
        returnArr[count] = list.get(count); // add the list value to the array
        count++; // move to the next element in the list and array
    }

    return returnArr; // return the ordered, unique array
}

---

Method to find the mode:

public static String findMode(int[] intSet)
{
    Arrays.sort(intSet); // needs to be sorted

    int[] noDupSet = noDups(intSet);
    int[] modePositions = new int[noDupSet.length];

    String modes = "modes: no modes."; boolean isMode = false;

    int pos = 0;
    for (int i = 0; i < intSet.length-1; i++)
    {
        if (intSet[i] != intSet[i + 1]) {
            modePositions[pos]++;
            pos++;
        }
        else {
            modePositions[pos]++;
        }
    }
    modePositions[pos]++;

    for (int modeNum = 0; modeNum < modePositions.length; modeNum++)
    {
        if (modePositions[modeNum] > 1 && modePositions[modeNum] != intSet.length)
            isMode = true;
    }

    List<Integer> MODES = new ArrayList<Integer>();

    int maxModePos = 0;
    if (isMode) {
        for (int i = 0; i< modePositions.length;i++)
        {
            if (modePositions[maxModePos] < modePositions[i]) {
                maxModePos = i;
            }
        }

        MODES.add(maxModePos);
        for (int i = 0; i < modePositions.length;i++)
        {
            if (modePositions[i] == modePositions[maxModePos] && i != maxModePos)
                MODES.add(i);
        }

        // THIS LIMITS THERE TO BE ONLY TWO MODES
        // TAKE THIS IF STATEMENT OUT IF YOU WANT MORE         
        if (MODES.size() > 2) {
            modes = "modes: no modes.";
        }
        else {
            modes = "mode(s): ";
            for (int m : MODES)
            {
                modes += noDupSet[m] + ", ";
            }
        }
    }
    return modes.substring(0,modes.length() - 2);
}

---

Testing the methods:

public static void main(String args[]) 
{
    int[] set = {4, 4, 5, 4, 3, 3, 3};
    int[] set2 = {4, 4, 5, 4, 3, 3};
    System.out.println(findMode(set)); // mode(s): 3, 4
    System.out.println(findMode(set2)); // mode(s): 4
}

Answer 4

There is a logic error in the last part of constructing the modes array. The original code reads modes[j++] = array[i]; . Instead, it should be modes[j++] = i . In other words, we need to add that number to the modes whose occurrence count is equal to the maximum occurrence count