How to calculate the median of an array?
JavaArraysJava Problem Overview
I'm trying to calculate the total, mean and median of an array thats populated by input received by a textfield. I've managed to work out the total and the mean, I just can't get the median to work. I think the array needs to be sorted before I can do this, but I'm not sure how to do this. Is this the problem, or is there another one that I didn't find? Here is my code:
import java.applet.Applet;
import java.awt.Graphics;
import java.awt.*;
import java.awt.event.*;
public class whileloopq extends Applet implements ActionListener
{
Label label;
TextField input;
int num;
int index;
int[] numArray = new int[20];
int sum;
int total;
double avg;
int median;
public void init ()
{
label = new Label("Enter numbers");
input = new TextField(5);
add(label);
add(input);
input.addActionListener(this);
index = 0;
}
public void actionPerformed (ActionEvent ev)
{
int num = Integer.parseInt(input.getText());
numArray[index] = num;
index++;
if (index == 20)
input.setEnabled(false);
input.setText("");
sum = 0;
for (int i = 0; i < numArray.length; i++)
{
sum += numArray[i];
}
total = sum;
avg = total / index;
median = numArray[numArray.length/2];
repaint();
}
public void paint (Graphics graf)
{
graf.drawString("Total = " + Integer.toString(total), 25, 85);
graf.drawString("Average = " + Double.toString(avg), 25, 100);
graf.drawString("Median = " + Integer.toString(median), 25, 115);
}
}
Java Solutions
Solution 1 - Java
The Arrays class in Java has a static sort function, which you can invoke with Arrays.sort(numArray).
Arrays.sort(numArray);
double median;
if (numArray.length % 2 == 0)
median = ((double)numArray[numArray.length/2] + (double)numArray[numArray.length/2 - 1])/2;
else
median = (double) numArray[numArray.length/2];
Solution 2 - Java
Sorting the array is unnecessary and inefficient. There's a variation of the QuickSort (QuickSelect) algorithm which has an average run time of O(n); if you sort first, you're down to O(n log n). It actually finds the nth smallest item in a list; for a median, you just use n = half the list length. Let's call it quickNth (list, n).
The concept is that to find the nth smallest, choose a 'pivot' value. (Exactly how you choose it isn't critical; if you know the data will be thoroughly random, you can take the first item on the list.)
Split the original list into three smaller lists:
- One with values smaller than the pivot.
- One with values equal to the pivot.
- And one with values greater than the pivot.
You then have three cases:
- The "smaller" list has >= n items. In that case, you know that the nth smallest is in that list. Return quickNth(smaller, n).
- The smaller list has < n items, but the sum of the lengths of the smaller and equal lists have >= n items. In this case, the nth is equal to any item in the "equal" list; you're done.
- n is greater than the sum of the lengths of the smaller and equal lists. In that case, you can essentially skip over those two, and adjust n accordingly. Return quickNth(greater, n - length(smaller) - length(equal)).
Done.
If you're not sure that the data is thoroughly random, you need to be more sophisticated about choosing the pivot. Taking the median of the first value in the list, the last value in the list, and the one midway between the two works pretty well.
If you're very unlucky with your choice of pivots, and you always choose the smallest or highest value as your pivot, this takes O(n^2) time; that's bad. But, it's also very unlikely if you choose your pivot with a decent algorithm.
Sample code:
import java.util.*;
public class Utility {
/****************
* @param coll an ArrayList of Comparable objects
* @return the median of coll
*****************/
public static <T extends Number> double median(ArrayList<T> coll, Comparator<T> comp) {
double result;
int n = coll.size()/2;
if (coll.size() % 2 == 0) // even number of items; find the middle two and average them
result = (nth(coll, n-1, comp).doubleValue() + nth(coll, n, comp).doubleValue()) / 2.0;
else // odd number of items; return the one in the middle
result = nth(coll, n, comp).doubleValue();
return result;
} // median(coll)
/*****************
* @param coll a collection of Comparable objects
* @param n the position of the desired object, using the ordering defined on the list elements
* @return the nth smallest object
*******************/
public static <T> T nth(ArrayList<T> coll, int n, Comparator<T> comp) {
T result, pivot;
ArrayList<T> underPivot = new ArrayList<>(), overPivot = new ArrayList<>(), equalPivot = new ArrayList<>();
// choosing a pivot is a whole topic in itself.
// this implementation uses the simple strategy of grabbing something from the middle of the ArrayList.
pivot = coll.get(n/2);
// split coll into 3 lists based on comparison with the pivot
for (T obj : coll) {
int order = comp.compare(obj, pivot);
if (order < 0) // obj < pivot
underPivot.add(obj);
else if (order > 0) // obj > pivot
overPivot.add(obj);
else // obj = pivot
equalPivot.add(obj);
} // for each obj in coll
// recurse on the appropriate list
if (n < underPivot.size())
result = nth(underPivot, n, comp);
else if (n < underPivot.size() + equalPivot.size()) // equal to pivot; just return it
result = pivot;
else // everything in underPivot and equalPivot is too small. Adjust n accordingly in the recursion.
result = nth(overPivot, n - underPivot.size() - equalPivot.size(), comp);
return result;
} // nth(coll, n)
public static void main (String[] args) {
Comparator<Integer> comp = Comparator.naturalOrder();
Random rnd = new Random();
for (int size = 1; size <= 10; size++) {
ArrayList<Integer> coll = new ArrayList<>(size);
for (int i = 0; i < size; i++)
coll.add(rnd.nextInt(100));
System.out.println("Median of " + coll.toString() + " is " + median(coll, comp));
} // for a range of possible input sizes
} // main(args)
} // Utility
Solution 3 - Java
If you want to use any external library here is Apache commons math library using you can calculate the Median.
For more methods and use take look at the API documentation
import org.apache.commons.math3.*;
.....
......
........
//calculate median
public double getMedian(double[] values){
Median median = new Median();
double medianValue = median.evaluate(values);
return medianValue;
}
.......
- For more on evaluate method AbstractUnivariateStatistic#evaluate
Update
Calculate in program
Generally, median is calculated using the following two formulas given here
> If n is odd then Median (M) = value of ((n + 1)/2)th item term.
If n is even then Median (M) = value of [((n)/2)th item term + ((n)/2 + 1)th item term ]/2
In your program you have numArray, first you need to sort array using Arrays#sort
Arrays.sort(numArray);
int middle = numArray.length/2;
int medianValue = 0; //declare variable
if (numArray.length%2 == 1)
medianValue = numArray[middle];
else
medianValue = (numArray[middle-1] + numArray[middle]) / 2;
Solution 4 - Java
Arrays.sort(numArray);
return (numArray[size/2] + numArray[(size-1)/2]) / 2;
Solution 5 - Java
Arrays.sort(numArray);
int middle = ((numArray.length) / 2);
if(numArray.length % 2 == 0){
int medianA = numArray[middle];
int medianB = numArray[middle-1];
median = (medianA + medianB) / 2;
} else{
median = numArray[middle + 1];
}
EDIT: I initially had medianB setting to middle+1 in the even length arrays, this was wrong due to arrays starting count at 0. I have updated it to use middle-1 which is correct and should work properly for an array with an even length.
Solution 6 - Java
You can find good explanation at https://www.youtube.com/watch?time_continue=23&v=VmogG01IjYc
The idea it to use 2 Heaps viz one max heap and mean heap.
class Heap {
private Queue<Integer> low = new PriorityQueue<>(Comparator.reverseOrder());
private Queue<Integer> high = new PriorityQueue<>();
public void add(int number) {
Queue<Integer> target = low.size() <= high.size() ? low : high;
target.add(number);
balance();
}
private void balance() {
while(!low.isEmpty() && !high.isEmpty() && low.peek() > high.peek()) {
Integer lowHead= low.poll();
Integer highHead = high.poll();
low.add(highHead);
high.add(lowHead);
}
}
public double median() {
if(low.isEmpty() && high.isEmpty()) {
throw new IllegalStateException("Heap is empty");
} else {
return low.size() == high.size() ? (low.peek() + high.peek()) / 2.0 : low.peek();
}
}
}
Solution 7 - Java
Try sorting the array first. Then after it's sorted, if the array has an even amount of elements the mean of the middle two is the median, if it has a odd number, the middle element is the median.
Solution 8 - Java
Use Arrays.sort and then take the middle element (in case the number n of elements in the array is odd) or take the average of the two middle elements (in case n is even).
public static long median(long[] l)
{
Arrays.sort(l);
int middle = l.length / 2;
if (l.length % 2 == 0)
{
long left = l[middle - 1];
long right = l[middle];
return (left + right) / 2;
}
else
{
return l[middle];
}
}
Here are some examples:
@Test
public void evenTest()
{
long[] l = {
5, 6, 1, 3, 2
};
Assert.assertEquals((3 + 4) / 2, median(l));
}
@Test
public oddTest()
{
long[] l = {
5, 1, 3, 2, 4
};
Assert.assertEquals(3, median(l));
}
And in case your input is a Collection, you might use Google Guava to do something like this:
public static long median(Collection<Long> numbers)
{
return median(Longs.toArray(numbers)); // requires import com.google.common.primitives.Longs;
}
Solution 9 - Java
I was looking at the same statistics problems. The approach you are thinking it is good and it will work. (Answer to the sorting has been given)
But in case you are interested in algorithm performance, I think there are a couple of algorithms that have better performance than just sorting the array, one (QuickSelect) is indicated by @bruce-feist's answer and is very well explained.
[Java implementation: https://discuss.leetcode.com/topic/14611/java-quick-select ]
But there is a variation of this algorithm named median of medians, you can find a good explanation on this link: http://austinrochford.com/posts/2013-10-28-median-of-medians.html
Java implementation of this:
Solution 10 - Java
I faced a similar problem yesterday. I wrote a method with Java generics in order to calculate the median value of every collection of Numbers; you can apply my method to collections of Doubles, Integers, Floats and returns a double. Please consider that my method creates another collection in order to not alter the original one. I provide also a test, have fun. ;-)
public static <T extends Number & Comparable<T>> double median(Collection<T> numbers){
if(numbers.isEmpty()){
throw new IllegalArgumentException("Cannot compute median on empty collection of numbers");
}
List<T> numbersList = new ArrayList<>(numbers);
Collections.sort(numbersList);
int middle = numbersList.size()/2;
if(numbersList.size() % 2 == 0){
return 0.5 * (numbersList.get(middle).doubleValue() + numbersList.get(middle-1).doubleValue());
} else {
return numbersList.get(middle).doubleValue();
}
}
JUnit test code snippet:
/**
* Test of median method, of class Utils.
*/
@Test
public void testMedian() {
System.out.println("median");
Double expResult = 3.0;
Double result = Utils.median(Arrays.asList(3.0,2.0,1.0,9.0,13.0));
assertEquals(expResult, result);
expResult = 3.5;
result = Utils.median(Arrays.asList(3.0,2.0,1.0,9.0,4.0,13.0));
assertEquals(expResult, result);
}
Usage example (consider the class name is Utils):
List<Integer> intValues = ... //omitted init
Set<Float> floatValues = ... //omitted init
.....
double intListMedian = Utils.median(intValues);
double floatSetMedian = Utils.median(floatValues);
Note: my method works on collections, you can convert arrays of numbers to list of numbers as pointed here
Solution 11 - Java
And nobody paying attention when list contains only one element (list.size == 1). All your answers will crash with index out of bound exception, because integer division returns zero (1 / 2 = 0). Correct answer (in Kotlin):
MEDIAN("MEDIAN") {
override fun calculate(values: List<BigDecimal>): BigDecimal? {
if (values.size == 1) {
return values.first()
}
if (values.size > 1) {
val valuesSorted = values.sorted()
val mid = valuesSorted.size / 2
return if (valuesSorted.size % 2 != 0) {
valuesSorted[mid]
} else {
AVERAGE.calculate(listOf(valuesSorted[mid - 1], valuesSorted[mid]))
}
}
return null
}
},
Solution 12 - Java
As @Bruce-Feist mentions, for a large number of elements, I'd avoid any solution involving sort if performance is something you are concerned about. A different approach than those suggested in the other answers is Hoare's algorithm to find the k-th smallest of element of n items. This algorithm runs in O(n).
public int findKthSmallest(int[] array, int k)
{
if (array.length < 10)
{
Arrays.sort(array);
return array[k];
}
int start = 0;
int end = array.length - 1;
int x, temp;
int i, j;
while (start < end)
{
x = array[k];
i = start;
j = end;
do
{
while (array[i] < x)
i++;
while (x < array[j])
j--;
if (i <= j)
{
temp = array[i];
array[i] = array[j];
array[j] = temp;
i++;
j--;
}
} while (i <= j);
if (j < k)
start = i;
if (k < i)
end = j;
}
return array[k];
}
And to find the median:
public int median(int[] array)
{
int length = array.length;
if ((length & 1) == 0) // even
return (findKthSmallest(array, array.length / 2) + findKthSmallest(array, array.length / 2 + 1)) / 2;
else // odd
return findKthSmallest(array, array.length / 2);
}
Solution 13 - Java
public static int median(int[] arr) {
int median = 0;
java.util.Arrays.sort(arr);
for (int i=0;i<arr.length;i++) {
if (arr.length % 2 == 1) {
median = Math.round(arr[arr.length/2]);
} else {
median = (arr[(arr.length/2)] + arr[(arr.length/2)-1])/2;
}
}
return median;
}
Solution 14 - Java
Check out the Arrays.sort methods:
http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html
You should also really abstract finding the median into its own method, and just return the value to the calling method. This will make testing your code much easier.
Solution 15 - Java
public int[] data={31, 29, 47, 48, 23, 30, 21
, 40, 23, 39, 47, 47, 42, 44, 23, 26, 44, 32, 20, 40};
public double median()
{
Arrays.sort(this.data);
double result=0;
int size=this.data.length;
if(size%2==1)
{
result=data[((size-1)/2)+1];
System.out.println(" uneven size : "+result);
}
else
{
int middle_pair_first_index =(size-1)/2;
result=(data[middle_pair_first_index+1]+data[middle_pair_first_index])/2;
System.out.println(" Even size : "+result);
}
return result;
}
Solution 16 - Java
package arrays;
public class Arraymidleelement {
static public double middleArrayElement(int [] arr)
{
double mid;
if(arr.length%2==0)
{
mid=((double)arr[arr.length/2]+(double)arr[arr.length/2-1])/2;
return mid;
}
return arr[arr.length/2];
}
public static void main(String[] args) {
int arr[]= {1,2,3,4,5,6};
System.out.println( middleArrayElement(arr));
}
}