Question

I am attempting to find the smallest and largest numbers, stored as doubles, in an Array. This is the array:

double[] heightArray = {1.4,1.9,1.31,1.2};

In attempting to find the smallest and largest element, I have used the following method:

public static void studentHeights(double[] heightArray) {
    double total = 0;
    double average;
    double maximum = 0;
    double minimum = 0;

    for (int loop = 0; loop < heightArray.length; loop++) {
        System.out.println(heightArray[loop]);
        total += heightArray[loop];
        if (heightArray[loop] > maximum) {
            maximum = heightArray[loop];
        }
        if (heightArray[loop] < minimum) {
            minimum = heightArray[loop];
        }

    }
    average = total / heightArray.length;
    System.out.println("The maximum height is: " + maximum);
    System.out.println("The minimum height is: " + minimum);
    System.out.println("The average height is: " + average);

}

The console outputs:

1.4
1.9
1.31
1.2
The maximum height is: 1.9
The minimum height is: 0.0
The average height is: 1.4525

The largest (maximum) element is correct, as is the average. However, the smallest (minimum) element is being recorded as 0.0. This must be because none of the elements are less than 0, but I can't think how I can calculate the smallest element another way.

Answer 1

Set your minimum to be something very large instead of 0

double minimum = 0;

Such as the largest possible value, conveniently defined in a constant.

double minimum = Double.MAX_VALUE;

You may want to use the MIN_VALUE when initializing the maximum value

double maximum = Double.MIN_VALUE;

Answer 2

You can initialize the minimum variable with the first element of the array instead of the value 0,

for example

double minimum = heightArray[0];

If there are negative numbers in the array your maximum logic will not work as excepted, consider initialize the maximum variable also

double maximum = heightArray[0];

Answer 3

If you are interested in an alternative method you can use Collections.max() and Collections.min() for this task:

List<Double> lst = Arrays.asList(heightArray); // array to list
System.out.println(Collections.min(lst)); // print max value
System.out.println(Collections.max(lst)); // print min value
System.out.println(lst.stream().mapToDouble(e->e.doubleValue()).average().getAsDouble()); //average

Why Collections.max() or Collections.min() ? Because it can be used in any type of primitive or Wrapper Class Objects.

Links:

Answer 4

In Java 8 be more easier, you have multiple choices to go with

Option 1

double max = Arrays.stream(heightArray).max().getAsDouble();
double min = Arrays.stream(heightArray).min().getAsDouble();
double avg = Arrays.stream(heightArray).average().getAsDouble();

Option 2

You can use DoubleStream.of(double... values)

double max = DoubleStream.of(heightArray).max().getAsDouble();
double min = DoubleStream.of(heightArray).min().getAsDouble();
double avg = DoubleStream.of(heightArray).average().getAsDouble();

Option 3

You can use DoubleSummaryStatistics

DoubleSummaryStatistics stats = DoubleStream.of(heightArray).summaryStatistics();

double min = stats.getMin();
double max = stats.getMax();
double avg = stats.getAverage();

Answer 5

Or you could just sort() your array:

// Initialize unsorted double array.
double HeightsArray[] = {1.4, 1.9, 1.31, 1.2};

// Sorting the array.
Arrays.sort(HeightsArray);

System.out.println("Smallest number: " + HeightsArray[0]);
System.out.println("Largest number: " + HeightsArray[HeightsArray.length - 1]);

Will output:

Smallest number: 1.2
Largest number: 1.9

Answer 6

The problem, as you've already identified, is that you want to start your iteration with a value larger than any other value in your array. Ideally, we would start with the largest value available to store in a double. Fortunately, we have just such a value: Double.MAX_VALUE.

Likewise, you could change your maximum search to start with - Double.MAX_VALUE so that your function continues working if passed a list of entirely negative numbers.

Finding min, max and average of elements in an Array

6 个答案: