Average of even numbers gives back wrong value

Question

I am trying to get closer to understanding streams thus I'm doing some basic exercises. In this one, I'd like to calculate the average of the odd numbers. I wrote this algorithm to do so, but it gives back an incorrect result (8.0). I've tried to debug it but I couldn't find what it does actually.

List<Integer> numbers = Arrays.asList(1, 3, -2, -4, -7, -3, -8, 12, 19, 6, 9, 10, 14);

OptionalDouble result = numbers.stream()
                               .filter(i -> i % 2 == 1)
                               .mapToDouble(i -> i).average();
if (result.isPresent()) {
   System.out.println(result);
} else {
   System.out.println("Error");
}

What is my code doing now? How should I fix it to do what it's supposed to do?

your title says you want an average of even numbers and your description says the opposite. what do you want? — Ousmane D., Jan 02 '19 at 14:29

score 11 · Accepted Answer · answered Jan 02 '19 at 14:30

(i -> i % 2 == 1)

This is only true for positive odd numbers, because in Java the % operator returns a negative number (or zero) if its first operand is negative.

If you want to retain only even numbers, it should be:

(i -> i % 2 == 0)

If you want to retain all odd numbers (positive and negative), you can use:

(i -> i % 2 != 0)

Naman · Answer 2 · 2019-01-02T14:46:35.093

3

What is my code doing now?

You're performing a modulo operation on negative number.

How should I fix it to do what it's supposed to do?

You can use Math.abs to validate the absolute value is odd or not in your list:

OptionalDouble result = numbers.stream()
                .filter(i -> Math.abs(i) % 2 == 1) // verify id the absolute value is odd
                .mapToDouble(i->i).average();

edited Jan 02 '19 at 14:46

answered Jan 02 '19 at 14:32

Naman

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Ousmane D. · Answer 3 · 2019-01-02T15:11:57.033

In addition to what @khelwood mentioned regarding modulo with negative numbers.

It's important to know that, the filter intermediate operation does not remove elements nor does any stream operation, instead, filter returns a new stream in which all the elements satisfying the provided predicate i.e. i -> i % 2 == 1 are present.

i -> i % 2 == 1 is saying "only keep the element represented as i if it's an odd number".

if you want even then you should do i -> i % 2 == 0, reads as "only keep the element represented as i is it's an even number".

On another note, if you're running on JDK9 you can use ifPresentOrElse to simply the isPresent() check.

 numbers.stream()
        .filter(i -> i % 2 == 0)
        .mapToDouble(i -> i)
        .average()
        .ifPresentOrElse(System.out::println,
                        () -> System.out.println("Error"));

Average of even numbers gives back wrong value

3 Answers3