带有使用者或过滤器的Java8的List的条件，哪种方式更好

Question

I have tried this

    List<Integer> numbers = Arrays.asList(1, 1, 0, -1, -1);
    List<Integer> positiveNum2 = new ArrayList<>();
    List<Integer> negativeNum2 = new ArrayList<>();
    List<Integer> zeroNumbers2 = new ArrayList<>();
    List<Integer> positiveNumbers = numbers.stream().filter(number -> number > 0).collect(Collectors.toList());
    List<Integer> negativeNumbers = numbers.stream().filter(number -> number < 0).collect(Collectors.toList());
    List<Integer> zeroNumbers = numbers.stream().filter(number -> number.equals(0)).collect(Collectors.toList());
    positiveNumbers.forEach(System.out::println);
    negativeNumbers.forEach(System.out::println);
    zeroNumbers.forEach(System.out::println);
    System.out.println("*********with Consumer******************");
    Consumer<Integer> determineNumber = number -> {
        if (number > 0) {
            positiveNum2.add(number);
        } else if (number < 0) {
            negativeNum2.add(number);
        } else {
            zeroNumbers2.add(number);
        }

    };
    numbers.forEach(determineNumber);
    positiveNum2.forEach(System.out::println);
    negativeNum2.forEach(System.out::println);
    zeroNumbers2.forEach(System.out::println);

but I dont know which one is better, I think the forEach with the Consumer, but the Consumer does three validations, therefore, does not have single responsability

Answer 1

I would generally prefer your Consumer based solution as it at least encapsulates the whole operation in one call/stream. But I think you don't yet take full advantage of the functional approach of streams.

You could use a simple stream/collect to achieve this sorting:

numbers .stream()
        .collect(Collectors.groupingBy(Math::signum));

will result in a map like this:

{1.0=[1], 0.0=[0, 0], -1.0=[-1, -2, -1]}

This approach avoids side effects (ie does not modify lists outside of the stream's scope) and can hence be extracted more easily and is easily run in parallel.

Answer 2

The consumer only takes one iteration, so it's more efficient, and it does have a single responsibility: dividing the numbers over three lists based on their sign.

For this case, I'd probably use a simple for loop though.

If you really want to over-engineer things, you can define an enum to represent the sign, and use a grouping collector to group into a map:

import java.util.Arrays;
import java.util.List;
import java.util.Map;

import static java.util.stream.Collectors.groupingBy;

public class Test {
    public static void main(String[] args) {
        List<Integer> numbers = Arrays.asList(1, 1, 0, -1, -1);

        Map<Sign, List<Integer>> map = numbers
                .stream()
                .collect(groupingBy(i -> i > 0
                                    ? Sign.POSITIVE
                                    : i < 0 
                                        ? Sign.NEGATIVE
                                        : Sign.ZERO));

        System.out.println(map);
    }
}

enum Sign {POSITIVE, NEGATIVE, ZERO}

This yields the following output:

{ZERO=[0], POSITIVE=[1, 1], NEGATIVE=[-1, -1]}

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^{Note: If you want peak performance for future lookups in your map, you can use an EnumMap instead. Take a look at this answer to see how.}