Sorting List of Arrays by different elements using Java 8 lambdas

Question

If I have a List<String[]> in which each String[] is as follows: {FirstName, LastName, Income, City} how would I go about using Java 8 lambdas to sort the List by a certain value such as income or first name?

Why don't you turn the String[] into an object, e.g. Person, and have the Person class implement Comparable? — Robin Topper
– Robin Topper, Commented May 3, 2017 at 7:30
Also if income is a number putting it in a string will make it not sortable (it will sort alphabetically). — Lluis Martinez
– Lluis Martinez, Commented May 3, 2017 at 7:32

Matt Goodrich · Accepted Answer · 2017-05-03 07:38:14Z

5

Here's a couple examples. Replace x in the first two examples below with the index of the field you'd like to use for sorting.

Collections.sort(personList, (p1, p2) -> p1[x].compareTo(p2[x]));

or

personList.sort((p1, p2) -> p1[x].compareTo(p2[x]);

Also, I agree with @Robin Topper's comment. If lambdas are required (and you wanted to sort by first name), you could use:

Collections.sort(personList, (p1, p2) -> p1.getFirstName().compareTo(p2.getFirstName()));

or

personList.sort((p1, p2) -> p1.getFirstName().compareTo(p2.getFirstName()));

Also consider using the comparable implementation from Robin's comment and a data-structure allowing sorting.

edited May 3, 2017 at 7:38

answered May 3, 2017 at 7:33

Matt Goodrich

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5 Comments

Robin Topper Over a year ago

My suggested Person class would use getters and setters instead of public fields ;-)

shmosel Over a year ago

Or Comparator.comparing(Person::getFirstName).

Holger Over a year ago

Likewise, Comparator.comparing(a -> a[x]) for the array case. In either case, Comparator.comparing ensures that both sides of the comparison use the same function, which avoids copy&paste errors and such alike.

Hunter Tipton Over a year ago

Thank you it works wonderfully for the strings, is there any way I could parse the income element to an integer and sort it by that?

Hunter Tipton Over a year ago

I actually managed to figure it out. I used Collections.sort(relative, (p1, p2) -> Integer.valueOf(p2[2]).compareTo(Integer.valueOf(p1[2])));

Eugene · Accepted Answer · 2017-05-03 10:20:45Z

2

If you can rely on the order in this array, then as simple as:

List<String[]> list = Arrays.asList(new String[] { "eugene", "test", "300", "LA" }, 
                              new String[] { "hunter", "test2", "25", "CA" });

 List<String[]> sorted = list.stream()
            .sorted(Comparator.comparingLong(s -> Long.parseLong(s[2])))
            .collect(Collectors.toList());

    sorted.forEach(s -> System.out.println(Arrays.toString(s)));

But the general advice to create an Object from those fields is much better.

edited May 3, 2017 at 10:20

answered May 3, 2017 at 7:38

Eugene

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4 Comments

Robin Topper Over a year ago

You don't need to create a stream for sorting, e.g. list.sort(Comparator.comparing(s -> Long.parseLong(s[2])))

Eugene Over a year ago

@RobinTopper yup I know, the other answer shows that. thx anyway

Holger Over a year ago

List.of(…) requires Java 9. In Java 8, you have to stay with Arrays.asList(…). But if you use List.of, you can’t use List.sort, so using a stream is justified.

Holger Over a year ago

It might be beneficial to use Comparator.comparingLong(s -> Long.parseLong(s[2])), by the way, to avoid boxing overhead.

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