Sort a list of strings in java in alphabetical order

Question

I have a list with some strings in this format:

List<String> ids = new ArrayList<>();

ids.add("B-7");
ids.add("B-5");
ids.add("A-3");
ids.add("B-8");
ids.add("B-1");
ids.add("B-6");
ids.add("B-2");
ids.add("B-3");
ids.add("B-10");
ids.add("A-1");
ids.add("B-4");
ids.add("B-9");
ids.add("A-2");

I need to sort it to have this output, (iterating over the list):

A-1
A-2
A-3
B-1
B-2
B-3
B-4
B-5
B-6
B-7
B-8
B-9
B-10

I am using:

List<String> sortedIds = ids.stream().sorted().collect(Collectors.toList());

But instead, my output:

A-1
A-2
A-3
B-1
B-10   -- Error
B-2
B-3
B-4
B-5
B-6
B-7
B-8
B-9

This is where the documentation comes in handy. If you look at [`Stream#sorted()`](https://docs.oracle.com/en/java/javase/15/docs/api/java.base/java/util/stream/Stream.html#sorted()), notice that it says "sorted according to natural order." As you've demonstrated, you don't want natural order, you want to sort it a different way. Java is way ahead of you; it's got an [overloaded method](https://docs.oracle.com/en/java/javase/15/docs/api/java.base/java/util/stream/Stream.html#sorted(java.util.Comparator)) which lets you specify a custom Comparator. — Charlie Armstrong, Apr 09 '21 at 22:41
[Sort on a string that may contain a number](https://stackoverflow.com/q/104599/3890632) — khelwood, Apr 09 '21 at 22:43

Unmitigated · Accepted Answer · 2021-04-09T22:48:45.287

5

You can create a custom Comparator using Comparator.comparing and Comparator.thenComparing.

List<String> sortedIds = ids.stream().sorted(
    Comparator.comparing((String s) -> s.substring(0, s.indexOf('-')))
    .thenComparingInt(s -> Integer.parseInt(s.substring(s.indexOf('-') + 1))))
    .collect(Collectors.toList());

edited Apr 09 '21 at 22:48

answered Apr 09 '21 at 22:46

Unmitigated

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1

By the way, that last line can be [`.toList()`](https://docs.oracle.com/en/java/javase/16/docs/api/java.base/java/util/stream/Stream.html#toList()) in Java 16 and later. – Basil Bourque Apr 10 '21 at 05:44

score 4 · Answer 2 · answered Apr 09 '21 at 22:40

The default Comparator is going to operate in lexicographical order. You need to compare the String parts and Integer parts separately. Something like

Collections.sort(ids, (a, b) -> {
    String[] at = a.split("-");
    String[] bt = b.split("-");
    int c = at[0].compareTo(bt[0]);
    if (c != 0) {
        return c;
    }
    return Integer.valueOf(Integer.parseInt(at[1])).compareTo(Integer.parseInt(bt[1]));
});

Yassin Hajaj · Answer 3 · 2021-04-09T22:42:59.383

What you'll need is a customer Comparator<String> to use inside of the sorted() intermediate operation

List<String> sorted = ids.stream().sorted((o1, o2) -> {
    String[] first = o1.split("-");
    String[] second = o2.split("-");

    int lettersComparison = first[0].compareTo(second[0]);
    if (lettersComparison != 0) {
        return lettersComparison;
    }

    Integer firstNumber = Integer.valueOf(first[1]);
    Integer secondNumber = Integer.valueOf(second[1]);
    return firstNumber.compareTo(secondNumber);
}).toList();

Which outputs

[A-1, A-2, A-3, B-1, B-2, B-3, B-4, B-5, B-6, B-7, B-8, B-9, B-10]

That said, if you want to just sort the existing List, I would suggest not to go through a Stream for that because it has an overhead in terms of performance and creation of new object.

You can use list.sort(comparator) and use the same Comparator<String> as hereabove

ids.sort((o1, o2) -> {
    String[] first = o1.split("-");
    String[] second = o2.split("-");

    int lettersComparison = first[0].compareTo(second[0]);
    if (lettersComparison != 0) {
        return lettersComparison;
    }

    Integer firstNumber = Integer.valueOf(first[1]);
    Integer secondNumber = Integer.valueOf(second[1]);
    return firstNumber.compareTo(secondNumber);
});

score 0 · Answer 4 · answered Apr 09 '21 at 22:47

0

Try this.

List<String> ids = Arrays.asList(
    "B-7", "B-5", "A-3", "B-8", "B-1", "B-6", "B-2",
    "B-3", "B-10", "A-1", "B-4", "B-9", "A-2");

Collections.sort(ids,
    Comparator.comparingInt(String::length)
    .thenComparing(Function.identity()));

System.out.println(ids);

output

[A-1, A-2, A-3, B-1, B-2, B-3, B-4, B-5, B-6, B-7, B-8, B-9, B-10]

answered Apr 09 '21 at 22:47

Very smart ! But what if `B-05` ? – Yassin Hajaj Apr 09 '21 at 22:50

score 0 · Answer 5 · answered Apr 09 '21 at 23:36

Here is yet another way. I tend to split first and then glue together after sorting rather than continually splitting inside a comparator.

List<String> result = ids.stream().map(s -> s.split("-"))
        .sorted(Comparator.comparing((String[] a) -> a[0])
                .thenComparingInt(
                        a -> Integer.parseInt(a[1])))
        .map(a -> String.join(",", a))
        .collect(Collectors.toList());

Sort a list of strings in java in alphabetical order

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