Convert a list of string into a comma separated string without duplications

Question

I have a list of strings in Java. I would like to to insert it into a string with commas, something like:

Input: [option1,option2,option3,option1,option4]
Output: option1,option2,option3,option4

I saw this topic and I really liked this one:

        StringBuilder result = new StringBuilder();
        String delimiter= "";
        for (String i : list) {
            result.append(delimiter).append(i);
            delimiter = ",";
        }
        return result.toString();

But what would be the best way to add support for unique only items? How can I return a string without any duplicates? I could remove duplication from list first but it feels like there is a better solution here.

Would be nice if that restriction was mentioned in the question. — Andreas
– Andreas, Commented Jun 1, 2020 at 18:12

azro · Accepted Answer · 2020-06-01 17:45:16Z

3

The best to avoid duplicates is to use a Set

TreeSet for alphabetical order
LinkedHashSet to keep insertion order (initial order)

StringBuilder result = new StringBuilder();
String delimiter= "";
for (String i : new LinkedHashSet<String>(list)) {
    result.append(delimiter).append(i);
    delimiter = ",";
}
return result.toString();

But you can just do String.join(',', new LinkedHashSet<String>(list))

Back to List

List<String> inputList = (unique) ? new ArrayList<>(new HashSet<>(list)) : list;

Case-insensitive

Set<String> check = new HashSet<String>();
StringBuilder result = new StringBuilder();
String delimiter= "";
for (String i : list) {
    if(!check.contains(i.toLowerCase())){
        result.append(delimiter).append(i);
        delimiter = ",";
        check.add(i.toLowerCase());
    }
}
return result.toString();

edited Jun 1, 2020 at 17:45

answered Jun 1, 2020 at 17:23

azro

54.2k9 gold badges39 silver badges75 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Andreas Over a year ago

Reordering the values is probably not appropriate.

abuka123 Over a year ago

Using HashSet will sort alphabetically the values?

abuka123 Over a year ago

Also, I'm trying to add support to a flag. I would use something like List<String> inputList = (unique) ? new HashSet<String>(list) : list; But it looks like I can't convert HashSet to List. Is there a way to easily support the unique flag?

Andreas Over a year ago

Using LinkedHashSet would retain original order (without using "external set"). Using TreeSet would order the values alphabetically.

Andreas Over a year ago

@abuka123 Collection<String> inputList = (unique ? new LinkedHashSet<>(list) : list); --- No need to copy the values twice to get a List. The for loop works for Collections. It actually works for anything Iterable.

|

deHaar · Accepted Answer · 2020-06-01 17:31:03Z

0

You can stream() the list and collect only distinct values. Afterwards, you can just String.join(...) them to a comma separated String, like this:

public static void main(String[] args) {
    // example values
    List<String> list = new ArrayList<>();
    list.add("option1");
    list.add("option2");
    list.add("option3");
    list.add("option4");
    list.add("option1");
    list.add("option3");
    list.add("option5");

    // stream the list and collect only distinct values in a new list
    List<String> distinctOptions = list.stream()
                                        .distinct()
                                        .collect(Collectors.toList());
    // then create a comma separated String from that list
    String commaSeparatedUniqueStrings = String.join(",", distinctOptions);
    // and print it
    System.out.println(commaSeparatedUniqueStrings);
}

The output is

option1,option2,option3,option4,option5

answered Jun 1, 2020 at 17:31

deHaar

18.7k11 gold badges48 silver badges57 bronze badges

2 Comments

Andreas Over a year ago

Why build a list? Why not just use the stream to build the desired result string? See my comment.

deHaar Over a year ago

@Andreas this is just an example and may of course be optimized... Sure, there's no real need for an intermediate list, but it makes the example more readable and easier to understand. But it is not optimal, I admit.

Community · Accepted Answer · 2020-06-20 09:12:55Z

Extra requirements from comments:

Limited to Java 7

Support unique flag

Handle case-sensitive

Optional: Alphabetical sort may be allowed

Sorted

If sorting the values is ok, handling case-sensitive is easiest done using a TreeSet with a case-insensitive Comparator, like the String.CASE_INSENSITIVE_ORDER, or by using a Collator for full language support.

static String toString(List<String> list, boolean unique) {
    Collection<String> dataToProcess = list;
    if (unique) {
        dataToProcess = new TreeSet<>(String.CASE_INSENSITIVE_ORDER);
        dataToProcess.addAll(list);
    }
    
    StringBuilder result = new StringBuilder();
    String delimiter= "";
    for (String s : dataToProcess) {
        result.append(delimiter).append(s);
        delimiter = ",";
    }
    return result.toString();
}

Test

List<String> list = Arrays.asList("option1", "option3", "OPTION1", "OPTION2", "option4", "option2");
System.out.println(toString(list, false));
System.out.println(toString(list, true));

Output

option1,option3,OPTION1,OPTION2,option4,option2
option1,OPTION2,option3,option4

Notice how the "unique" result is sorted, but the non-unique result is not.

Using Collator

For better language support, use a Collator:

    if (unique) {
        Collator collator = Collator.getInstance(/*Locale.GERMANY*/);
        collator.setStrength(Collator.SECONDARY);
        dataToProcess = new TreeSet<>(collator);
        dataToProcess.addAll(list);
    }

Unsorted

To do it without sorting the values, we'll keep using the TreeSet but build a new list.

    if (unique) {
        Collator collator = Collator.getInstance(/*Locale.GERMANY*/);
        collator.setStrength(Collator.SECONDARY);
        Set<String> set = new TreeSet<>(collator);
        dataToProcess = new ArrayList<>();
        for (String s : list)
            if (set.add(s))
                dataToProcess.add(s);
    }

Output

option1,option3,OPTION1,OPTION2,option4,option2
option1,option3,OPTION2,option4

Collectives™ on Stack Overflow

Convert a list of string into a comma separated string without duplications

3 Answers 3

8 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

8 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related