phoenix
phoenix

Reputation: 3621

Converting ArrayList of Characters to a String?

How to convert an ArrayList<Character> to a String in Java?

The List.toString method returns it as [a,b,c] string - I want to get rid of the brackets (etcetera) and store it as abc.

Upvotes: 43

Views: 94829

Answers (11)

Angelin Nadar
Angelin Nadar

Reputation: 9300

I consider this an easy and smart way

 // given list containing the chars
 List<Character> arr = Arrays.asList('a', 'b', 'c'); 
 
//convert list to string
 String output = arr.toString().replaceAll("[ ,]","")
                       .substring(1, tmpArr.length()-1);

Explanation:

// convert to string.
String str = arr.toString(); // result is "[a ,b ,c]"

// get rid of the start and the end char i.e '[' & ']'
str = str.substring(1, tmpArr.length()-1);  //result is "a ,b ,c"

// replace <space> and ','
str = str.replaceAll(" ,","") "ABC"

Upvotes: -1

Vineet Reynolds
Vineet Reynolds

Reputation: 76719

You can iterate through the list and create the string.

String getStringRepresentation(ArrayList<Character> list)
{    
    StringBuilder builder = new StringBuilder(list.size());
    for(Character ch: list)
    {
        builder.append(ch);
    }
    return builder.toString();
}

Setting the capacity of the StringBuilder to the list size is an important optimization. If you don't do this, some of the append calls may trigger an internal resize of the builder.

As an aside, toString() returns a human-readable format of the ArrayList's contents. It is not worth the time to filter out the unnecessary characters from it. It's implementation could change tomorrow, and you will have to rewrite your filtering code.

Upvotes: 42

waggledans
waggledans

Reputation: 1281

Here a possible one-line solution using Java8 streams.

a) List of Character objects to String :

String str = chars.stream()
                  .map(e->e.toString())
                  .reduce((acc, e) -> acc  + e)
                  .get();

b) array of chars (char[] chars)

String str = Stream.of(chars)
                   .map(e->new String(e))
                   .reduce((acc, e) -> acc  + e)
                   .get();

UPDATE (following comment below):

a) List of Character objects to String :

String str = chars.stream()
                  .map(e->e.toString())
                  .collect(Collectors.joining());

b) array of chars (char[] chars)

String str = Stream.of(chars)
                   .map(e->new String(e))
                   .collect(Collectors.joining());

Note that the map(e->e.toString()) step in the above solutions will create a temporary string for each character in the list. The strings immediately become garbage. So, if the performance of the conversion is a relevant concern, you should consider using the StringBuilder approach instead.

Upvotes: 30

hahahaha123567
hahahaha123567

Reputation: 61

a tiny complement to @waggledans 's answer

a) List of Character objects to String :

String str = chars.stream().map(e->e.toString()).collect(Collectors.joining());

which e->e.toString() can be replaced by Object::toString

String str = chars.stream().map(Object::toString).collect(Collectors.joining());

Upvotes: 0

Muhammad Soliman
Muhammad Soliman

Reputation: 23866

Using join of a Joiner class:

// create character list and initialize 
List<Character> arr = Arrays.asList('a', 'b', 'c');   
String str = Joiner.on("").join(arr);
System.out.println(str);

Use toString then remove , and spaces

import com.google.common.base.Joiner; 

....
<Character> arr = Arrays.asList('h', 'e', 'l', 'l', 'o'); 
// remove [] and spaces 
String str = arr.toString() 
          .substring(1, 3 * str.size() - 1) //3 bcs of commas ,
          .replaceAll(", ", ""); 
System.out.println(str);

Or by using streams:

import java.util.stream.Collectors; 
...
// using collect and joining() method 
String str =  arr.stream().map(String::valueOf).collect(Collectors.joining()); 

Upvotes: 1

amit mannur
amit mannur

Reputation: 21

 private void countChar() throws IOException {
    HashMap hashMap = new HashMap();
    List list = new ArrayList();
    list = "aammit".chars().mapToObj(r -> (char) r).collect(Collectors.toList());
    list.stream().forEach(e -> {
        hashMap.computeIfPresent(e, (K, V) -> (int) V + 1);
        hashMap.computeIfAbsent(e, (V) -> 1);
    });

    System.out.println(hashMap);

}

Upvotes: -3

Leonid Sivko
Leonid Sivko

Reputation: 65

You can do it using toString() and RegExp without any loops and streams:

List<Character> list = Arrays.asList('a', 'b', 'c'); String s = list.toString().replaceAll("[,\\s\\[\\]]", "");

Upvotes: 0

Shiva
Shiva

Reputation: 727

How about this, Building the list

List<Character> charsList = new ArrayList<Character>();
charsList.add('h');
charsList.add('e');
charsList.add('l');
charsList.add('l');
charsList.add('o');

Actual code to get String from List of Character:

String word= new String();
for(char c:charsList){
word= word+ c; 
}
System.out.println(word);

Still learning if there is a misake point out.

Upvotes: 2

Lucas Kauffman
Lucas Kauffman

Reputation: 6891

I would say :

public String arayListToString(ArrayList arrayList){

StringBuffer b = new StringBuffer();

for(String s : arrayList){
   b.append(s);
   b.append(",");
}

return b.toString();
}

Upvotes: -2

Tomasz Nurkiewicz
Tomasz Nurkiewicz

Reputation: 340963

Assuming you have a following list:

final ArrayList<Character> charsList = new ArrayList<Character>();
charsList.add('h');
charsList.add('e');
charsList.add('l');
charsList.add('l');
charsList.add('o');

This will yield hello (I am using org.apache.commons.lang.ArrayUtils helper class):

final Character[] charactersArray =
    charsList.toArray(new Character[charsList.size()]);
final char[] charsArray = ArrayUtils.toPrimitive(charactersArray);
System.out.println(String.valueOf(charsArray));

Upvotes: 1

mkro
mkro

Reputation: 1892

Many solutions available. You can iterate over the chars and append to a StringBuilder, then when finished appending, call .toString on the StringBuilder.

Or use something like commons-lang StringUtils.join from the apache commons-lang project.

Upvotes: 0

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