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.
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11 Answers
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.
answered Jun 12, 2011 at 21:39
Vineet Reynolds
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2 Comments
Line
What is the advantage of using StringBuilder and than toString method instead of simply use String class?
Stephen C
@Line - it is more efficient to use a
StringBuilder. Appending N characters using naive string concatenation is O(N^2). And if you are talking about using ArrayList.toString() and filtering, the filtering step is another O(N) pass over the string ... compared with single pass approach ^^^.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.
1 Comment
Abhishek Thapliyal
In my case i have to typecast String str = (String)(chars.stream().map(e->e.toString()).collect(Collectors.joining()));
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.
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));
answered Jun 12, 2011 at 21:40
Tomasz Nurkiewicz
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2 Comments
mre
why use a third-party library when the standard library will suffice?
Tomasz Nurkiewicz
Is there a standard method that does the same thing as
ArrayUtils#toPrimitive? I know the problem can be solved with different approaches, I am asking about this specific method which I have chosen to avoid explicit loop. Besides, I can't imagine no commons-lang on CLASSPATH nowadays...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());
Comments
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.
Comments
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\\[\\]]", "");
1 Comment
shmosel
This is a terrible idea.
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());
Comments
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"
1 Comment
Ry-
Doesn’t work if any characters are
, or space.I would say :
public String arayListToString(ArrayList arrayList){
StringBuffer b = new StringBuffer();
for(String s : arrayList){
b.append(s);
b.append(",");
}
return b.toString();
}
Comments
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);
}
1 Comment
Northern Poet
This is not an answer to the question