Java: Convert List≪String≫ to a Join()D String

Java: convert List<String> to a join()d String

String.join

With Java 8 you can do this without any third party library.

If you want to join a Collection of Strings you can use the String.join() method:

List<String> list = Arrays.asList("foo", "bar", "baz");
String joined = String.join(" and ", list); // "foo and bar and baz"

Collectors.joining

If you have a Collection with another type than String you can use the Stream API with the joining Collector:

List<Person> list = Arrays.asList(
  new Person("John", "Smith"),
  new Person("Anna", "Martinez"),
  new Person("Paul", "Watson ")
);

String joinedFirstNames = list.stream()
  .map(Person::getFirstName)
  .collect(Collectors.joining(", ")); // "John, Anna, Paul"

The StringJoiner class may also be useful.

Best way to concatenate List of String objects?

Your approach is dependent on Java's ArrayList#toString() implementation.

While the implementation is documented in the Java API and very unlikely to change, there's a chance it could. It's far more reliable to implement this yourself (loops, StringBuilders, recursion whatever you like better).

Sure this approach may seem "neater" or more "too sweet" or "money" but it is, in my opinion, a worse approach.

Using streams to convert a list of objects into a string obtained from the toString method

One simple way is to append your list items in a StringBuilder

List<Integer> list = new ArrayList<>();
list.add(1);
list.add(2);
list.add(3);

StringBuilder b = new StringBuilder();
list.forEach(b::append);

System.out.println(b);

you can also try:

String s = list.stream().map(e -> e.toString()).reduce("", String::concat);

Explanation: map converts Integer stream to String stream, then its reduced as concatenation of all the elements.

Note: This is normal reduction which performs in O(n²)

for better performance use a StringBuilder or mutable reduction similar to F. Böller's answer.

String s = list.stream().map(Object::toString).collect(Collectors.joining(","));

Ref: Stream Reduction

Best way to convert an ArrayList to a string

Java 8 introduces a String.join(separator, list) method; see Vitalii Federenko's answer.

Before Java 8, using a loop to iterate over the ArrayList was the only option:

DO NOT use this code, continue reading to the bottom of this answer to see why it is not desirable, and which code should be used instead:

ArrayList<String> list = new ArrayList<String>();
list.add("one");
list.add("two");
list.add("three");

String listString = "";

for (String s : list)
{
    listString += s + "\t";
}

System.out.println(listString);

In fact, a string concatenation is going to be just fine, as the javac compiler will optimize the string concatenation as a series of append operations on a StringBuilder anyway. Here's a part of the disassembly of the bytecode from the for loop from the above program:

   61:  new #13; //class java/lang/StringBuilder
   64:  dup
   65:  invokespecial   #14; //Method java/lang/StringBuilder."<init>":()V
   68:  aload_2
   69:  invokevirtual   #15; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   72:  aload   4
   74:  invokevirtual   #15; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   77:  ldc #16; //String \t
   79:  invokevirtual   #15; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   82:  invokevirtual   #17; //Method java/lang/StringBuilder.toString:()Ljava/lang/String;

As can be seen, the compiler optimizes that loop by using a StringBuilder, so performance shouldn't be a big concern.

(OK, on second glance, the StringBuilder is being instantiated on each iteration of the loop, so it may not be the most efficient bytecode. Instantiating and using an explicit StringBuilder would probably yield better performance.)

In fact, I think that having any sort of output (be it to disk or to the screen) will be at least an order of a magnitude slower than having to worry about the performance of string concatenations.

Edit: As pointed out in the comments, the above compiler optimization is indeed creating a new instance of StringBuilder on each iteration. (Which I have noted previously.)

The most optimized technique to use will be the response by Paul Tomblin, as it only instantiates a single StringBuilder object outside of the for loop.

Rewriting to the above code to:

ArrayList<String> list = new ArrayList<String>();
list.add("one");
list.add("two");
list.add("three");

StringBuilder sb = new StringBuilder();
for (String s : list)
{
    sb.append(s);
    sb.append("\t");
}

System.out.println(sb.toString());

Will only instantiate the StringBuilder once outside of the loop, and only make the two calls to the append method inside the loop, as evidenced in this bytecode (which shows the instantiation of StringBuilder and the loop):

   // Instantiation of the StringBuilder outside loop:
   33:  new #8; //class java/lang/StringBuilder
   36:  dup
   37:  invokespecial   #9; //Method java/lang/StringBuilder."<init>":()V
   40:  astore_2

   // [snip a few lines for initializing the loop]
   // Loading the StringBuilder inside the loop, then append:
   66:  aload_2
   67:  aload   4
   69:  invokevirtual   #14; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   72:  pop
   73:  aload_2
   74:  ldc #15; //String \t
   76:  invokevirtual   #14; //Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
   79:  pop

So, indeed the hand optimization should be better performing, as the inside of the for loop is shorter and there is no need to instantiate a StringBuilder on each iteration.

Convert List<String> to delimited String

If you don't mind using the StringUtils library provided by apache, you could do:

// Output is "a,b,c"
StringUtils.join(["a", "b", "c"], ','); 

https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html

Join a list of object's properties into a String

To retrieve a String consisting of all the ID's separated by the delimiter "," you first have to map the Person ID's into a new stream which you can then apply Collectors.joining on.

String result = personList.stream().map(Person::getId)
                          .collect(Collectors.joining(","));

if your ID field is not a String but rather an int or some other primitive numeric type then you should use the solution below:

String result = personList.stream().map(p -> String.valueOf(p.getId()))
                          .collect(Collectors.joining(","));

---

Related Topics