Best way to convert an ArrayList to a string

Question

I have an ArrayList that I want to output completely as a String. Essentially I want to output it in order using the toString of each element separated by tabs. Is there any fast way to do this? You could loop through it (or remove each element) and concatenate it to a String but I think this will be very slow.

Answer 1

In Java 8 or later:

String listString = String.join(", ", list);

In case the list is not of type String, a joining collector can be used:

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

Answer 2

If you happen to be doing this on Android, there is a nice utility for this called TextUtils which has a .join(String delimiter, Iterable) method.

List<String> list = new ArrayList<String>();
list.add("Item 1");
list.add("Item 2");
String joined = TextUtils.join(", ", list);

Obviously not much use outside of Android, but figured I'd add it to this thread...

Answer 3

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.

Answer 4

Download the Apache Commons Lang and use the method

 StringUtils.join(list)

 StringUtils.join(list, ", ") // 2nd param is the separator.

You can implement it by yourself, of course, but their code is fully tested and is probably the best possible implementation.

I am a big fan of the Apache Commons library and I also think it's a great addition to the Java Standard Library.

Answer 5

This is a pretty old question, but I figure I might as well add a more modern answer - use the Joiner class from Guava:

String joined = Joiner.on("\t").join(list);

Answer 6

Changing List to a readable and meaningful String is really a common question that every one may encounter.

Case 1. If you have apache's StringUtils in your class path (as from rogerdpack and Ravi Wallau):

import org.apache.commons.lang3.StringUtils;
String str = StringUtils.join(myList);

Case 2 . If you only want to use ways from JDK(7):

import java.util.Arrays;
String str = Arrays.toString(myList.toArray()); 

Just never build wheels by yourself, dont use loop for this one-line task.

Answer 7

If you were looking for a quick one-liner, as of Java 5 you can do this:

myList.toString().replaceAll("\\[|\\]", "").replaceAll(", ","\t")

Additionally, if your purpose is just to print out the contents and are less concerned about the "\t", you can simply do this:

myList.toString()

which returns a string like

[str1, str2, str3]

If you have an Array (not ArrayList) then you can accomplish the same like this:

 Arrays.toString(myList).replaceAll("\\[|\\]", "").replaceAll(", ","\t")

Answer 8

Loop through it and call toString. There isn't a magic way, and if there were, what do you think it would be doing under the covers other than looping through it? About the only micro-optimization would be to use StringBuilder instead of String, and even that isn't a huge win - concatenating strings turns into StringBuilder under the covers, but at least if you write it that way you can see what's going on.

StringBuilder out = new StringBuilder();
for (Object o : list)
{
  out.append(o.toString());
  out.append("\t");
}
return out.toString();

Answer 9

Most Java projects often have apache-commons lang available. StringUtils.join() methods is very nice and has several flavors to meet almost every need.

public static java.lang.String join(java.util.Collection collection,
                                    char separator)


public static String join(Iterator iterator, String separator) {
    // handle null, zero and one elements before building a buffer 
    Object first = iterator.next();
    if (!iterator.hasNext()) {
        return ObjectUtils.toString(first);
    }
    // two or more elements 
    StringBuffer buf = 
        new StringBuffer(256); // Java default is 16, probably too small 
    if (first != null) {
        buf.append(first);
    }
    while (iterator.hasNext()) {
        if (separator != null) {
            buf.append(separator);
        }
        Object obj = iterator.next();
        if (obj != null) {
            buf.append(obj);
        }
    }
    return buf.toString();
}

Parameters:

collection - the Collection of values to join together, may be null

separator - the separator character to use

Returns: the joined String, null if null iterator input

Since: 2.3

Answer 10

Android has a TextUtil class you can use:

String implode = TextUtils.join("\t", list);

Answer 11

For this simple use case, you can simply join the strings with comma. If you use Java 8:

String csv = String.join("\t", yourArray);

otherwise commons-lang has a join() method:

String csv = org.apache.commons.lang3.StringUtils.join(yourArray, "\t");

Answer 12

In Java 8 it's simple. See example for list of integers:

String result = Arrays.asList(1,2,3).stream().map(Object::toString).reduce((t, u) -> t + "\t" + u).orElse("");

Or multiline version (which is simpler to read):

String result = Arrays.asList(1,2,3).stream()
    .map(Object::toString)
    .reduce((t, u) -> t + "\t" + u)
    .orElse("");

Update - a shorter version

String result = Arrays.asList(1,2,3).stream()
                .map(Object::toString)
                .collect(Collectors.joining("\t"));

Answer 13

An elegant way to deal with trailing separation characters is to use Class Separator

StringBuilder buf = new StringBuilder();
Separator sep = new Separator("\t");
for (String each: list) buf.append(sep).append(each);
String s = buf.toString();

The toString method of Class Separator returns the separater, except for the first call. Thus we print the list without trailing (or in this case) leading separators.

Answer 14

In case you happen to be on Android and you are not using Jack yet (e.g. because it's still lacking support for Instant Run), and if you want more control over formatting of the resulting string (e.g. you would like to use the newline character as the divider of elements), and happen to use/want to use the StreamSupport library (for using streams on Java 7 or earlier versions of the compiler), you could use something like this (I put this method in my ListUtils class):

public static <T> String asString(List<T> list) {
    return StreamSupport.stream(list)
            .map(Object::toString)
            .collect(Collectors.joining("\n"));
}

And of course, make sure to implement toString() on your list objects' class.

Answer 15

It's an O(n) algorithm either way (unless you did some multi-threaded solution where you broke the list into multiple sublists, but I don't think that is what you are asking for).

Just use a StringBuilder as below:

StringBuilder sb = new StringBuilder();

for (Object obj : list) {
  sb.append(obj.toString());
  sb.append("\t");
}

String finalString = sb.toString();

The StringBuilder will be a lot faster than string concatenation because you won't be re-instantiating a String object on each concatenation.

Answer 16

In one Line : From [12,0,1,78,12] to 12 0 1 78 12

String srt= list.toString().replaceAll("\\[|\\]|,","");

Answer 17

List<String> stringList = getMyListOfStrings();
StringJoiner sj = new StringJoiner(" ");
stringList.stream().forEach(e -> sj.add(e));
String spaceSeparated = sj.toString()

You pass to the new StringJoiner the char sequence you want to be used as separator. If you want to do a CSV: new StringJoiner(", ");

Answer 18

If you don't want the last \t after the last element, you have to use the index to check, but remember that this only "works" (i.e. is O(n)) when lists implements the RandomAccess.

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

StringBuilder sb = new StringBuilder(list.size() * apprAvg); // every apprAvg > 1 is better than none
for (int i = 0; i < list.size(); i++) {
    sb.append(list.get(i));
    if (i < list.size() - 1) {
        sb.append("\t");
    }
}
System.out.println(sb.toString());

Answer 19

If you're using Eclipse Collections, you can use the makeString() method.

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

Assert.assertEquals(
    "one\ttwo\tthree",
    ArrayListAdapter.adapt(list).makeString("\t"));

If you can convert your ArrayList to a FastList, you can get rid of the adapter.

Assert.assertEquals(
    "one\ttwo\tthree",
    FastList.newListWith("one", "two", "three").makeString("\t"));

Note: I am a committer for Eclipse Collections.

Answer 20

The below code may help you,

List list = new ArrayList();
list.add("1");
list.add("2");
list.add("3");
String str = list.toString();
System.out.println("Step-1 : " + str);
str = str.replaceAll("[\\[\\]]", "");
System.out.println("Step-2 : " + str);

Output:

Step-1 : [1, 2, 3]
Step-2 : 1, 2, 3

Answer 21

May not be the best way, but elegant way.

Arrays.deepToString(Arrays.asList("Test", "Test2")

import java.util.Arrays;

    public class Test {
        public static void main(String[] args) {
            System.out.println(Arrays.deepToString(Arrays.asList("Test", "Test2").toArray()));
        }
    }

Output

[Test, Test2]

Answer 22

Would the following be any good:

List<String> streamValues = new ArrayList<>();
Arrays.deepToString(streamValues.toArray()));   

Answer 23

For seperating using tabs instead of using println you can use print

ArrayList<String> mylist = new ArrayList<String>();

mylist.add("C Programming");
mylist.add("Java");
mylist.add("C++");
mylist.add("Perl");
mylist.add("Python");

for (String each : mylist)
{       
    System.out.print(each);
    System.out.print("\t");
}

Answer 24

I see quite a few examples which depend on additional resources, but it seems like this would be the simplest solution: (which is what I used in my own project) which is basically just converting from an ArrayList to an Array and then to a List.

    List<Account> accounts = new ArrayList<>();

   public String accountList() 
   {
      Account[] listingArray = accounts.toArray(new Account[accounts.size()]);
      String listingString = Arrays.toString(listingArray);
      return listingString;
   }

Answer 25

You can use a Regex for this. This is as concise as it gets

System.out.println(yourArrayList.toString().replaceAll("\\[|\\]|[,][ ]","\t"));

Answer 26

This is quite an old conversation by now and apache commons are now using a StringBuilder internally: http://commons.apache.org/lang/api/src-html/org/apache/commons/lang/StringUtils.html#line.3045

This will as we know improve performance, but if performance is critical then the method used might be somewhat inefficient. Whereas the interface is flexible and will allow for consistent behaviour across different Collection types it is somewhat inefficient for Lists, which is the type of Collection in the original question.

I base this in that we are incurring some overhead which we would avoid by simply iterating through the elements in a traditional for loop. Instead there are some additional things happening behind the scenes checking for concurrent modifications, method calls etc. The enhanced for loop will on the other hand result in the same overhead since the iterator is used on the Iterable object (the List).

Answer 27

How about this function:

public static String toString(final Collection<?> collection) {
    final StringBuilder sb = new StringBuilder("{");
    boolean isFirst = true;
    for (final Object object : collection) {
        if (!isFirst)
            sb.append(',');
        else
            isFirst = false;
        sb.append(object);
    }
    sb.append('}');
    return sb.toString();
}

it works for any type of collection...