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Best way to convert an ArrayList to a string

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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.

or remove each element Be careful, removing elements from an ArrayList list is a big NO-NO, as your "loop" will take quadratic time due to shifting elements (provided that you loop from first to last element).

V
Vitalii Fedorenko

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(", "));

I can't believe it took until Java 8 to add this.
This answer should be up-voted more! No hacks, no libraries & no loops.
this doesn't work for what the OP said. String.join takes as second argument Iterable. So it works if you have a List that you want to display, but if you have List it will not call toString() in it as wanted by the OP
Why the hell this is not the first (and accepted) answer.. I always have to scroll down until I know it's there.. elegant and concise
Requires Android API 26+
J
JJ Geewax

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...


Since TextUtils.join takes an Object[], I'd assume that it's calling toString() on each token. Does PatientDetails implement toString()? If that doesn't solve the problem, you may be stuck doing something with the Separator class.
org.apache.lang3.StringUtils does virtually the same, so your answer was absolutely of much use to me outside of Android :)
The popularity of this answer (currently the highest voted one) shows you that many java questions stem from Android development
I can bet you were sent to this Earth to solve my problem :)
You saved my life. <3
m
mic

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.


Consider using a StringBuilder class instead of just appending strings, for performance reasons.
THe compiler will optimize the string concatenation but you will be creating a new StringBuilder object each time the loop is executed.
-1 for the use of +=, in this case it is a performance killer. Always use StringBuilder for repeated appending to a string.
This solution adds an extra "\t" at the end of the string which is often undesired, especially when you want to create a comma separated list or similar. I think the other solutions posted here, using Apache Commons or Guava, are superior.
This is not the best answer. Look below at Vitalii for Java or Geewax for Android
R
Ravi Wallau

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.


Unfortunately, the denizens of SO prefer to reinvent the wheel.
Also in Apache Commons Lang. Usage would look like StringUtils.join(list.toArray(),"\t")
This particular wheel is hardly worth an extra dependency.
@SevaAlekseyev I think that's debatable. If you add this dependency right away you will make use of its classes much more often than not. Instead of using "if string != null && string.trim() != "" you will use StringUtils.isNotEmpty... You will use ObjectUtils.equals() so that you don't have to check for null everywhere. But if you wait for the one dependency that justifies using these libraries, you might never actually use them.
It's also nice that it won't add an additional tab at the end!
C
Community

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);

This is my favorite solution. :-> +1 for guava.
Ctrl+F'd for Guava as well. Thanks!
List list = new ArrayList(); list.add("Hello"); list.add("Hai"); list.add("Hi"); System.out.println(list.toString().substring(1, list.toString().length()-1).replaceAll(",", "\t"));
The beautiful thing with the Guava Joiner is that it can also join a Map using Joiner.on(", ").withKeyValueSeparator(": ").join(myMap)
D
Dimitri Hautot

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.


I didn't notice your Case 2 solution before posting my own. Yours is actually much better and saves the extra step. I would definitely recommend it over some of the answers that seem to require extra tools.
Arrays.toString(myList.toArray()) - the easiest and the most straightforward solution
I think this should be considered as the best and most efficient solution
Arrays.toString(list.toArray()) is easy to write, but highly inefficient.
r
rogerdpack

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")

I think this is actually more efficient that the methods posted above. Such a shame it's shown on the bottom. It might not be elegant but I strongly believe your solution runs O(n) while the others theoretically run in O(n^2)(since Java Strings are immutable you basically create a new String on every merge, and it becomes worse as the resulting String becomes larger)
If your text is very large you end up eating your computers resources.
This method is 7-8 times more slower than using StringBuilder.
@user3790827 You are looking at a very superficial level. If I tell you to search for a person in m households in n cities in x states, you would say it's O(mnx) or O(n^3), but I can rephrase it to say "look for this person in this country," and you would immediately tell me O(n). On top of that, all algorithms here are generally O(n), there is no looping within a loop.
l
luvieere

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();

Thanks! this is what I ending up doing :)
For big arrays this is definitely not micro-optimizing. The automatic conversion to use StringBuilder is done separately for each string concatenation, which doesn't help in the case of a loop. It is OK if you concatenate a large number of elements in one expression.
Using the StringBuilder explicitly is a big deal if you are concatenating let's say 50 000 strings making ~ 1 MB result string - it can be a difference of 1 min compared to 1 s execution time.
B
Brian

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


This is very nice since the join(...) method has variants for both arrays and collections.
I needed to create collections of strings with rarely populated elements, with random numbers. At the end I needed a string, not an array not array to string. This is my code : Collections.shuffle(L); StringBuilder sb = new StringBuilder(); L.forEach(e -> sb.append(e)); return sb.toString();
k
kehers

Android has a TextUtil class you can use http://developer.android.com/reference/android/text/TextUtils.html

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

This solution is better than String.join, which requires API 26+
K
Khader M A

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");

a
amra

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"));

This might be only a line of code, but to me that doesn't look simple. For me traversing the List and parsing its elements into the String is simpler than this.
This was my first impression too. But after I get used to it I find it awesome. For example you can add operations like filter. In the end it's much easier for me to read the code.
In Java 8: String.join("\t", array)
The last comment works only if the content is a String as well. If its a list of Integers you have a problem
All 3 of those requires API level 24 on android.
a
akuhn

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.


J
Jason Plank

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.


J
Javad

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.


Using reduce is extremely inefficient for String concatenation. You should do it the Java 8 way return StreamSupport.stream(list).map(Object::toString).collect(joining("\n")) which internally uses StringJoiner. There is no reason you couldn't do this with streamsupport also.
Another thing is that your code would fail horribly (due to the Optional#get) if it gets passed an empty list.
@StefanZobel Thanks for pointing out the efficiency and edge case issues. Modified the code.
P
Pratyush Pranjal

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

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

1) There's no reason to use code like this anymore (as of 4 years ago!). 2) This answer doesn't add anything to the Q&A.
This works perfectly. For comma separated string use: String srt= list.toString().replaceAll("\[|\]","");
佚名

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());

S
Srinivasan.S

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

M
Mohan Narayanaswamy

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]


I think this is the best way if "best" means code readability. I haven't looked at the code, but it's probably just as efficient as the StringBuilder method (it probably uses a StringBuilder under the hood), though not as customizable in the output.
B
Beezer

Would the following be any good:

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

C
Craig P. Motlin

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.


f
fmsf
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(", ");


P
Paul Rooney

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");
}

E
Elliander

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;
   }

A
Alex VI

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).


V
Vikrant Goel

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

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

a
android developer

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...