java comma - Best way to convert an ArrayList to a string




separated android (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;
   }

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.


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).


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

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]


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.


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());

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(", "));

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.


Would the following be any good:

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

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);

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.


Basically, using a loop to iterate over the ArrayList is 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.



ArrayList class (Java Docs) extends AbstractList class, which extends AbstractCollection class which contains a toString() method (Java Docs). So you simply write

listName.toString();

Java developers have already figured out the most efficient way and have given you that in a nicely packaged and documented method. Simply call that method.


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");

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.


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

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

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


Download the Jakarta Commons Lang and use the method

 StringUtils.join(list)

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 Jakarta Commons library and I also think it's a great addition to the Java Standard Library.


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")

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("");

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...


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();

What is happening is that stock_list.toArray() is creating an Object[] rather than a String[] and hence the typecast is failing1.

The correct code would be:

  String [] stockArr = stockList.toArray(new String[stockList.size()]);

or even

  String [] stockArr = stockList.toArray(new String[0]);

For more details, refer to the javadocs for the two overloads of List.toArray.

The latter version uses the zero-length array to determine the type of the result array. (Surprisingly, it is faster to do this than to preallocate ... at least, for recent Java releases. See https://.com/a/4042464/139985 for details.)

From a technical perspective, the reason for this API behavior / design is that an implementation of the List<T>.toArray() method has no information of what the <T> is at runtime. All it knows is that the raw element type is Object. By contrast, in the other case, the array parameter gives the base type of the array. (If the supplied array is big enough to hold the list elements, it is used. Otherwise a new array of the same type and a larger size is allocated and returned as the result.)


1 - In Java, an Object[] is not assignment compatible with a String[]. If it was, then you could do this:

    Object[] objects = new Object[]{new Cat("fluffy")};
    Dog[] dogs = (Dog[]) objects;
    Dog d = dogs[0];     // Huh???

This is clearly nonsense, and that is why array types are not generally assignment compatible.







java string arraylist