How can I convert ArrayList<Object> to ArrayList<String>?

Question

ArrayList<Object> list = new ArrayList<Object>();
list.add(1);
list.add("Java");
list.add(3.14);
System.out.println(list.toString());

I tried:

ArrayList<String> list2 = (String)list;

But it gave me a compile error.

Why do you want to do that? I mean its not safe. Have a look at the example. — athspk
– athspk, Commented Jan 3, 2011 at 1:00
You're putting non-strings into the list. What is your expectation of what happens if you force them to be strings? — Ben Zotto
– Ben Zotto, Commented Jan 3, 2011 at 1:01
I assume you meant: ArrayList<String> list2 = (ArrayList<String>)list; — Costi Ciudatu
– Costi Ciudatu, Commented Jan 3, 2011 at 1:05

BalusC · Accepted Answer · 2022-10-26 09:52:36Z

183

Since this is actually not a list of strings, the easiest way is to loop over it and convert each item into a new list of strings yourself:

List<String> strings = list.stream()
   .map(object -> Objects.toString(object, null))
   .toList();

Or when you're not on Java 16 yet:

List<String> strings = list.stream()
   .map(object -> Objects.toString(object, null))
   .collect(Collectors.toList());

Or when you're not on Java 8 yet:

List<String> strings = new ArrayList<>(list.size());
for (Object object : list) {
    strings.add(Objects.toString(object, null));
}

Or when you're not on Java 7 yet:

List<String> strings = new ArrayList<String>(list.size());
for (Object object : list) {
    strings.add(object != null ? object.toString() : null);
}

Note that you should be declaring against the interface (java.util.List in this case), not the implementation.

edited Oct 26, 2022 at 9:52

answered Jan 3, 2011 at 1:01

BalusC

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7 Comments

user166390 Over a year ago

I like how null values are preserved this this implementation.

user219882 Over a year ago

If you use String.valueOf(object), you won't have to do the object != null ? object.toString() : null thing

BalusC Over a year ago

@Tomas: it would add a string with value "null" instead of the literal null. It depends further on the business requitements whether that's desireable or not. For me, it wouldn't have been acceptable at all as it introduces a potentially major bug.

BalusC Over a year ago

@Tomas: "null" is not null at all. The null has a special meaning in Java. It makes no sense to be forced to do if (item.equals("null")) when checking for that afterwards.

BalusC Over a year ago

@Tomas: you're only defeating the strong typing nature of Java this way. But as said, it depends on the business requirements whether that's desireable. For me, that's simply unacceptable. Do whatever you want and the time will teach :)

|

athspk · Accepted Answer · 2011-01-05 11:48:48Z

16

It's not safe to do that!
Imagine if you had:

ArrayList<Object> list = new ArrayList<Object>();
list.add(new Employee("Jonh"));
list.add(new Car("BMW","M3"));
list.add(new Chocolate("Twix"));

It wouldn't make sense to convert the list of those Objects to any type.

edited Jan 5, 2011 at 11:48

answered Jan 3, 2011 at 1:03

athspk

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1 Comment

BalusC Over a year ago

Unless they have Object#toString() method overridden. However, the whole type conversion from X to String for other purposes than human presentation does indeed not make much sense.

hellrocker · Accepted Answer · 2016-07-05 03:41:54Z

16

You can use wildcard to do this as following

ArrayList<String> strList = (ArrayList<String>)(ArrayList<?>)(list);

answered Jul 5, 2016 at 3:41

hellrocker

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Comments

scadge · Accepted Answer · 2016-03-10 12:46:22Z

15

Using Java 8 you can do:

List<Object> list = ...;
List<String> strList = list.stream()
                           .map( Object::toString )
                           .collect( Collectors.toList() );

edited Mar 10, 2016 at 12:46

answered Aug 17, 2015 at 8:03

scadge

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Comments

Per Lindberg · Accepted Answer · 2014-05-21 07:48:45Z

14

If you want to do it the dirty way, try this.

@SuppressWarnings("unchecked")
public ArrayList<String> convert(ArrayList<Object> a) {
   return (ArrayList) a;
}

Advantage: here you save time by not iterating over all objects.

Disadvantage: may produce a hole in your foot.

answered May 21, 2014 at 7:48

Per Lindberg

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Comments

Emil · Accepted Answer · 2011-01-03 05:37:06Z

12

Using guava:

List<String> stringList=Lists.transform(list,new Function<Object,String>(){
    @Override
    public String apply(Object arg0) {
        if(arg0!=null)
            return arg0.toString();
        else
            return "null";
    }
});

answered Jan 3, 2011 at 5:37

Emil

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Comments

husayt · Accepted Answer · 2013-07-17 13:17:28Z

8

Here is another alternative using Guava

List<Object> lst ...    
List<String> ls = Lists.transform(lst, Functions.toStringFunction());

answered Jul 17, 2013 at 13:17

husayt

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Comments

fastcodejava · Accepted Answer · 2011-01-03 01:13:55Z

5

Your code ArrayList<String> list2 = (String)list; does not compile because list2 is not of type String. But that is not the only problem.

answered Jan 3, 2011 at 1:13

fastcodejava

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Comments

Ankush soni · Accepted Answer · 2015-10-06 13:09:16Z

0

Using Java 8 lambda:

ArrayList<Object> obj = new ArrayList<>();
obj.add(1);
obj.add("Java");
obj.add(3.14);

ArrayList<String> list = new ArrayList<>();
obj.forEach((xx) -> list.add(String.valueOf(xx)));

answered Oct 6, 2015 at 13:09

Ankush soni

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Comments

Just Me · Accepted Answer · 2015-12-01 18:42:44Z

0

With Java Generics Takes a list of X and returns a list of T that extends or implements X, Sweet!

    // the cast is is actually checked via the method API
@SuppressWarnings("unchecked")
public static <T extends X, X> ArrayList<T> convertToClazz(ArrayList<X> from, Class<X> inClazz, Class<T> outClazz) {
    ArrayList<T> to = new ArrayList<T>();
    for (X data : from) {
        to.add((T) data);
    }
    return to;
}

answered Dec 1, 2015 at 18:42

Just Me

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Comments

S Krishna · Accepted Answer · 2019-07-30 13:43:07Z

0

A simple solution:

List<Object> lst  =listOfTypeObject;   
ArrayList<String> aryLst = new ArrayList<String>();
    for (int i = 0; i < lst.size(); i++) {
        aryLst.add(lst.get(i).toString());
    }

Note: this works when the list contains all the elements of datatype String.

answered Jul 30, 2019 at 13:43

S Krishna

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