Java ArrayList<Integer> to ArrayList<String> & vice versa

Question

I'm wondering how to convert ArrayList<String> to ArrayList<Integer>.

I have this code:

ArrayList<String> sNumbers = new ArrayList<String>();
ArrayList<Integer> numbers = new ArrayList<Integer>();
numbers.add(1);
numbers.add(2);
numbers.add(3);

sNumbers.add(numbers.toString());
System.out.println(sNumbers);//for testing

I'm getting this:

[[1, 2, 3]]

And that's my question: How can I reverse this procces (Get one line of numbers and put it in numbers ArrayList)

It should be like this:

sNumbers
0 [1, 2, 3] --> 

numbers
0 [1]
1 [2]
2 [3]
(without brackets)

Mena · Accepted Answer · 2015-04-16 12:13:20Z

10

Invoking toString on a List<Integer> will return a String representation of the List (i.e. "[1,2,3]"), not a List<String> containing a String representation of each of its contents.

You must fill your Strings manually:

for (Integer i: numbers) {
    sNumbers.add(String.valueOf(i));
}

Note

There is probably some fancier stuff you could do with Java 8 in terms of declarative vs imperative coding - unfortunately i don't have the jdk installed at the moment.

Hopefully someone else will pick this up and answer for Java 8.

answered Apr 16, 2015 at 12:13

Mena

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7 Comments

Jump3r Over a year ago

This is correct, but I meant something else. When Array numbers is changing, I save it to sNumbers. Going your way I don't have any separator between 2 arrays. I just wanted to avoid 2 dimensional Arrays

Anthony Grist Over a year ago

The way I read the question, they're more than happy with getting the string representation of the ArrayList (i.e. a single String: [1, 2, 3]), they just want to be able to reverse that process; take the String [1, 2, 3] and create an ArrayList<Integer> with three elements, 1, 2, and 3. This doesn't seem like it answers the question being asked at all (to me, your understanding obviously differs).

Jump3r Over a year ago

Yes, I agree. When I getting new content of numbers (after saving numbers in sNumbers I cleaning it) I want to put it to sNumbers Array, then I want to pick one row from array. I can paste all code if you want.

Mena Over a year ago

Although I agree Austin's answer seems to specifically respond to the question details, the general scope is, and I quote, to "to convert ArrayList<String> to ArrayList<Integer>". Beri's answer and mine address this with what should be considered as best practices for Java, given the limited scope. I cannot think of any real life use case where Austin's answer could be preferable.

Mena Over a year ago

Consider the String representation of a List is already iterating its items to produce the result. Parsing it back to build another List (of different parametrized type) is pointless, costly, and would denote a serious misconception of how Java collections and String representation of objects actually work, unless I'm missing an academical point here.

|

Community · Accepted Answer · 2017-05-23 11:43:51Z

6

As @mena mentioned you can do it with:

pure Java:

List<String> sNumbers = new ArrayList<String>();
for (Integer i: numbers) {
    sNumbers.add(String.valueOf(i));
}

JDK 8 - map:

List<String> sNumbers = numbers.stream().map(
    n -> n.toString()).collect(Collectors.toList()
);

Or use guava library, it's similar to JDK8.

I assume you can find more examples here: Convert List<String> to List<Integer> directly

edited May 23, 2017 at 11:43

CommunityBot

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answered Apr 16, 2015 at 12:18

Beri

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1 Comment

Mena Over a year ago

Plus 1 for the Java 8 part!

score 2 · Accepted Answer · 2015-04-16 12:33:19Z

2

Heres a simple algorithm for turning the string representation of an array back into an array

public static ArrayList<Integer> arrayStringToIntegerArrayList(String arrayString){
    String removedBrackets = arrayString.substring(1, arrayString.length() - 1);
    String[] individualNumbers = removedBrackets.split(",");
    ArrayList<Integer> integerArrayList = new ArrayList<>();
    for(String numberString : individualNumbers){
        integerArrayList.add(Integer.parseInt(numberString.trim()));
    }
    return integerArrayList;
}

You would call it with

System.out.println(arrayStringToIntegerArrayList(sNumbers.get(0)));

and it will return an ArrayList<Integer> with the values [1, 2, 3]

edited Apr 16, 2015 at 12:33

answered Apr 16, 2015 at 12:27

user2033671

Comments

Loki · Accepted Answer · 2015-04-16 12:24:37Z

1

Just loop over the array

for(int i : numbers ) sNumbers.add(String.valueOf(i));

edited Apr 16, 2015 at 12:24

Loki

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answered Apr 16, 2015 at 12:15

Shayan Ghosh

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Comments

Loco234 · Accepted Answer · 2015-04-16 12:30:33Z

0

You could use the following loops:

//number 1
for(Integer number : numbers){
            sNumbers.add(String.valueOf(number));
    }

//number 2
for(int i = 0; i < numbers.size(); i++){
    sNumbers.add(String.valueOf(numbers.get(i)));
}

//number 3
int j = 0;
while(numbers.size() > j){
    sNumbers.add(String.valueOf(numbers.get(j)));
    j++;
}

//number 4  
Iterator<Integer> iterator = numbers.iterator();
while(iterator.hasNext()){
    sNumbers.add(String.valueOf(iterator.next()));
}

answered Apr 16, 2015 at 12:30

Loco234

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