"Merge" multiple elements of the same type in an ArrayList

Question

I have a class, called ADVAObject, with a variable "type" and "value":

public class ADVAObject {
   String type = "";
   String value = "";
}

Then, I did some ArrayList<ADVAObject> pattern = new ArrayList<ADVAObject>();. After some processing, I ended up with this list using:

for(ADVAObject a : pattern) {
   System.out.println(a.type + " : " + a.value);
}

The list:

STRING : V
STRING : A
STRING : R
STRING : I
STRING : A
STRING : B
STRING : L
STRING : E
SPACE : null
SPECIAL : null
SPACE : null
NUMBER : 3
NUMBER : 4
SPACE : null
DIV : null
SPACE : null
NUMBER : 5
NUMBER : 6
SPACE : null
QUESTION : null

As you can see, instead of getting "VARIABLE", I get it separated. This isn't a problem, for now. "For now"? Yes. You see, I wanted to keep them separated to do additional processing, but now I need to "merge", as said in the title of this question, all repeated elements which have a value (which aren't null). So:

//This:
STRING : AB
STRING : CD
STRING : EF

//Turns into this:
STRING : ABCDEF

//However, ...:

//This:

SPACE : null
SPACE : null

//Can't turn into this:

SPACE : null

//.

I have tried creating a seperate array and regexing it all inside some loops, but it didn't... work as expected.

beastlyCoder · Accepted Answer · 2020-06-29 19:49:51Z

1

What you should look into, instead of an arraylist, could be a HashMap, with a linkedlist, or an arraylist as it's parameter as follows:

EDIT: Changed ArrayList<String> to ArrayList<ADVAObject> to align with the class you are using

HashMap<String, ArrayList<ADVAObject>> keyValues = new HashMap<>();

This way you could map each value even if there are duplicates, for example, STRING could be the key, and the values of that would be: ABCDEF.

Note: to handle the duplicate values seen, you could just have a conditional statement that only adds one of the duplicate values

Hope this helps :)

edited Jun 29, 2020 at 19:49

answered Jun 29, 2020 at 19:27

beastlyCoder

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3 Comments

Carlos de Almeida Over a year ago

Hello, turns out, although better than what I had before, both your answers don't work, because A, they change the order of the elements, eg. instead of STRING : a; NUMBER: 7; DIV : null, it may change to DIV: null; STRING: a; NUMBER: 7. The second problem is, if I have NUMBER : 5; SPACE : NULL; NUMBER : 5, I don't want to combine the first NUMBER with the second NUMBER, because they are separated by a SPACE.

beastlyCoder Over a year ago

those are things you can handle with conditional statements like my answer was saying, for the sorting you can do something like this: geeksforgeeks.org/sorting-hashmap-according-key-value-java

beastlyCoder Over a year ago

and with the numbers, you could say "if there is a space between them, count them as seperate entries"

Unmitigated · Accepted Answer · 2020-06-29 20:01:47Z

0

You can do some simple List processing with a for loop. See the below code in action here.

final List<ADVAObject> result = new ArrayList<>(pattern.size());
for (int i = 0, size = pattern.size(); i < size; i++) {
    final ADVAObject curr = pattern.get(i);
    if (i == 0 || !pattern.get(i - 1).type.equals(curr.type)) {
        result.add(curr);
    } else {
        result.get(result.size() - 1).value += curr.value;
    }
}

edited Jun 29, 2020 at 20:01

answered Jun 29, 2020 at 19:30

Unmitigated

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2 Comments

beastlyCoder Over a year ago

Don't know if this is quite what he's asking for, he wants one value, to be pointed to many other values, that's why I suggested using a HashMap with a string as the key, and it assigned to an arraylist of strings

Carlos de Almeida Over a year ago

Hello, turns out, although better than what I had before, both your answers don't work, because A, they change the order of the elements, eg. instead of STRING : a; NUMBER: 7; DIV : null, it may change to DIV: null; STRING: a; NUMBER: 7. The second problem is, if I have NUMBER : 5; SPACE : NULL; NUMBER : 5, I don't want to combine the first NUMBER with the second NUMBER, because they are separated by a SPACE.

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