Remove in Array list of array lists

Question

public Point{
int x, y;

public Point(int x, int y){
    int x = x;
    int y = y;
    // setters
    // getters
    // toString
}
}

ArrayList<Point> points = new ArrayList<>();

// add points by using points.add() then use this loop 
ArrayList<ArrayList<Point>> A = new ArrayList<>();
    for (int i = 0; i < 3; i++) {
            A.add(points);

        }

I have an arraylist of arraylists that each of them have the same points For example when you print this is the output

ArrayList<ArrayList<Point>> A = [[(0,10), (20,3), (2,5), (2,8)], [(0,10), (20,3), (2,5), (2,8)], [(0,10), (20,3), (2,5), (2,8)]]

I want to remove all the points that their x value is not equal to the index of the array list before printing

A.get(0) = [(0,10)]
A.get(1) = []
A.get(2) = [(2,5),(2,8)]

so the output should be

[[(0,10)], [], [(2,5), (2,8)]]

I tried to use remove method and for loop and other ways but they didn't work.

for(int i = 0; i < list.size();i++){
            if(list.get(i).x != 0){
                list.remove(i--);
            }
        }

Is there anyone who can help?

Your problem is here: A.add(points); You are adding THE SAME array 3 times. It's not 3 copies of the array, it's the same array. That's why when you follow my answer you are getting empty arrays. — LLL, Jun 06 '22 at 14:42
This is how you can do a copy: https://stackoverflow.com/questions/5785745/make-copy-of-an-array — LLL, Jun 06 '22 at 14:46
"*I have an arraylist of arraylists that each of them have the same points For example when you print this is the output `ArrayList> A = [[(0,10), (20,3), (2,5), (2,8)], [(0,10), (20,3), (2,5), (2,8)], [(0,10), (20,3), (2,5), (2,8)]]`*" are you sure that this is what you *have* or do you just *assume* that you have it? — Pshemo, Jun 06 '22 at 15:13

score 0 · Answer 1 · edited Jun 07 '22 at 00:48

0

You are probably looking for something like this:

int index = 0;
for (ArrayList<Point> list : A) {
     list.removeIf(point -> point.x != index);
     index++;
}

index variable could be removed as well but I didn't want to oversimplify the code, to preserve clarity.

edited Jun 07 '22 at 00:48

WJS

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answered Jun 06 '22 at 14:03

LLL

1,777
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1

Maybe you could update your answer to show how you are trying to do that – LLL Jun 06 '22 at 14:40
Won't even compile. Nasty thing about effective final values in lambdas. – WJS Jun 06 '22 at 16:28

score 0 · Answer 2 · answered Jun 06 '22 at 14:58

import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;

public class teste {

    public static void main(String[] args) {

        ArrayList<Point> points = new ArrayList<>();

        points.add(new Point(0, 10));
        points.add(new Point(9, 10));
        points.add(new Point(20, 3));


        System.out.println(points);
        //solution 1
        List<Point> list = points.stream().filter(o -> !o.equals(9, 10)).collect(Collectors.toList());
        System.out.println(list);
        
        //solution 2
        List<Point> list2 = new ArrayList<>();
        points.stream().forEach(o -> {
            if (o.equals(9, 10)) {
                list2.add(null);
            } else {
                list2.add(o);
            }
        });

        System.out.println(list2);
    }



    public static class Point {
        int x, y;

        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }

        public String toString() {
            return "(" + x + ", " + y + ")";
        }

        public boolean equals(int x, int y) {
            return this.x == x && this.y == y;
        }

    }

}


[(0, 10), (9, 10), (20, 3)]
[(0, 10), (20, 3)]
[(0, 10), null, (20, 3)]

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). — Community, Jun 06 '22 at 18:09

WJS · Answer 3 · 2022-06-06T16:30:46.080

Here is one way. To account for change in indexed position when removing items, start at the end and move backward.

iterate over the list of lists.
for each inner list, iterate backwards and remove the unwanted points

for (int i = 0; i < lists.size(); i++) {
    List<Point> points = lists.get(i);
     for (int k = points.size()-1; k >= 0; k--) {
         if (points.get(k).x != i) {
             points.remove(k);
         }
     }
}

Given the current data, here is the output.

[[(0, 10)], [], [(2, 5), (2, 8)]]

Here is a stream version if you're interested. This doesn't explicitly remove anything. It just filters out the unwanted values.

stream the indices of the outer list
using map, stream the inner list.
filter out the unwanted points
and repackage into a list of lists

List<List<Point>> result = IntStream.range(0,lists.size())
              .mapToObj(i->lists.get(i).stream()
              .filter(point->point.x == i)
              .toList()).toList();

Didn't understand at first. Please check out my updated answer. — WJS, Jun 06 '22 at 16:14

Remove in Array list of array lists

3 Answers3