Pushing different values in 2d array to a unique array

Question

I have a 2d array (which will always have a random number of rows) as follows:

arr = [['tag4', 'example', 'project1'],
       ['tag1', 'example', 'project2'],
       ['tag3', 'example', 'project2'],
       ['tag2', 'example', 'project3']];

Taking note that this array will always be sorted by the values in the 3rd column and the # of rows in this array will not be consistent(random)... how can I group each different tags of a project in a unique array?

Example desired outcome:

arr1 = [['tag4', 'example', 'project1']];

arr2 = [['tag1', 'example', 'project2'],
       ['tag3', 'example', 'project2']];

arr3 = [['tag2', 'exmaple', 'project3']];

Also, how could I keep track of all the unique arrays created? I need to know because I need to implement these arrays in another function.

Are the different number of values for the third field restricted to just these three values, or is it also dynamic? Because if it is dynamic, trying to break these out into their own variables is going to add more complexity to the logic than it is worth. — Taplar
– Taplar, Commented Aug 26, 2020 at 21:15

Jairo Nava Magaña · Accepted Answer · 2020-08-26 23:09:35Z

In this case use the function every() but you can use map() or foreach(). Everything is quite straightforward, the previous index is maintained and in each iteration it is checked if it matches the current one, in case of being different an empty array is added. in this way the row will always be added to the last cell of the result array

//original array
let arr = [
    ['tag4', 'example', 'project1', 'foo'],
    ['tag1', 'example', 'project2'],
    ['tag3', 'example', 'project2'],
    ['tag2', 'example', 'project3', 'bar']
];

// Final array
let result = [];
// last identifier
let last = null;

// loop through every item
arr.every(row => {
    if(last !== row[2]) {
        result.push([]);
    }
    // Always insert the row in the last cell of the result array
    result[result.length - 1].push(row);
    last = row[2];
    return true;
});

console.log(result);

Ele · Accepted Answer · 2020-08-26 21:13:57Z

0

An alternative is to keep the grouped arrays by the project value.

Assuming there are always three indexes.

let arr = [['tag4', 'example', 'project1'],       ['tag1', 'example', 'project2'],       ['tag3', 'example', 'project2'],       ['tag2', 'example', 'project3']],
    result = arr.reduce((a, [tag, desc, project]) => {
      (a[project] || (a[project] = [])).push([tag, desc, project]);
      return a;
    }, Object.create(null));

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Aug 26, 2020 at 21:13

Ele

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