Grouping elements of a 1D array to form a Jagged 2D array in Javascript

Question

My goal is to select elements of a 1D array,B, grouping them into subarrays based on their positions in the 1D array B. Their positions (indices) in array B are provided in the 2D array indices.

const B = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.01, 1.1, 1.2, 1.5];
const indices =  [ [ 3, 4, 5 ], [ 7, 8 ], [ 10, 11 ] ];

//Getting the elements from array B but flattened
const arrs = [];
for (let i = 0; i < indices.length; i++) {
    for (let j = 0; j < indices[i].length; j++) {
        arrs.push(B[indices[i][j]]);
    }
}
console.log("arrs", arrs)

//Converting the flattened array to 2D
var newArr = [];
let k = indices.length
while (k--) {
    newArr.push(arrs.splice(0, indices[k].length));
}

console.log("newArr", newArr);

What I have done is to use a nested for-loop to try to get the desired output, but the array arrs is flattened. Then, I converted the flattened array arrs to a 2D array newArr. Is there a more elegant and straight-forward approach?

Nick Parsons · Accepted Answer · 2019-12-17 01:05:19Z

1

You could use .map() on your inner arrays and use each element as an index for your B array. This still requires a nested loop as you need to go through each inner array, and each element in that inner array:

const B = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.01, 1.1, 1.2, 1.5];
const indices =  [ [ 3, 4, 5 ], [ 7, 8 ], [ 10, 11 ] ];

const newArr = indices.map(
  arr => arr.map(idx => B[idx])
);

console.log("newArr", newArr);

answered Dec 17, 2019 at 1:05

Nick Parsons

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Robby Cornelissen · Accepted Answer · 2019-12-17 01:06:20Z

1

A double map() operation will do it:

const array = [0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.01, 1.1, 1.2, 1.5];
const indices =  [ [ 3, 4, 5 ], [ 7, 8 ], [ 10, 11 ] ];

const result = indices.map(a => a.map(i => array[i]));

console.log(result);

answered Dec 17, 2019 at 1:06

Robby Cornelissen

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