Add array to another array with keeping the same amount of indices-CodePudding

I have array A and I am trying to concat it to array B while array B keeps the same amount of indices.

for example:

const array_A = [1, 2, 3, 4];
const array_B = [0, 0, 0, 0, 0, 0, 0];

the result should look like this

const result = [1, 2, 3, 4, 0, 0, 0];

I tried this approach

const result = array_A.concat(array_B);
console.log(result);

but then I got array of 11 indices which I only want array of 7 indices.

[1, 2, 3, 4, 0, 0, 0, 0, 0, 0, 0];

CodePudding user response：

You could map over the second array and check if the first array has a value at that index.

array_B.map((n, i) => array_A[i] ?? n)

If the value at an index can be null / undefined, then the ?? operator will ignore it and use the array_B value. In that case, you can check if array_A has that specific index using the in opeartor

array_B.map((n, i) => i in array_A ? array_A[i] : n)

Here's a snippet:

const array_A = [1, 2, 3, 4],
      array_B = [0, 0, 0, 0, 0, 0, 0],
      output1 = array_B.map((n, i) => array_A[i] ?? n),
      output2 = array_B.map((n, i) => i in array_A ? array_A[i] : n)
      
console.log(...output1)
console.log(...output2)

CodePudding user response：

You could do something like below:

const array_A = [1, 2, 3, 4];
const array_B = [0, 0, 0, 0, 0, 0, 0];

for (let i = array_A.length - 1; i >= 0; i--) {
  array_B.pop()
  array_B.unshift(array_A[i])
}

console.log(array_B)

Poping out last element of second array ie. array_B using pop() then inserting element from array_A to array_B from front using unshift() method. Must Loop with respect to element present in array_A

CodePudding user response：

you can use for or forEach loop on Array_A then you can replace value of Array_B with value of Array_A

const A = [1, 2, 3, 4];
const B = [0, 0, 0, 0, 0, 0, 0];

A.forEach((val, i) => {
  B[i] = val;
})

console.log(B) // [ 1, 2, 3, 4, 0, 0, 0 ]