What would be the best way to do this?

JavaScript
1

Hi All,

What would be the best way to return nested arrays filtered with element removed?

function remove(arr, element) {
  return arr.filter(el => el !== element);
}

const arr = [[3, 4, 5], [6, 7, 8], [9, 0, 6]];

const removeItem = remove(arr, 6);
console.log(removeItem.toString());

Desired output:

[[3, 4, 5], [7, 8], [9, 0]]
2

Consider that arr is an array of arrays, so el is an array. How do you check if an array contain something?

3

This doesn’t return desired output.

el.includes(element)

I might be wrong though on the .includes as to wether it’s the correct method to check for element.

6

Hi Randell,

Yes it is.

Iterate Through All an Array’s Items Using For Loops in the Data Structures section.

https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/basic-data-structures/iterate-through-all-an-arrays-items-using-for-loops

I wanted to know how to return the nested arrays without the (element) instead of removing the whole nested array.

8

Yes thanks for that Randell. Always love your input and appreciate it very much.

Anyways, this has got way more complicated than it should be. I managed to fix this but i got the feeling it can be so much more cleaner code than what i came up with. How can this be refactored/beautified?

const remove = (arr, elem) => {
  let newArr = [];

  for (let i = 0; i < arr.length; i++) {
    let index = arr[i].indexOf(elem);

    if (elem === arr[i][index]) {
      arr[i].splice(index, 1);
    }

    newArr.push(arr[i]);
  }

  return newArr.filter(item => {
    return !item.includes(elem);
  });
};

const arr = [[3, 4, 5, 6], [6, 7, 8], [9, 0, 6]];
console.log(remove(arr, 6));
10 
const remove = (arrs, element) => {
  return arrs.filter((arr) => {
    return arr.every((number) => number !== element);
  });
};

You were almost there,. but you were checking an array against an element.

12

Yeah., I have a tendency for selective reading… lol. Good catch @camperextraordinaire