Using a Array.splice is fine as long as it's a single value. Suppose I have an array with 10 elements and I want to remove element 2,4 and 8, using Array.splice(index,1) in a for loop is a bad idea because the index of each element changes after each splice operation.

How can I remove specific array items and then re-arrange the array accordingly

var array = ["Apple", "Banana", "Peach", "Pumpkin", "Tomato", "Mango", "Guava"];

remove(array, 4,5); //Is there a lodash function for this?

//desired result --> ["Apple", "Banana", "Peach", "Guava"]

Using a Array.splice is fine as long as it's a single value. Suppose I have an array with 10 elements and I want to remove element 2,4 and 8, using Array.splice(index,1) in a for loop is a bad idea because the index of each element changes after each splice operation.

How can I remove specific array items and then re-arrange the array accordingly

var array = ["Apple", "Banana", "Peach", "Pumpkin", "Tomato", "Mango", "Guava"];

remove(array, 4,5); //Is there a lodash function for this?

//desired result --> ["Apple", "Banana", "Peach", "Guava"]

• javascript
• arrays
• lodash

• actually it is not clear, what do you want, just to splice an element at an index or remove more than one element. – Nina Scholz Commented Jun 30, 2017 at 17:15
• 2 Using array.splice works fine,just loop backwards... – Nick is tired Commented Jun 30, 2017 at 17:18

3 Answers 3

There's a Lodash method _.pullAt that pulls elements out of the array (in place) that are specified in an array:

_.pullAt(array, [indexes])

Removes elements from array corresponding to indexes and returns an array of removed elements.

Thus, apply it like this:

var array = ["Apple", "Banana", "Peach", "Pumpkin", "Tomato", "Mango", "Guava"];
_.pullAt(array, [4, 5]);

And if you want to pull out non-adjacent items (from the documentation):

var array = ['a', 'b', 'c', 'd'];
var pulled = _.pullAt(array, [1, 3]);

console.log(array);
// => ['a', 'c']

This will remove the elements at index 4 and 5 in place, and also return the removed element (if you need it).

You could, of course with vanilla JavaScript, look backwards as Nick A. mentioned and remove items from the back. The reason this works is because removing from the back will not make a changing length an issue. Consider this example where I want to remove the 1st and 3rd elements:

[1, 3, 7, 9]

If I were to loop 'forwards', I would remove element at index 1 (3), but the length would change to 3 and there would be no element at index 3! If I were to loop backwards though, I would remove the element at index 3 first (9). Then the length would be 3, but that doesn't affect any elements before it, the element at index 1 is still at index 1 thus removing from the back works.

You can loop reverse :

var array = ["Apple", "Banana", "Peach", "Pumpkin", "Tomato", "Mango", "Guava"],
removeFromIndex = [4,5];    
for (var i = removeFromIndex.length -1; i >= 0; i--)
    array.splice(removeFromIndex[i],1);

JSFIDDLE DEMO

The approach you need to take is, removing the elements in a backward manner (not reversing the array, as that has the same underlying problem of decrementing the indices of 'not yet checked' elements upon removal.

You can use the reduceRight method, which acts very much like Reduce, but instead, it iterates an array from (right to left).

Here is an example of how this could be done:

var array = ["Apple", "Banana", "Peach", "Pumpkin", "Tomato", "Mango", "Guava"];

let indexToRemove = [4, 5]; //We're looking to remove "Tomato" & "Mango"

array.reduceRight((_, elem, index) => {
    if (indexToRemove.includes(index)) { //Our condition(s)
        array.splice(index, 1);
    }
});

console.log(array); //Result: [ 'Apple', 'Banana', 'Peach', 'Pumpkin', 'Guava' ]