JavaScript how to remove elements from one array which contains in second

Question

I have misunderstood how to remove an element in an array which has a matching element to another array.

Example:

const items = ['first', 'second', 'third']
const secondItems = ['first', 'second']

Expected:

console.log(items) | 'third'
console.log(secondItems) | 'first', 'second

Hope some example how to achieve that. Tried many times with two forEach or filter and checking statement but I always get the wrong result.

https://stackoverflow.com/questions/7837456/how-to-compare-arrays-in-javascript — Pooshonk, Sep 06 '18 at 21:32
https://codereview.stackexchange.com/questions/112300/array-diff-in-javascript — Muhammad Omer Aslam, Sep 06 '18 at 21:32

score 2 · Answer 1 · answered Sep 06 '18 at 21:35

Simply use Array.prototype.filter() method:

const items = ['first', 'second', 'third']
const secondItems = ['first', 'second']

console.log(items.filter(i => !secondItems.includes(i)))
console.log(secondItems.filter(i => items.includes(i)))

I don't exactly understand this expectation of yours though:

console.log(secondItems) | 'first', 'second

Can you elaborate on that in the comments?

score 1 · Accepted Answer · answered Sep 06 '18 at 21:32

This could be solved by the following:

const items = ['first', 'second', 'third'];
const secondItems = ['first', 'second'];


const filteredBySecondItems = items.filter(function(item) {

  // For item in items array, check if there is a match
  // in secondItems. If there is no match, allow it
  // to be included in the returned list
  return secondItems.indexOf(item) === -1 

})

console.log(filteredBySecondItems) // [ 'third' ]

score 1 · Answer 3 · answered Sep 06 '18 at 21:36

1

items.filter(x => !secondItems.includes(x))

Just tried in the browser console, it works.

answered Sep 06 '18 at 21:36

mp92

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JavaScript how to remove elements from one array which contains in second

3 Answers3