Remove all elements contained in another array

Use the Array.filter() method:

myArray = myArray.filter( function( el ) {  return toRemove.indexOf( el ) < 0;} );

Small improvement, as browser support for Array.includes() has increased:

myArray = myArray.filter( function( el ) {  return !toRemove.includes( el );} );

Next adaptation using arrow functions:

myArray = myArray.filter( ( el ) => !toRemove.includes( el ) );

The filter method should do the trick:

const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];const toRemove = ['b', 'c', 'g'];// ES5 syntaxconst filteredArray = myArray.filter(function(x) {   return toRemove.indexOf(x) < 0;});

If your toRemove array is large, this sort of lookup pattern can be inefficient. It would be more performant to create a map so that lookups are O(1) rather than O(n).

const toRemoveMap = toRemove.reduce(  function(memo, item) {    memo[item] = memo[item] || true;    return memo;  },  {} // initialize an empty object);const filteredArray = myArray.filter(function (x) {  return toRemoveMap[x];});// or, if you want to use ES6-style arrow syntax:const toRemoveMap = toRemove.reduce((memo, item) => ({  ...memo,  [item]: true}), {});const filteredArray = myArray.filter(x => toRemoveMap[x]);

ECMAScript 6 sets can permit faster computing of the elements of one array that aren't in the other:

const myArray = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];const toRemove = new Set(['b', 'c', 'g']);const difference = myArray.filter( x => !toRemove.has(x) );console.log(difference); // ["a", "d", "e", "f"]