Dividing an array by filter function

Question

With ES6 you can make use of the spread syntax with reduce:

function partition(array, isValid) {
  return array.reduce(([pass, fail], elem) => {
    return isValid(elem) ? [[...pass, elem], fail] : [pass, [...fail, elem]];
  }, [[], []]);
}

const [pass, fail] = partition(myArray, (e) => e > 5);

Or on a single line:

const [pass, fail] = a.reduce(([p, f], e) => (e > 5 ? [[...p, e], f] : [p, [...f, e]]), [[], []]);

You can use lodash.partition

var users = [
  { 'user': 'barney',  'age': 36, 'active': false },
  { 'user': 'fred',    'age': 40, 'active': true },
  { 'user': 'pebbles', 'age': 1,  'active': false }
];

_.partition(users, function(o) { return o.active; });
// → objects for [['fred'], ['barney', 'pebbles']]

// The `_.matches` iteratee shorthand.
_.partition(users, { 'age': 1, 'active': false });
// → objects for [['pebbles'], ['barney', 'fred']]

// The `_.matchesProperty` iteratee shorthand.
_.partition(users, ['active', false]);
// → objects for [['barney', 'pebbles'], ['fred']]

// The `_.property` iteratee shorthand.
_.partition(users, 'active');
// → objects for [['fred'], ['barney', 'pebbles']]

or ramda.partition

R.partition(R.contains('s'), ['sss', 'ttt', 'foo', 'bars']);
// => [ [ 'sss', 'bars' ],  [ 'ttt', 'foo' ] ]

R.partition(R.contains('s'), { a: 'sss', b: 'ttt', foo: 'bars' });
// => [ { a: 'sss', foo: 'bars' }, { b: 'ttt' }  ]

I came up with this little guy. It uses for each and all that like you described, but it looks clean and succinct in my opinion.

//Partition function
function partition(array, filter) {
  let pass = [], fail = [];
  array.forEach((e, idx, arr) => (filter(e, idx, arr) ? pass : fail).push(e));
  return [pass, fail];
}

//Run it with some dummy data and filter
const [lessThan5, greaterThanEqual5] = partition([0,1,4,3,5,7,9,2,4,6,8,9,0,1,2,4,6], e => e < 5);

//Output
console.log(lessThan5);
console.log(greaterThanEqual5);

You can use reduce for it:

function partition(array, callback){
  return array.reduce(function(result, element, i) {
    callback(element, i, array) 
      ? result[0].push(element) 
      : result[1].push(element);
    
        return result;
      }, [[],[]]
    );
 };

Update. Using ES6 syntax you also can do that using recursion (updated to avoid creating new arrays on every iteration):

function partition([current, ...tail], f, left = [], right = []) {
    if(current === undefined) {
        return [left, right];
    }
    if(f(current)) {
        left.push(current);
        return partition(tail, f, left, right);
    }
    right.push(current);
    return partition(tail, f, left, right);
}

This sounds very similar to Ruby's Enumerable#partition method.

If the function can't have side-effects (i.e., it can't alter the original array), then there's no more efficient way to partition the array than iterating over each element and pushing the element to one of your two arrays.

That being said, it's arguably more "elegant" to create a method on Array to perform this function. In this example, the filter function is executed in the context of the original array (i.e., this will be the original array), and it receives the element and the index of the element as arguments (similar to jQuery's each method):

Array.prototype.partition = function (f){
  var matched = [],
      unmatched = [],
      i = 0,
      j = this.length;

  for (; i < j; i++){
    (f.call(this, this[i], i) ? matched : unmatched).push(this[i]);
  }

  return [matched, unmatched];
};

console.log([1, 2, 3, 4, 5].partition(function (n, i){
  return n % 2 == 0;
}));

//=> [ [ 2, 4 ], [ 1, 3, 5 ] ]

In filter function you can push your false items into another variable outside function:

var bad = [], good = [1,2,3,4,5];
good = good.filter(function (value) { if (value === false) { bad.push(value) } else { return true});

Of course value === false need to be real comparasion ;)

But it do almost that same operation like forEach. I think you should use forEach for better code readability.