Simplest code for array intersection in javascript

Question

What's the simplest, library-free code for implementing array intersections in javascript? I want to write

intersection([1,2,3], [2,3,4,5])

and get

[2, 3]

Solution 1

Use a combination of Array.prototype.filter and Array.prototype.includes:

const filteredArray = array1.filter(value => array2.includes(value));

For older browsers, with Array.prototype.indexOf and without an arrow function:

var filteredArray = array1.filter(function(n) {
    return array2.indexOf(n) !== -1;
});

NB! Both .includes and .indexOf internally compares elements in the array by using ===, so if the array contains objects it will only compare object references (not their content). If you want to specify your own comparison logic, use Array.prototype.some instead.

Solution 2

Destructive seems simplest, especially if we can assume the input is sorted:

/* destructively finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *  State of input arrays is undefined when
 *  the function returns.  They should be 
 *  (prolly) be dumped.
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length, b.length)
 */
function intersection_destructive(a, b)
{
  var result = [];
  while( a.length > 0 && b.length > 0 )
  {  
     if      (a[0] < b[0] ){ a.shift(); }
     else if (a[0] > b[0] ){ b.shift(); }
     else /* they're equal */
     {
       result.push(a.shift());
       b.shift();
     }
  }

  return result;
}

Non-destructive has to be a hair more complicated, since we’ve got to track indices:

/* finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length(), b.length())
 */
function intersect_safe(a, b)
{
  var ai=0, bi=0;
  var result = [];

  while( ai < a.length && bi < b.length )
  {
     if      (a[ai] < b[bi] ){ ai++; }
     else if (a[ai] > b[bi] ){ bi++; }
     else /* they're equal */
     {
       result.push(a[ai]);
       ai++;
       bi++;
     }
  }

  return result;
}

Solution 3

If your environment supports ECMAScript 6 Set, one simple and supposedly efficient (see specification link) way:

function intersect(a, b) {
  var setA = new Set(a);
  var setB = new Set(b);
  var intersection = new Set([...setA].filter(x => setB.has(x)));
  return Array.from(intersection);
}

Shorter, but less readable (also without creating the additional intersection Set):

function intersect(a, b) {
  var setB = new Set(b);
  return [...new Set(a)].filter(x => setB.has(x));
}

Note that when using sets you will only get distinct values, thus new Set([1, 2, 3, 3]).size evaluates to 3.

Solution 4

Using Underscore.js or lodash.js

_.intersection( [0,345,324] , [1,0,324] )  // gives [0,324]

Solution 5

// Return elements of array a that are also in b in linear time:
function intersect(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

// Example:
console.log(intersect([1,2,3], [2,3,4,5]));

I recommend above succinct solution which outperforms other implementations on large inputs. If performance on small inputs matters, check the alternatives below.

Alternatives and performance comparison:

See the following snippet for alternative implementations and check https://jsperf.com/array-intersection-comparison for performance comparisons.

function intersect_for(a, b) {
  const result = [];
  const alen = a.length;
  const blen = b.length;
  for (let i = 0; i < alen; ++i) {
    const ai = a[i];
    for (let j = 0; j < blen; ++j) {
      if (ai === b[j]) {
        result.push(ai);
        break;
      }
    }
  } 
  return result;
}

function intersect_filter_indexOf(a, b) {
  return a.filter(el => b.indexOf(el) !== -1);
}

function intersect_filter_in(a, b) {
  const map = b.reduce((map, el) => {map[el] = true; return map}, {});
  return a.filter(el => el in map);
}

function intersect_for_in(a, b) {
  const result = [];
  const map = {};
  for (let i = 0, length = b.length; i < length; ++i) {
    map[b[i]] = true;
  }
  for (let i = 0, length = a.length; i < length; ++i) {
    if (a[i] in map) result.push(a[i]);
  }
  return result;
}

function intersect_filter_includes(a, b) {
  return a.filter(el => b.includes(el));
}

function intersect_filter_has_this(a, b) {
  return a.filter(Set.prototype.has, new Set(b));
}

function intersect_filter_has_arrow(a, b) {
  const set = new Set(b);
  return a.filter(el => set.has(el));
}

function intersect_for_has(a, b) {
  const result = [];
  const set = new Set(b);
  for (let i = 0, length = a.length; i < length; ++i) {
    if (set.has(a[i])) result.push(a[i]);
  }
  return result;
}

Results in Firefox 53:

• Ops/sec on large arrays (10,000 elements):

  filter + has (this)               523 (this answer)
  for + has                         482
  for-loop + in                     279
  filter + in                       242
  for-loops                          24
  filter + includes                  14
  filter + indexOf                   10
  

• Ops/sec on small arrays (100 elements):

  for-loop + in                 384,426
  filter + in                   192,066
  for-loops                     159,137
  filter + includes             104,068
  filter + indexOf               71,598
  filter + has (this)            43,531 (this answer)
  filter + has (arrow function)  35,588
  

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