32 Sets (Set)

Before ES6, JavaScript didn’t have a data structure for sets. Instead, two work-arounds were used:

• The keys of an object were used as a set of strings.
• Arrays were used as sets of arbitrary values. The downside is that checking membership (if an Array contains a value), is slower.

Since ES6, JavaScript has the data structure Set, which can contain arbitrary values and performs membership checks quickly.

32.1 Using Sets

32.1.1 Creating Sets

There are three common ways of creating Sets.

First, you can use the constructor without any parameters to create an empty Set:

const emptySet = new Set();
assert.equal(emptySet.size, 0);

Second, you can pass an iterable (e.g. an Array) to the constructor. The iterated values become elements of the new Set:

const set = new Set(['red', 'green', 'blue']);

Third, the .add() method adds elements to a Set and is chainable:

const set = new Set()
.add('red')
.add('green')
.add('blue');

32.1.2 Adding, removing, checking membership

.add() adds an element to a Set.

const set = new Set();
set.add('red');

.has() checks if an element is a member of a Set.

assert.equal(set.has('red'), true);

.delete() removes an element from a Set.

assert.equal(set.delete('red'), true); // there was a deletion
assert.equal(set.has('red'), false);

32.1.3 Determining the size of a Set and clearing it

.size contains the number of elements in a Set.

const set = new Set()
  .add('foo')
  .add('bar');
assert.equal(set.size, 2)

.clear() removes all elements of a Set.

set.clear();
assert.equal(set.size, 0)

32.1.4 Iterating over Sets

Sets are iterable and the for-of loop works as you’d expect:

const set = new Set(['red', 'green', 'blue']);
for (const x of set) {
  console.log(x);
}
// Output:
// 'red'
// 'green'
// 'blue'

As you can see, Sets preserve insertion order. That is, elements are always iterated over in the order in which they were added.

Given that Sets are iterable, you can use spreading (...) to convert them to Arrays:

const set = new Set(['red', 'green', 'blue']);
const arr = [...set]; // ['red', 'green', 'blue']

32.2 Examples of using Sets

32.2.1 Removing duplicates from an Array

Converting an Array to a Set and back, removes duplicates from the Array:

assert.deepEqual(
  [...new Set([1, 2, 1, 2, 3, 3, 3])],
  [1, 2, 3]);

32.2.2 Creating a set of Unicode characters (code points)

Strings are iterable and can therefore be used as parameters for new Set():

assert.deepEqual(
  new Set('abc'),
  new Set(['a', 'b', 'c']));

32.3 What Set elements are considered equal?

As with Map keys, Set elements are compared similarly to ===, with the exception of NaN being equal to itself.

> const set = new Set([NaN, NaN, NaN]);
> set.size
1
> set.has(NaN)
true

As with ===, two different objects are never considered equal (and there is no way to change that, at the moment):

> const set = new Set();

> set.add({});
> set.size
1

> set.add({});
> set.size
2

32.4 Missing Set operations

Sets are missing several common operations. Such an operation can usually be implemented by:

• Converting the input Sets to Arrays, by spreading into Array literals.
• Performing the operation on Arrays.
• Converting the result to a Set and returning it.

32.4.1 Union (a ∪ b)

Computing the union of two Sets a and b means creating a Set that contains the elements of both a and b.

const a = new Set([1,2,3]);
const b = new Set([4,3,2]);
// Use spreading to concatenate two iterables
const union = new Set([...a, ...b]);

assert.deepEqual([...union], [1, 2, 3, 4]);

32.4.2 Intersection (a ∩ b)

Computing the intersection of two Sets a and b means creating a Set that contains those elements of a that are also in b.

const a = new Set([1,2,3]);
const b = new Set([4,3,2]);
const intersection = new Set(
  [...a].filter(x => b.has(x)));

assert.deepEqual([...intersection], [2, 3]);

32.4.3 Difference (a \ b)

Computing the difference between two Sets a and b means creating a Set that contains those elements of a that are not in b. This operation is also sometimes called minus (−).

const a = new Set([1,2,3]);
const b = new Set([4,3,2]);
const difference = new Set(
  [...a].filter(x => !b.has(x)));

assert.deepEqual([...difference], [1]);

32.4.4 Mapping over Sets

Sets don’t have a method .map(). But we can borrow the one that Arrays have:

const set = new Set([1, 2, 3]);
const mappedSet = new Set([...set].map(x => x * 2));

// Convert mappedSet to an Array to check what’s inside it
assert.deepEqual([...mappedSet], [2, 4, 6]);

32.4.5 Filtering Sets

We can’t directly .filter() Sets, so we need to use the corresponding Array method:

const set = new Set([1, 2, 3, 4, 5]);
const filteredSet = new Set([...set].filter(x => (x % 2) === 0));

assert.deepEqual([...filteredSet], [2, 4]);

32.5 Quick reference: Set<T>

32.5.1 Constructor

• new Set<T>(values?: Iterable<T>) ^[ES6]

  If you don’t provide the parameter values, then an empty Set is created. If you do, then the iterated values are added as elements to the Set. For example:

  const set = new Set(['red', 'green', 'blue']);

32.5.2 Set<T>.prototype: single Set elements

• .add(value: T): this ^[ES6]

  Adds value to this Set. This method returns this, which means that it can be chained.

  const set = new Set(['red']);
  set.add('green').add('blue');
  assert.deepEqual([...set], ['red', 'green', 'blue']);

• .delete(value: T): boolean ^[ES6]

  Removes value from this Set. Returns true if something was deleted and false, otherwise.

  const set = new Set(['red', 'green', 'blue']);
  assert.equal(set.delete('red'), true); // there was a deletion
  assert.deepEqual([...set], ['green', 'blue']);

• .has(value: T): boolean ^[ES6]

  Checks whether value is in this Set.

  const set = new Set(['red', 'green']);
  assert.equal(set.has('red'), true);
  assert.equal(set.has('blue'), false);

32.5.3 Set<T>.prototype: all Set elements

• get .size: number ^[ES6]

  Returns how many elements there are in this Set.

  const set = new Set(['red', 'green', 'blue']);
  assert.equal(set.size, 3);

• .clear(): void ^[ES6]

  Removes all elements from this Set.

  const set = new Set(['red', 'green', 'blue']);
  assert.equal(set.size, 3);
  set.clear();
  assert.equal(set.size, 0);

32.5.4 Set<T>.prototype: iterating and looping

• .values(): Iterable<T> ^[ES6]

  Returns an iterable over all elements of this Set.

  const set = new Set(['red', 'green']);
  for (const x of set.values()) {
    console.log(x);
  }
  // Output:
  // 'red'
  // 'green'

• [Symbol.iterator](): Iterable<T> ^[ES6]

  Default way of iterating over Sets. Same as .values().

  const set = new Set(['red', 'green']);
  for (const x of set) {
    console.log(x);
  }
  // Output:
  // 'red'
  // 'green'

• .forEach(callback: (value: T, key: T, theSet: Set<T>) => void, thisArg?: any): void ^[ES6]

  Feeds each element of this Set to callback(). value and key both contain the current element. This redundancy was introduced so that this callback has the same type signature as the callback of Map.prototype.forEach().

  You can specify the this of callback via thisArg. If you omit it, this is undefined.

  const set = new Set(['red', 'green']);
  set.forEach(x => console.log(x));
  // Output:
  // 'red'
  // 'green'

32.5.5 Symmetry with Map

The following two methods mainly exist so that Sets and Maps have similar interfaces. Each Set element is handled as if it were a Map entry whose key and value are both the element.

• Set.prototype.entries(): Iterable<[T,T]> ^[ES6]
• Set.prototype.keys(): Iterable<T> ^[ES6]

.entries() enables you to convert a Set to a Map:

const set = new Set(['a', 'b', 'c']);
const map = new Map(set.entries());
assert.deepEqual(
  [...map.entries()],
  [['a','a'], ['b','b'], ['c','c']]);