Set.prototype.delete()

Baseline Widely available

This feature is well established and works across many devices and browser versions. It’s been available across browsers since July 2015.

The delete() method of Set instances removes the specified value from this set, if it is in the set.

Try it

const set = new Set();
set.add({ x: 10, y: 20 }).add({ x: 20, y: 30 });

// Delete any point with `x > 10`.
set.forEach((point) => {
  if (point.x > 10) {
    set.delete(point);
  }
});

console.log(set.size);
// Expected output: 1

Syntax

setInstance.delete(value)

Parameters

value

The value to remove from the Set object. Objects are compared by reference, not by value.

Return value

true if a value in the Set object has been removed successfully. false if the value is not found in the Set.

Examples

Using delete()

js
const mySet = new Set();
mySet.add("foo");

console.log(mySet.delete("bar")); // false; no "bar" element found to be deleted.
console.log(mySet.delete("foo")); // true; successfully removed.

console.log(mySet.has("foo")); // false; the "foo" element is no longer present.

Deleting an object from a set

Because objects are compared by reference, you have to delete them by checking individual properties if you don't have a reference to the original object.

js
const setObj = new Set(); // Create a new set.

setObj.add({ x: 10, y: 20 }); // Add object in the set.

setObj.add({ x: 20, y: 30 }); // Add object in the set.

// Delete any point with `x > 10`.
setObj.forEach((point) => {
  if (point.x > 10) {
    setObj.delete(point);
  }
});