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javascript instance method

Set.prototype.delete()

View on MDN ↗

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

Interactive exampleOpen the canonical MDN source to run this embedded demo.
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()

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.

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);
  }
});

Specifications

SpecificationsStandards references are available on the canonical MDN page.

Browser compatibility

Browser compatibilityCompatibility data is available on the canonical MDN page.

See also