Set.prototype.delete()

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

{{InteractiveExample("JavaScript Demo: Set.prototype.delete()")}}

```js interactive-example 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

```js-nolint
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#

{{Specifications}}

Browser compatibility#

{{Compat}}

See also#

• {{jsxref("Set")}}
• {{jsxref("Set.prototype.add()")}}
• {{jsxref("Set.prototype.clear()")}}
• {{jsxref("Set.prototype.has()")}}