Set.prototype.delete()
The delete() method removes a specified value from a
Set object, if it is in the set.
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Syntax
delete(value)
Parameters
value-
The value to remove from
Set.
Return value
Returns true if value was already in
Set; otherwise false.
Examples
Using the delete() method
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
| Specification |
|---|
| ECMAScript Language Specification # sec-set.prototype.delete |
Browser compatibility
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