Set.prototype.delete()

BaselineWidely available

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

Thedelete() method ofSet instances removes a specified value from this set, if it is in the set.

Try it

const set1 = new Set();set1.add({ x: 10, y: 20 }).add({ x: 20, y: 30 });// Delete any point with `x > 10`.set1.forEach((point) => {  if (point.x > 10) {    set1.delete(point);  }});console.log(set1.size);// Expected output: 1

Syntax

setInstance.delete(value)

Parameters

value: The value to remove fromSet.

Return value

Returnstrue ifvalue was already inSet; otherwisefalse.

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);  }});