removeDuplicates()
takes in an array of objects and returns a new array without any duplicate objects (based on the id property).
const allTests = [
{name: 'Test1', id: '1'},
{name: 'Test3', id: '3'},
{name: 'Test2', id: '2'},
{name: 'Test2', id: '2'},
{name: 'Test3', id: '3'}
];
function removeDuplicates(array) {
let uniq = {};
return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true))
}
removeDuplicates(allTests);
Expected outcome:
[
{name: 'Test1', id: '1'},
{name: 'Test3', id: '3'},
{name: 'Test2', id: '2'}
];
First, we set the value of variable uniq to an empty object.
Next, we filter through the array of objects. Filter creates a new array with all elements that pass the test implemented by the provided function.
return array.filter(obj => !uniq[obj.id] && (uniq[obj.id] = true));
Above, we use the short-circuiting functionality of &&. If the left side of the && evaluates to true, then it returns the value on the right of the &&. If the left side is false, it returns what is on the left side of the &&.
For each object(obj) we check uniq for a property named the value of obj.id (In this case, on the first iteration it would check for the property '1'.) We want the opposite of what it returns (either true or false) which is why we use the ! in !uniq[obj.id]
. If uniq has the id property already, it returns true which evaluates to false (!) telling the filter function NOT to add that obj. However, if it does not find the obj.id property, it returns false which then evaluates to true (!) and returns everything to the right of the &&, or (uniq[obj.id] = true). This is a truthy value, telling the filter method to add that obj to the returned array, and it also adds the property {1: true} to uniq. This ensures that any other obj instance with that same id will not be added again.
aaaaa.aaaa.push(...)
:) – dazitothings.thing
. This unnecessarily complicates both the question and the answer. – mikemaccana