How can I remove duplicates from an array someArray like below based on the name property given the condition that if name is the same for two elements but for one them the type is new, the original one (without type new) will be retained?
someArray = [{id: 1, name:"apple"}, {id: 2, name:"mango"}, {id: 3, name:"apple", type: "new"}, {id: 4, name:"orange"}, {id: 5, name:"orange", type: "new"}, {id: 6, name: "pineapple", type: "new"}]
to
[{id: 1, name:"apple"}, {id: 2, name: "mango"}, {id: 4, name:"orange"}, {id: 6, name: "pineapple", type: "new"}]
You can use Map to club values by name and in case there are two values with same name just use the one without type = "new"
let someArray = [{id: 3, name:"apple", type: "new"}, {id: 1, name:"apple"}, {id: 2, name:"mango"}, {id: 4, name:"orange"}, {id: 5, name:"orange", type: "new"}, {id: 6, name: "pineapple", type: "new"}]
function getUnique(arr){
let mapObj = new Map()
arr.forEach(v => {
let prevValue = mapObj.get(v.name)
if(!prevValue || prevValue.type === "new"){
mapObj.set(v.name, v)
}
})
return [...mapObj.values()]
}
console.log(getUnique(someArray))
You can use Array.prototype.reduce and filter out the items that satisfy the condition.
const
input = [
{ id: 1, name: "apple" },
{ id: 2, name: "mango" },
{ id: 3, name: "apple", type: "new" },
{ id: 4, name: "orange" },
{ id: 5, name: "orange", type: "new" },
{ id: 6, name: "pineapple", type: "new" },
],
output = Object.values(
input.reduce((r, o) => {
if (!r[o.name] || (r[o.name].type === "new" && o.type !== "new")) {
r[o.name] = o;
}
return r;
}, {})
);
console.log(output);
You can also do it using the Spread Syntax.
const
input = [
{ id: 1, name: "apple" },
{ id: 2, name: "mango" },
{ id: 3, name: "apple", type: "new" },
{ id: 4, name: "orange" },
{ id: 5, name: "orange", type: "new" },
{ id: 6, name: "pineapple", type: "new" },
],
output = Object.values(
input.reduce(
(r, o) =>
!r[o.name] || (r[o.name].type === "new" && o.type !== "new")
? { ...r, [o.name]: o }
: r,
{}
)
);
console.log(output);
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