I have a source list of objects of 4 properties with some duplicate values, want to select distinct objects of 2 properties without using a foreach loop.
Source list:
let Students= [
{Id: 1, Name: 'A', Class: 'I', Subject: 'Math'},
{Id: 2, Name: 'B', Class: 'II', Subject: 'Bengali'},
{Id: 1, Name: 'A', Class: 'I', Subject: 'Science'},
{Id: 2, Name: 'B', Class: 'II', Subject: 'English'}
];
I want an output list like:
[
{Id: 1, Name: 'A'},
{Id: 2, Name: 'B'}
]
Simple Solution to return Distinct data for Object Array
this.studentOptions = this.students.filter((item, i, arr) => arr.findIndex((t) => t.Id=== item.Id) === i);
This solution produces the result in the form you asked for. Objects need to be stringified before they get added to the Set, otherwise equal ones are not recognized (see https://stackoverflow.com/a/29759699/2358409).
const set = new Set(Students.map(s => JSON.stringify({ Id: s.Id, Name: s.Name })));
const identifiers = Array.from(set).map(i => JSON.parse(i));
If you want to avoid stringify, you can do the following.
const set = new Set(Students.map(s => s.Id + ';' + s.Name));
const identifiers = Array.from(set).map(i => ({ Id: i.split(';')[0], Name: i.split(';')[1] }));
The same can be done in a cleaner way using Map.
const map = new Map();
Students.map(s => map.set(s.Id, { Id: s.Id, Name: s.Name }));
const identifiers = Array.from(map.values());
Since we're in the world of typescript, make use of the types provided by it. Create a class Student, this will be useful for easier management of your objects.
export class Student {
Id: number;
Name: string;
Class?: string;
Subject?: string;
constructor(Id: number, Name: string) {
this.Id = Id;
this.Name = Name;
}
}
Create an array to hold the distinct objects:
distinctStud: Student[] = [];
Now you may:
from(this.Students)
.pipe(
map(std => new Student(std.Id, std.Name)),
distinct(x => x.Id && x.Name)
)
.subscribe(val => this.distinctStud.push(val));
Array.from(new Set(yourArray.map((item: any) => item.id)))
You can try this like newbie way
let st=[];
Students.map((data)=>{
let isExist=!!st.find((data2)=>{
return data2.Id==data.Id});
if(!isExist)
st.push(data);
});
Using rxjs's map and distinct operators we can achieve this. Please check the code below
var Students= [
{Id: 1, Name: 'A', Class: 'I', Subject: 'Math'},
{Id: 2, Name: 'B', Class: 'II', Subject: 'Bengali'},
{Id: 1, Name: 'A', Class: 'I', Subject: 'Science'},
{Id: 2, Name: 'B', Class: 'II', Subject: 'English'}
];
const sourceObj = from(Students);
const mappedObj=sourceObj.pipe(map((x)=>{return {Id: x.Id, Name:x.Name}}));
const distictObj = mappedObj.pipe(distinct((x)=>x.Id && x.Name));
console.log(distictObj.subscribe(x=>console.log(x)));Please check this StackBlitz.
Thanks.
