how to make an array unique using reduce method of javascript

Question

I have created one example for you to understand the problem.

let arr = [];
arr.push({
  module_id: 41,
  name: 'first'
}, {
  module_id: 41,
  name: 'second',
  important: true,
}, {
  module_id: 43,
  name: 'third'
});

const lookup = arr.reduce((a, e) => {
  a[e.module_id] = ++a[e.module_id] || 0;
  return a;
}, {});
console.log('lookup is', lookup);

let unique = [];
arr.filter((e) => {

  if (lookup[e.module_id] === 0) {
    unique.push(e)
  }
})

console.log('unique', unique)

first, I have an empty array arr and I am pushing 3 objects in it.
Notice, the module_name. There are two repeatings and I'd like to use the second one with the property name important.
I have used reduce here for finding out which one is repeating based on module_name. it'll return 1 with the key of 41 and 0 for 43. I want to use both but I don't want to duplicate into my unique array. Currently, I'll push only those elements which are unique, in our case module_id: 43.
Now how can I get the duplicate value (with important property)?

Does this answer your question? How to get distinct values from an array of objects in JavaScript? — Mark Baijens
– Mark Baijens, Commented Dec 21, 2021 at 11:26

siva ganesh siva ganesh · Accepted Answer · 2021-12-21 12:01:40Z

3

let arr = [];
arr.push(
  {
    module_id: 41,
    name: 'first',
  },
  {
    module_id: 41,
    name: 'second',
    important: true,
  },
  {
    module_id: 43,
    name: 'third',
  }
);


const result = arr.reduce(
  (acc, curr) =>
    acc.find((v) => v.module_id === curr.module_id) ? acc : [...acc, curr],
  []
);

console.log(result);

answered Dec 21, 2021 at 12:01

siva ganesh siva ganesh

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2 Comments

Nina Scholz Over a year ago

this does not return the object with important property.

MD. RAKIB HASAN Over a year ago

Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.

Nina Scholz · Accepted Answer · 2021-12-21 11:32:11Z

1

You could take a Map and get only distinct first values or the ones with important flag.

const
    array = [{ module_id: 41, name: 'first' }, { module_id: 41, name: 'second', important: true }, { module_id: 43, name: 'third' }],
    result = Array.from(array
        .reduce(
            (m, o) => m.set(o.module_id, !o.important && m.get(o.module_id) || o),
            new Map
        )
        .values()
    );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

answered Dec 21, 2021 at 11:32

Nina Scholz

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Comments

Timsib Adnap · Accepted Answer · 2021-12-21 11:33:59Z

1

try

let unique = [];
arr.filter((e) => {

  if (lookup[e.module_id] === 0) {
    unique.push(e)
  }
  else if (e.important) {
    unique.push(e)
  }
})

answered Dec 21, 2021 at 11:33

Timsib Adnap

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Comments

Manav.J · Accepted Answer · 2023-09-13 09:08:54Z

0

const arr = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5];
const uniqueNums = arr.reduce( (all,item) => {
  if(all.indexOf(item) === -1) {
    all.push(item)
  }
  return all
},[])

answered Sep 13, 2023 at 9:08

Manav.J

1

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