filtering an array of objects based on another array containing keys in javascript

Question

I have an array of Objects as shown below:

people = [
{id: "1", name: "abc", gender: "m", age:"15" },
{id: "2", name: "a", gender: "m", age:"25" },
{id: "3", name: "efg", gender: "f", age:"5" },
{id: "4", name: "hjk", gender: "m", age:"35" },
{id: "5", name: "ikly", gender: "m", age:"41" },
{id: "6", name: "ert", gender: "f", age:" 30" },
{id: "7", name: "qwe", gender: "f", age:" 31" },
{id: "8", name: "bdd", gender: "m", age:" 78" },
 ]

And I have another array containing the values I want from the object:

filter = ['name','age']

I want to filter the array of people to return an array of objects that contains only values in that filter. So the Output should look like this:

    people = [
{ name: "abc",  age:"15" },
{ name: "a", age:"25" },
{ name: "efg",  age:"5" },
{ name: "hjk",  age:"35" },
{ name: "ikly",  age:"41" },
{ name: "ert",  age:" 30" },
{ name: "qwe",  age:" 31" },
{ name: "bdd", age:" 78" },
 ]

This is similar to the question here: filtering an array of objects based on another array in javascript

Use map method.

Devang
– Devang

2020-05-19 08:20:40 +00:00
Commented May 19, 2020 at 8:20 — Devang
– Devang, Commented May 19, 2020 at 8:20

Nina Scholz · Accepted Answer · 2020-05-19 08:19:32Z

2

You could map the object and take the keys for new entries.

var people = [{ id: "1", name: "abc", gender: "m", age: "15" }, { id: "2", name: "a", gender: "m", age: "25" }, { id: "3", name: "efg", gender: "f", age: "5" }, { id: "4", name: "hjk", gender: "m", age: "35" }, { id: "5", name: "ikly", gender: "m", age: "41" }, { id: "6", name: "ert", gender: "f", age: " 30" }, { id: "7", name: "qwe", gender: "f", age: " 31" }, { id: "8", name: "bdd", gender: "m", age: " 78" }],
    keys = ['name', 'age'],
    result = people.map(o => Object.fromEntries(keys.map(k => [k, o[k]])));

console.log(result);

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answered May 19, 2020 at 8:19

Nina Scholz

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Comments

MorKadosh · Accepted Answer · 2020-05-19 08:19:27Z

0

One way is to use map with reduce:

const result = people.map(item => filter.reduce((acc, cur) => {
  acc[cur] = item[cur];
  return acc;
}, {}))

answered May 19, 2020 at 8:19

MorKadosh

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Comments

StepUp · Accepted Answer · 2020-05-19 08:24:15Z

We can create entries by Object.entries, then filter our object by filter array using filter method and then just map desired elements which can be created by Object.fromEntries:

people = people.map(s => ({ ...Object.fromEntries(Object.entries(s)
    .filter(([k, v]) => filter.some(f => f == k)))}));

let people = [
    {id: "1", name: "abc", gender: "m", age:"15" },
    {id: "2", name: "a", gender: "m", age:"25" },
    {id: "3", name: "efg", gender: "f", age:"5" },
    {id: "4", name: "hjk", gender: "m", age:"35" },
    {id: "5", name: "ikly", gender: "m", age:"41" },
    {id: "6", name: "ert", gender: "f", age:" 30" },
    {id: "7", name: "qwe", gender: "f", age:" 31" },
    {id: "8", name: "bdd", gender: "m", age:" 78" },
];

let filter = ['name','age'];
people = people.map(s => ({ ...Object.fromEntries(Object.entries(s)
    .filter(([k, v]) => filter.some(f => f ==k)))}));

console.log(people);

Devang · Accepted Answer · 2020-05-19 08:40:03Z

-1

Map method can resolve your issue.

const people = [
{id: "1", name: "abc", gender: "m", age:"15" },
{id: "2", name: "a", gender: "m", age:"25" },
{id: "3", name: "efg", gender: "f", age:"5" },
{id: "4", name: "hjk", gender: "m", age:"35" },
{id: "5", name: "ikly", gender: "m", age:"41" },
{id: "6", name: "ert", gender: "f", age:" 30" },
{id: "7", name: "qwe", gender: "f", age:" 31" },
{id: "8", name: "bdd", gender: "m", age:" 78" },
 ];
 
 const result = people.map(person => {
    return {
        name: person.name,
        age: person.age
    }
})

console.log(result);

edited May 19, 2020 at 8:40

answered May 19, 2020 at 8:25

Devang

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4 Comments

yunzen Over a year ago

You haven't used the filter array

Devang Over a year ago

Why to use filter array?

yunzen Over a year ago

Because it says so in the OP

yunzen Over a year ago

Nina changed the name of the filter array to keys. She did use the dynamic filtering the OP asker wants. You did a hard coded mapping. That's why your answer is not correct

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