JavaScript - filter in loop, create duplicate

Question

I want to filter an array with another array in order to know if there are new people.

const people = [
    { "name": "jerry" },
    { "name": "tom" },
    { "name": "alex" }
]
const newList = [
    { "name": "bran" },
    { "name": "jerry" },
    { "name": "john" }
]

const new_people = []

for (const pp of people) {

    let result = newList.filter(newL => newL.name != pp.name)

    if (result) {
        new_people.push(result)
    }
}

console.log(new_people)

This is the result:

[
  [ { name: 'bran' }, { name: 'john' } ],
  [ { name: 'bran' }, { name: 'jerry' }, { name: 'john' } ],
  [ { name: 'bran' }, { name: 'jerry' }, { name: 'john' } ]
]

But I'm looking for:

[ { name: 'bran' }, { name: 'john' } ]

I would like to avoid the loop because it makes duplicate in the result but I don't know how I can't do it without the loop.

Kazi Tajnur Islam · Accepted Answer · 2022-08-31 10:06:07Z

2

First make a temporary array of people name:

const peopleNames = people.map(pp => pp.name);

Then the peopleNames will be as follows:

['jerry', 'tom', 'alex']

Now filter the new people from the newList:

const newPeople = newList.filter(pp => !peopleNames.includes(pp.name));

The newPeople will be an array of objects that you are looking for.

[{name: 'bran'}, {name: 'john'}]

answered Aug 31, 2022 at 10:06

Kazi Tajnur Islam

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Comments

Sakil · Accepted Answer · 2022-08-31 10:04:25Z

1

const people = [
    { "name": "jerry" },
    { "name": "tom" },
    { "name": "alex" }
]
const newList = [
    { "name": "bran" },
    { "name": "jerry" },
    { "name": "john" }
]
const output = newList.filter(a => people.filter(x => x.name == a.name).length == 0);
console.log('By USing Filter', output);
//filter the newList and retain those object, which are not present in the people

//Way 2: By using some
//const output2 = newList.filter(a => !people.some(x => x.name == a.name));
//console.log('By Using Some', output2);

answered Aug 31, 2022 at 10:04

Sakil

6525 silver badges8 bronze badges

2 Comments

GrowyTobe Over a year ago

Thank you, way 2 is simple to understand. But for way 1, it is difficult to understand the following part: ".length == 0". I thought by removing ".length == 0" I would get [{ name: 'jerry' }] but I didn't

Sakil Over a year ago

@GrowyTobe, If you want to use Way 1 and expect to get the output i.e. present in both [{name: 'jerry'}] you can replace .length == 0 with .length > 0

prasad_ · Accepted Answer · 2022-08-31 12:04:46Z

1

You can use Array's reduce method to get the desired result:

const new_people = newList.reduce((accVal, e) => 
    (people.map(p => p.name).includes(e.name)) 
    ? accVal  
    : accVal.concat({ "name": e.name } ), 
[ ] )

answered Aug 31, 2022 at 12:04

prasad_

14.4k2 gold badges27 silver badges44 bronze badges

2 Comments

GrowyTobe Over a year ago

If I'm not wrong you use the conditional (ternary) operator in reduce method. Shouldn't accVal have the value of [{ name: 'jerry' }] as "jerry" is in both array?

prasad_ Over a year ago

Initially accVal will be an empty array ([ ]). So, when there is a match nothing happens. When there is no match the element from newList is added to the accVal. Hence the result will have the expected elements - [ { name: 'bran' }, { name: 'john' } ].

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