Filter Javascript array of objects with another array of objects

Question

I have list

var list = [
  {name: 'Hello', ads: true, pas: ['Av', 'Sd', 'Qr']},
  {name: 'Fre', ads: false, pas: ['Sd', 'Bc']},
  {name: 'Nat', ads: false, pas: ['Bc', 'Mo', 'Rr']},
  {name: 'Bor', ads: true, pas: ['Bo', 'Be', 'De']},
  {name: 'Arg', ads: true, pas: ['Ar', 'Na', 'Br']},
];

and filter of array

var filter = [
 'Sd','Be'
];

I have tried with filter but its not filtering.

list
  .filter(function(ls) {
    return ls.pas
      .filter(function(p){
        return filter.indexOf(p) > -1;
      })
  })

My expected output

[
  {name: 'Hello', ads: true, pas: ['Av', 'Sd', 'Qr']},
  {name: 'Fre', ads: false, pas: ['Sd', 'Bc']},
  {name: 'Bor', ads: true, pas: ['Bo', 'Be', 'De']}
];

i cant use jquery, its inside react, but i can use lodash, but i don't know how to use it. — Wimal Weerawansa
– Wimal Weerawansa, Commented Oct 17, 2016 at 11:41
why does the answer of the last question of you does not work for you? — Nina Scholz
– Nina Scholz, Commented Oct 17, 2016 at 11:42
last question only got one level of filtering, this collection contains multiple levels of array to be filter. — Wimal Weerawansa
– Wimal Weerawansa, Commented Oct 17, 2016 at 11:43

Nenad Vracar · Accepted Answer · 2016-10-17 11:42:40Z

3

You can use some() to check if any element from pas array of current object is inside filter array.

var list = [
  {name: 'Hello', ads: true, pas: ['Av', 'Sd', 'Qr']},
  {name: 'Fre', ads: false, pas: ['Sd', 'Bc']},
  {name: 'Nat', ads: false, pas: ['Bc', 'Mo', 'Rr']},
  {name: 'Bor', ads: true, pas: ['Bo', 'Be', 'De']},
  {name: 'Arg', ads: true, pas: ['Ar', 'Na', 'Br']},
];

var filter = [
 'Sd','Be'
];

var result = list.filter(function(e) {
  return e.pas.some(function(a) {
    return filter.indexOf(a) != -1
  })
})

console.log(result);

answered Oct 17, 2016 at 11:42

Nenad Vracar

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Comments

James Wakefield · Accepted Answer · 2016-10-17 11:52:54Z

0

Using es6 it looks quite nice:

const list = [
   {name: 'Hello', ads: true, pas: ['Av', 'Sd', 'Qr']},
   {name: 'Fre', ads: false, pas: ['Sd', 'Bc']},
   {name: 'Nat', ads: false, pas: ['Bc', 'Mo', 'Rr']},
   {name: 'Bor', ads: true, pas: ['Bo', 'Be', 'De']},
   {name: 'Arg', ads: true, pas: ['Ar', 'Na', 'Br']},
],
filter = ['Sd','Be']

list.filter(a=>a.pas.some(a=>filter.includes(a))

answered Oct 17, 2016 at 11:52

James Wakefield

5363 silver badges11 bronze badges

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