Filter an array with another array INCLUSIVELY

Question

I am having a hard time filtering an array by matching with all elements of another array, INCLUSIVELY. So for example:

var s = [
 {id: 1, area: ['foo', 'bar', 'other', 'again']},
 {id: 2, area: ['bar']},
 {id: 3, area: ['other']},
 {id: 4, area: ['foo']}
]

var areas = ['foo', 'bar']

Expected output should be:

[
 {id: 1, area: ['foo', 'bar', 'other', 'again']}
]

that is, each element in the expected result must contain ALL elements in the 'areas' array. This is what i tried but its returning an empty array so I think my function is wrong:

const filteredArray = s.filter(n => n.area.every(a => areas.includes(a)));

Nikita Madeev · Accepted Answer · 2020-07-28 07:20:39Z

4

You need to check that each element from areas is present in area field. In your example, you are doing the opposite, trying to check that each area field is present in areas.

const s = [
    { id: 1, area: ['foo', 'bar', 'other', 'again'] },
    { id: 2, area: ['bar'] },
    { id: 3, area: ['other'] },
    { id: 4, area: ['foo'] },
];
const areas = ['foo', 'bar'];

const result = s.filter(n => areas.every(a => n.area.includes(a)));

console.log(result);

answered Jul 28, 2020 at 7:20

Nikita Madeev

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2 Comments

Diego Over a year ago

thanks very much. That works for that array. but i just realized my array was slightly differently structured... The array area in s should actually be an array of objects like this: s = [ { id: 1, area: [{area: 'foo'}, {area: 'bar}', {area: 'other}', {area: 'again'}] }, { id: 2, area: [{area: 'bar}'] }, { id: 3, area: [{area: 'other'}] }, { id: 4, area: [{area: 'foo'}] }, ]; how would the function be in this case?

Nikita Madeev Over a year ago

@Diego Try it s.filter(obj => areas.every(filterArea => obj.area.map(areaObj => areaObj.area).includes(filterArea)));

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