CODEWITHSUNDEEP
javascriptmultidimensional-arrayfilter
Simon
Simon

Reputation: 11

Filter a Multidimensional Array

I have a multidimensional array like this

Owner[0].dog[0].typeOfDog = "shiba inu", Owner[0].dog[1].typeOfDog = "poodle", Owner[0].dog[2].typeOfDog = "samoyan", Owner[1].dog[0].typeOfDog = "poodle", Owner[1].dog[1].typeOfDog = "poodle", Owner[1].dog[2].typeOfDog = "samoyan", Owner[2].dog[0].typeOfDog = "poodle"

I want to create a variable that contains this exact data structure and returns the same list but without any poodles.

For example:

Owner[0].dog[0].typeOfDog = "shiba inu", Owner[0].dog[0].typeOfDog = "samoyan", Owner[1].dog[0].typeOfDog = "samoyan"

I managed to filter it out using Map and Filter but I am unable to keep the same structure. How would I do this?

owners.Map(owner => owner.dogs.filter(dog => dog.typeOfDog !== "poodle"));

This is returning an array of dogs that are not poodles but I would like to get a array of owners each of which have an array of dogs that are not poodles.

Upvotes: 1

Views: 106

Answers (3)

Mister Jojo
Mister Jojo

Reputation: 22320

you can do that...

const Owner = 
  [ { dog :
      [ { typeOfDog : 'shiba inu' }
      , { typeOfDog : 'poodle'    } 
      , { typeOfDog : 'samoyan'   } 
    ] }
  , { dog :
      [ { typeOfDog : 'poodle'  }
      , { typeOfDog : 'poodle'  } 
      , { typeOfDog : 'samoyan' } 
    ] }
  , { dog :
      [ { typeOfDog : 'poodle' }
    ] }
  ]
, NoPoodles = Owner
  .map(({dog})=>({dog: dog.filter(({typeOfDog})=>typeOfDog!=='poodle')}))
  .filter(({dog})=>dog.length)
  ;
  
console.log( 'result length =',  NoPoodles.length  )
console.log(  NoPoodles  )
.as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}

Upvotes: 0

Gohchi
Gohchi

Reputation: 469

You forgot to return an object instead of an array:

owners.map(owner => (
  Object.assign({}, owner, {
    dogs: owner.dogs.filter(dog => dog.typeOfDog !== "poodle")
  })
));

Upvotes: 0

Adrian Brand
Adrian Brand

Reputation: 21638

Use the spread operator to construct a new object and override the dog property.

const OwnersNoPoodles = Owner.map(o => ({...o, dog: o.dog.filter(d => d.typeOfDog !== 'poodle' )}));

Upvotes: 1

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