Remove elements from an array in Typescript if they are not present in a nested array

Question

Remove elements from an array in Typescript if they are not present in a nested array

I am struggling with filtering values from a single-dimensional array based on id matches in an array of objects.

array1 = [1, 3, 15, 16, 18];

array2 = [
       { id: 1, dinner : pizza },
       { id: 15, dinner : sushi },
       { id: 18, dinner : hummus }
]

My goal is to remove values from array1 that don't have a matching id in array2.

While I know how to filter two single-dimensional arrays, I'm having trouble adapting the code for this scenario involving an array of objects.

const array1 = array1.filter(id => array2.includes(id));

If anyone has any insight or suggestions, please share them. Thank you!

javascript arrays typescript

Answer 1

Answer №1

These two arrays consist of single dimension elements.

To eliminate numbers from array1 that do not match any of the ids in the objects within array2, you can combine the .some() and .filter() functions.

const array1 = [1, 3, 15,16,18];
const array2 = [
    { id: 1, dinner : 'pizza' },
    { id: 15, dinner : 'sushi' },
    { id: 18, dinner : 'hummus' }
]

const filteredArr = array1.filter(v => array2.some(o => v == o.id));

console.log(filteredArr);

Answer 2

These two arrays consist of single dimension elements.

To eliminate numbers from array1 that do not match any of the ids in the objects within array2, you can combine the .some() and .filter() functions.

const array1 = [1, 3, 15,16,18];
const array2 = [
    { id: 1, dinner : 'pizza' },
    { id: 15, dinner : 'sushi' },
    { id: 18, dinner : 'hummus' }
]

const filteredArr = array1.filter(v => array2.some(o => v == o.id));

console.log(filteredArr);

Answer 3

Answer №2

If you want to extract specific IDs and filter them, you can follow the below steps:

var array1 = [1, 3, 15, 16, 18];

var array2 = [
       { id: 1, dinner : "pizza" },
       { id: 15, dinner : "sushi" },
       { id: 18, dinner : "hummus" }
]

const Ids = array2.map(i=> i.id);
var res = array2.filter(i => Ids.includes(i.id)); 
var res2 = Ids.filter(i => array1.includes(i))// Or, just to get common Ids

console.log(res)
console.log(res2)

Alternatively, using reduce can help you achieve the same result without making two passes:

var array1 = [1, 3, 15, 16, 18];

var array2 = [
       { id: 1, dinner : "pizza" },
       { id: 15, dinner : "sushi" },
       { id: 18, dinner : "hummus" }
]

var res = array2.reduce((acc, {id})=>{
  if(array1.includes(id)){
   acc = [...acc, id]
  }
  return acc
},[]);

console.log(res)

Answer 4

If you want to extract specific IDs and filter them, you can follow the below steps:

var array1 = [1, 3, 15, 16, 18];

var array2 = [
       { id: 1, dinner : "pizza" },
       { id: 15, dinner : "sushi" },
       { id: 18, dinner : "hummus" }
]

const Ids = array2.map(i=> i.id);
var res = array2.filter(i => Ids.includes(i.id)); 
var res2 = Ids.filter(i => array1.includes(i))// Or, just to get common Ids

console.log(res)
console.log(res2)

Alternatively, using reduce can help you achieve the same result without making two passes:

var array1 = [1, 3, 15, 16, 18];

var array2 = [
       { id: 1, dinner : "pizza" },
       { id: 15, dinner : "sushi" },
       { id: 18, dinner : "hummus" }
]

var res = array2.reduce((acc, {id})=>{
  if(array1.includes(id)){
   acc = [...acc, id]
  }
  return acc
},[]);

console.log(res)

Remove elements from an array in Typescript if they are not present in a nested array

Answer №1

Answer №2

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