Get Object Properties By Predefined Property Names Array

Question

I am pretty sure about this kind of question answered before, but I couldn't make good search.

I have an array of object like;

[
  { prop1:"foo1", prop2:"baz1", prop3:"bar1", prop4:"qux1"}
  { prop1:"foo2", prop2:"baz2", prop3:"bar2", prop4:"qux2"}
  { prop1:"foo3", prop2:"baz3", prop3:"bar3", prop4:"qux3"}
]

And I have an array which is storing needed property names like;

[ "prop1", "prop2" ]

So I need to filter all of my objects in array with the property names which is given by another array. And the output will be like;

[   
 { prop1:"foo1", prop2:"baz1" }   
 { prop1:"foo2", prop2:"baz2" }  
 { prop1:"foo3", prop2:"baz3" } 
]

How can I do this in proper way ?

charlietfl · Accepted Answer · 2019-03-22 11:33:55Z

4

You can use map() and reduce()

const data = [
  { prop1:"foo1", prop2:"baz1", prop3:"bar1", prop4:"qux1"},
  { prop1:"foo2", prop2:"baz2", prop3:"bar2", prop4:"qux2"},
  { prop1:"foo3", prop2:"baz3", prop3:"bar3", prop4:"qux3"}
];

const props = [ "prop1", "prop2" ];

const res = data.map(e => props.reduce((a,c) => (a[c] = e[c] , a), {}));

console.log(res)

edited Mar 22, 2019 at 11:33

answered Mar 22, 2019 at 11:28

charlietfl

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3 Comments

a-d Over a year ago

Nice solution, could you please use more explicit variable names for the sake of readability? I understand the point of having a one liner, but reduce's params are obscure enough (upvoting anyway)

charlietfl Over a year ago

@axelduch a is accumulator, c is current element. Docs outline what arguments are

a-d Over a year ago

I was rather saying that for those unfamiliar with reduce. It doesn't have to be scary to them :) But fair enough - people curious enough about it will check the doc anyway

Nenad Vracar · Accepted Answer · 2019-03-22 11:26:26Z

4

You can create pick function with reduce method and then use it with map method.

const data = [{ prop1:"foo1", prop2:"baz1", prop3:"bar1", prop4:"qux1"},{ prop1:"foo2", prop2:"baz2", prop3:"bar2", prop4:"qux2"},{ prop1:"foo3", prop2:"baz3", prop3:"bar3", prop4:"qux3"}]
const props = [ "prop1", "prop2" ]

const pick = (o, p) => p.reduce((r, e) => Object.assign(r, {[e]: o[e]}), {})
const res = data.map(o => pick(o, props));
console.log(res)

answered Mar 22, 2019 at 11:26

Nenad Vracar

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Comments

Raeesaa · Accepted Answer · 2019-03-22 11:27:41Z

3

You can do something like below to achieve the result:

const data = [
  { prop1:"foo1", prop2:"baz1", prop3:"bar1", prop4:"qux1"},
  { prop1:"foo2", prop2:"baz2", prop3:"bar2", prop4:"qux2"},
  { prop1:"foo3", prop2:"baz3", prop3:"bar3", prop4:"qux3"},
]

const keys = [ "prop1", "prop2" ]

let result = data.map((record) => {
  let obj = {}
  keys.forEach((key) => {
    obj[key] = record[key]
  })
  return obj
})

console.log(result)

answered Mar 22, 2019 at 11:27

Raeesaa

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Comments

arizafar · Accepted Answer · 2019-03-22 11:54:35Z

1

let arr = [
  { prop1:"foo1", prop2:"baz1", prop3:"bar1", prop4:"qux1"},
  { prop1:"foo2", prop2:"baz2", prop3:"bar2", prop4:"qux2"},
  { prop1:"foo3", prop2:"baz3", prop3:"bar3", prop4:"qux3"}
]

let filter = [ "prop1", "prop2" ];

let out = [...arr].map(e => Object.keys(e).map(k => !filter.includes(k) ? delete e[k] :true) && e);
console.log(out)

answered Mar 22, 2019 at 11:54

arizafar

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