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Get Object Properties By Predefined Property Names Array

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

0

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 ?

javascript

share|improve this question

asked Mar 22 at 11:21

berkanberkan

1401111

add a comment | 

0

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 ?

javascript

share|improve this question

asked Mar 22 at 11:21

berkanberkan

1401111

add a comment | 

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 ?

javascript

share|improve this question

asked Mar 22 at 11:21

berkanberkan

1401111

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

[ "prop1", "prop2" ]

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

share|improve this question

asked Mar 22 at 11:21

berkanberkan

1401111

share|improve this question

asked Mar 22 at 11:21

berkanberkan

1401111

asked Mar 22 at 11:21

berkanberkan

1401111

asked Mar 22 at 11:21

berkanberkan

1401111

4 Answers
4

active

oldest

votes

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)

share|improve this answer

edited Mar 22 at 11:33

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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)
  
  – axelduch
  Mar 22 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @axelduch a is accumulator, c is current element. Docs outline what arguments are
  
  – charlietfl
  Mar 22 at 11:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  
  – axelduch
  Mar 22 at 11:37
  

  
  
  
  

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:27

User97User97

1,58211435

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:54

AZ_AZ_

942310

add a comment | 

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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)

share|improve this answer

edited Mar 22 at 11:33

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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)
  
  – axelduch
  Mar 22 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @axelduch a is accumulator, c is current element. Docs outline what arguments are
  
  – charlietfl
  Mar 22 at 11:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  
  – axelduch
  Mar 22 at 11:37
  

  
  
  
  

add a comment | 

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)

share|improve this answer

edited Mar 22 at 11:33

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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)
  
  – axelduch
  Mar 22 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @axelduch a is accumulator, c is current element. Docs outline what arguments are
  
  – charlietfl
  Mar 22 at 11:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  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
  
  – axelduch
  Mar 22 at 11:37
  

  
  
  
  

add a comment | 

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)

share|improve this answer

edited Mar 22 at 11:33

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

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)

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)

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)

share|improve this answer

edited Mar 22 at 11:33

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

answered Mar 22 at 11:28

charlietflcharlietfl

143k1391126

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)

share|improve this answer

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

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)

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)

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)

share|improve this answer

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

answered Mar 22 at 11:26

Nenad VracarNenad Vracar

73.8k126085

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)

share|improve this answer

answered Mar 22 at 11:27

User97User97

1,58211435

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:27

User97User97

1,58211435

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:27

User97User97

1,58211435

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)

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)

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)

share|improve this answer

answered Mar 22 at 11:27

User97User97

1,58211435

answered Mar 22 at 11:27

User97User97

1,58211435

answered Mar 22 at 11:27

User97User97

1,58211435

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)

share|improve this answer

answered Mar 22 at 11:54

AZ_AZ_

942310

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:54

AZ_AZ_

942310

add a comment | 

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)

share|improve this answer

answered Mar 22 at 11:54

AZ_AZ_

942310

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)

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)

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)

share|improve this answer

answered Mar 22 at 11:54

AZ_AZ_

942310

answered Mar 22 at 11:54

AZ_AZ_

942310

answered Mar 22 at 11:54

AZ_AZ_

942310