Destructure an object by an array of its keys

up vote
0
down vote

favorite

Say that we have an object like this one:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

And an array of some of its keys:

let arr_of_keys = ["a", "d", "e"]

Is it possible to destructure the object using the predefined keys in the array, something along the lines of:

let ...arr_of_keys = obj;

To finally end up with:

a = "John", d = "Joseph", e = "Roger"

edited yesterday

asked yesterday

Milan Velebit

6842619

2

No, because that would mean dynamic variable names, which are quite a code smell. (also, a is already defined)
– CertainPerformance
yesterday

2

I'm guessing you meant let arr_of_keys = ['a', 'd', 'e']? What you have is an array of references to some things.
– Andy
yesterday

@Andy Yeah, my bad, edited, thanks for the insight, completely missed it
– Milan Velebit
yesterday

Closest thing I found, stackoverflow.com/questions/17781472/…
– Adi
yesterday

add a comment |

up vote
0
down vote

favorite

Say that we have an object like this one:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

And an array of some of its keys:

let arr_of_keys = ["a", "d", "e"]

Is it possible to destructure the object using the predefined keys in the array, something along the lines of:

let ...arr_of_keys = obj;

To finally end up with:

a = "John", d = "Joseph", e = "Roger"

edited yesterday

asked yesterday

Milan Velebit

6842619

2

No, because that would mean dynamic variable names, which are quite a code smell. (also, a is already defined)
– CertainPerformance
yesterday

2

I'm guessing you meant let arr_of_keys = ['a', 'd', 'e']? What you have is an array of references to some things.
– Andy
yesterday

@Andy Yeah, my bad, edited, thanks for the insight, completely missed it
– Milan Velebit
yesterday

Closest thing I found, stackoverflow.com/questions/17781472/…
– Adi
yesterday

add a comment |

up vote
0
down vote

favorite

Say that we have an object like this one:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

And an array of some of its keys:

let arr_of_keys = ["a", "d", "e"]

Is it possible to destructure the object using the predefined keys in the array, something along the lines of:

let ...arr_of_keys = obj;

To finally end up with:

a = "John", d = "Joseph", e = "Roger"

edited yesterday

asked yesterday

Milan Velebit

6842619

Say that we have an object like this one:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

And an array of some of its keys:

let arr_of_keys = ["a", "d", "e"]

Is it possible to destructure the object using the predefined keys in the array, something along the lines of:

let ...arr_of_keys = obj;

To finally end up with:

a = "John", d = "Joseph", e = "Roger"

javascript ecmascript-6

edited yesterday

asked yesterday

Milan Velebit

6842619

edited yesterday

asked yesterday

Milan Velebit

6842619

edited yesterday

asked yesterday

Milan Velebit

6842619

asked yesterday

Milan Velebit

6842619

asked yesterday

Milan Velebit

6842619

2

No, because that would mean dynamic variable names, which are quite a code smell. (also, a is already defined)
– CertainPerformance
yesterday

2

I'm guessing you meant let arr_of_keys = ['a', 'd', 'e']? What you have is an array of references to some things.
– Andy
yesterday

@Andy Yeah, my bad, edited, thanks for the insight, completely missed it
– Milan Velebit
yesterday

Closest thing I found, stackoverflow.com/questions/17781472/…
– Adi
yesterday

add a comment |

2

No, because that would mean dynamic variable names, which are quite a code smell. (also, a is already defined)
– CertainPerformance
yesterday

2

I'm guessing you meant let arr_of_keys = ['a', 'd', 'e']? What you have is an array of references to some things.
– Andy
yesterday

@Andy Yeah, my bad, edited, thanks for the insight, completely missed it
– Milan Velebit
yesterday

Closest thing I found, stackoverflow.com/questions/17781472/…
– Adi
yesterday

No, because that would mean dynamic variable names, which are quite a code smell. (also, a is already defined)
– CertainPerformance
yesterday

I'm guessing you meant let arr_of_keys = ['a', 'd', 'e']? What you have is an array of references to some things.
– Andy
yesterday

@Andy Yeah, my bad, edited, thanks for the insight, completely missed it
– Milan Velebit
yesterday

Closest thing I found, stackoverflow.com/questions/17781472/…
– Adi
yesterday

add a comment |

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

You want a simple .reduce method like the one below:

var result = arr_of_keys.reduce(function(o,item)

 if(obj.hasOwnProperty(item))
 o[item] = obj[item]; 
 

 return o;
, );

Here's an example:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

Here's a JSFiddle runnable (since the built-in one returns a 503.)

edited yesterday

answered yesterday

Adriani6

3,8792923

1

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

1

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

add a comment |

up vote
1
down vote

Here is a possible helper function for the provided issue. I added a set conversion to remove possible duplications in order save resources. Also added simple error handling in form of console error messages.

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

answered yesterday

Zsolt Gulyás

337

add a comment |

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53224413%2fdestructure-an-object-by-an-array-of-its-keys%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
2
down vote

accepted

You want a simple .reduce method like the one below:

var result = arr_of_keys.reduce(function(o,item)

 if(obj.hasOwnProperty(item))
 o[item] = obj[item]; 
 

 return o;
, );

Here's an example:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

Here's a JSFiddle runnable (since the built-in one returns a 503.)

edited yesterday

answered yesterday

Adriani6

3,8792923

1

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

1

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

add a comment |

up vote
2
down vote

accepted

You want a simple .reduce method like the one below:

var result = arr_of_keys.reduce(function(o,item)

 if(obj.hasOwnProperty(item))
 o[item] = obj[item]; 
 

 return o;
, );

Here's an example:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

Here's a JSFiddle runnable (since the built-in one returns a 503.)

edited yesterday

answered yesterday

Adriani6

3,8792923

1

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

1

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

add a comment |

up vote
2
down vote

accepted

You want a simple .reduce method like the one below:

var result = arr_of_keys.reduce(function(o,item)

 if(obj.hasOwnProperty(item))
 o[item] = obj[item]; 
 

 return o;
, );

Here's an example:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

Here's a JSFiddle runnable (since the built-in one returns a 503.)

edited yesterday

answered yesterday

Adriani6

3,8792923

You want a simple .reduce method like the one below:

var result = arr_of_keys.reduce(function(o,item)

 if(obj.hasOwnProperty(item))
 o[item] = obj[item]; 
 

 return o;
, );

Here's an example:

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

Here's a JSFiddle runnable (since the built-in one returns a 503.)

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

let obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"
let arr_of_keys = ["a", "d", "e", "f"];

var result = arr_of_keys.reduce(function(o,item)

	if(obj.hasOwnProperty(item))
 	o[item] = obj[item]; 
 
 
 return o;
, );

console.log(result)

edited yesterday

answered yesterday

Adriani6

3,8792923

edited yesterday

answered yesterday

Adriani6

3,8792923

answered yesterday

Adriani6

3,8792923

answered yesterday

Adriani6

3,8792923

1

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

1

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

add a comment |

1

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

1

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

@dv reason for down-vote? The output matches the desired in OP.
– Adriani6
yesterday

it kinda doesn’t as it basically filters an object. OP said destructure which would create distinct variables.
– evolutionxbox
yesterday

@evolutionxbox "It kinda doesn't" isn't a valid reason. Just because the wording of a question states something it doesn't mean it's the way to go. Also you need to filter an object given you have two arrays to work with where values in one depend on the outcome - so at some point, a filter will take place regardless. In my answer there is no need for deconstructing an object.
– Adriani6
yesterday

I'm aware. I used the word "kinda" for that reason. I should have stated that I agree with your reasoning and the solution provided is better than what the OP was asking for.
– evolutionxbox
yesterday

add a comment |

up vote
1
down vote

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

answered yesterday

Zsolt Gulyás

337

add a comment |

up vote
1
down vote

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

answered yesterday

Zsolt Gulyás

337

add a comment |

up vote
1
down vote

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

answered yesterday

Zsolt Gulyás

337

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

const obj = a: "John", b: "Sarah", c: "Lara", d: "Joseph", e: "Roger"

const arr_of_keys = ["a", "d", "e"];


const customObjectDescructurer = (arrayOfDesiredPropKeys, object) => 

 const setOfDesiredPropKeys = new Set(arrayOfDesiredPropKeys);
 
 const filteredObject = [...setOfDesiredPropKeys].reduce(
 (filteredObject, desiredPropKey) => 
 if(object.hasOwnProperty(desiredPropKey))
 filteredObject[desiredPropKey] = object[desiredPropKey];
 else 
 console.error(`
 The given $desiredPropKey, does not exist in $object object.
 `);
 
 return filteredObject;
 , );

 return filteredObject;
 



const desiredKeys = customObjectDescructurer(arr_of_keys, obj);

console.log(desiredKeys);

answered yesterday

Zsolt Gulyás

337

answered yesterday

Zsolt Gulyás

337

answered yesterday

Zsolt Gulyás

337

answered yesterday

Zsolt Gulyás

337

add a comment |

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Pfthb