CODEWITHSUNDEEP
javascriptnested-object
Azima
Azima

Reputation: 4141

loop through nested object javascript

I am trying to loop through a following nested object and get an output as below:

const preference = {
    "ethnicity": {
        "value": "Newar",
        "rank": 1
    },
    "occupation": {
        "value": "Banker",
        "rank": 2
    }
}

I tried following:

let preferenceRank = {};
preference.map(pref => {
  preferenceRank[pref.rank] = pref; 
});

console.log(preferenceRank);

I get this error:

"TypeError: preference.map is not a function"...

Output required:

{
  1: "ethnicity",
  2: "occupation",
}

Upvotes: 4

Views: 13615

Answers (8)

prashant kumar singh
prashant kumar singh

Reputation: 1

const preference = {
    "ethnicity": {
        "value": "Gurung",
        "rank": 1
    },
    "occupation": {
        "value": "Banker",
        "rank": 2
    }
}
console.log({...Object.keys(preference)})

Upvotes: 0

CherryDT
CherryDT

Reputation: 29012

You can use Object.entries to get keys and values at once (as array of arrays [key, value]):

const preference = {
    "ethnicity": {
        "value": "Gurung",
        "rank": 1
    },
    "occupation": {
        "value": "Banker",
        "rank": 2
    }
}

const preferenceRank = {}
for (const [key, { rank }] of Object.entries(preference)) {
    preferenceRank[rank] = key
}

console.log(preferenceRank)

(By the way, in your code it doesn't make any sense to use map there, since you are not mapping the array to anything, and you ignore the return value of map. You probably wanted forEach instead or, as I used now, a for loop.)

2021 Update

There is now an easier way widely available, using Object.fromEntries, which does the opposite of Object.entries, thereby allowing us to express the whole thing as a mapping operation:

const preferenceRank = Object.fromEntries(
  Object.entries(preference).map(([key, { rank }]) => [rank, key])
)

Upvotes: 10

Sascha A.
Sascha A.

Reputation: 4616

You could map over the keys and add to a result-object the rank/key-objects.

const preference = {
    "ethnicity": {
        "value": "Gurung",
        "rank": 1
    },
    "occupation": {
        "value": "Banker",
        "rank": 2
    }
}

let res= {};
Object.keys(preference).map((el,key) => {
    res[preference[el].rank] = el;
});

console.log(res);

Upvotes: 2

Nina Scholz
Nina Scholz

Reputation: 386578

You could map the entries and build a new object.

const
    preference = { ethnicity: { value: "Gurung", rank: 1 }, occupation: { value: "Banker", rank: 2        } },
    result = Object.fromEntries(Object
        .entries(preference)
        .map(([k, { rank }]) => [rank, k])
    );

console.log(result);

Upvotes: 4

Barmar
Barmar

Reputation: 780994

Use Object.entries() to get an array of the keys and values of the object. You can then loop over that.

Use forEach if the loop is being done for side effect rather than using the values returned by the callback function.

const preference = {
    "ethnicity": {
        "value": "Gurung",
        "rank": 1
    },
    "occupation": {
        "value": "Banker",
        "rank": 2
    }
}

let preferenceRank = {};
Object.entries(preference).forEach(([pref, {rank}]) => {
  preferenceRank[rank] = pref; 
});

console.log(preferenceRank);

Upvotes: 5

ChrisG
ChrisG

Reputation: 2948

You can use the .entries() function to map over the object.

Object.entries(preference).reduce((out, [key, value]) => {
  out[value.rank] = key;
  return out;
},{});

Upvotes: 5

Derek Wang
Derek Wang

Reputation: 10194

This will work.

const preferenceRank = {};
Object.keys(preference).forEach((key) => {
  preferenceRank[preference[key]['rank']] = preference[key]['value'];
});

console.log(preferenceRank);

Upvotes: 2

Menawer
Menawer

Reputation: 883

map only works for arrays, you are dealing with an object, what you can is go through the keys of the objects by using

Object.keys(preference)

this will return to you the object keys in an array as the following ["ethnicity","occupation"]

then you can map through it if you want and do your code

Upvotes: 0

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