Javascript Array to Object How to convert an array into an object in javascript

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Originally posted on afewminutesofcode.com

To convert an array into an object we will create a function and give it 2 properties, an array and a key.

const convertArrayToObject = (array, key) => {};

We will then reduce the array, and create a unique property for each item based on the key we have passed in.

We also need to remember to set an initial Value, and also pass in the current value (...obj in the below).

const convertArrayToObject = (array, key) => {
  const initialValue = {};
  return array.reduce((obj, item) => {
    return {
      ...obj,
      [item[key]]: item,
    };
  }, initialValue);
};

So now if we log out our function (passing in the array and our key which is a unique identifier in this case the id property) we will see our array is now an object.

console.log(
  convertArrayToObject(
    [
      { id: 111, name: 'John', age: 29 },
      { id: 112, name: 'Sarah', age: 25 },
      { id: 122, name: 'Kate', age: 22 },
      { id: 123, name: 'Tom', age: 21 },
      { id: 125, name: 'Emma', age: 24 },
    ],
    'id',
  ),
);

returns

{
  111:{ id: 111, name: 'John', age: 29 },
  112:{ id: 112, name: 'Sarah', age: 25 },
  122:{ id: 122, name: 'Kate', age: 22 },
  123:{ id: 123, name: 'Tom', age: 21 },
  125:{ id: 125, name: 'Emma', age: 24 }
}

We can now easily look up data in our array by an id and use it as required.

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Comments

[Ravindra Singh]

can someone help how to get this output

a = [{id: 1, value: 1}, {id: 1, value: 2},
{id: 2, value: 20}, {id: 3, value: 30}]

output-->

{1: [{id: 1, value: 1}, {id: 1, value: 2}],
2: [{id: 2, value: 20}],3:[{id: 3, value: 30}]}

[Vinícius Hoyer]

I know that this post is from 2019, but I guess I will offer some more recent way of writing this:

const convert = (array, keyName) => Object.fromEntries(
  array.map(item => [item[keyName], item])
);

Or if you want to remove the key from the object:

const convert = (array, keyName) => Object.fromEntries(
  array.map(({ [keyName]: key, ...item }) => [key, item])
);

[rahulsachdeva02]

Nice !

how to convert something having more than one value of id like below :
for this example i understand there wont be more than one id but in my case there will be same value.

[
  { id: 111, name: 'John', age: 29 },
  { id: 111, name: 'Rob', age: 29 },
  { id: 111, name: 'Joe', age: 29 },
  { id: 112, name: 'Sarah', age: 25 },
  { id: 122, name: 'Kate', age: 22 },
  { id: 123, name: 'Tom', age: 21 },
  { id: 125, name: 'Emma', age: 24 },
],

to

{
111:[{ id: 111, name: 'John', age: 29 },{ id: 111, name: 'Rob', age: 29 },{ id: 111, name: 'Joe', age: 29 }]
112:{ id: 112, name: 'Sarah', age: 25 },
122:{ id: 122, name: 'Kate', age: 22 },
123:{ id: 123, name: 'Tom', age: 21 },
125:{ id: 125, name: 'Emma', age: 24 }
}

[vijayasantham-dev]

Thanks man. this really helpful for me

[Kukuster]

Here's type-friendly version in TypeScript

export function convertArrayToObject
<
T extends { [prop in string | number]: any },
K extends keyof Pick<T, {
    [Key in keyof T]: T[Key] extends string | number ? Key : never
}[keyof T]> = keyof Pick<T, {
    [Key in keyof T]: T[Key] extends string | number ? Key : never
}[keyof T]>,
A extends T[] = T[]
>
(array: readonly T[], key: K)
{
    const initialValue = {};
    return array.reduce((obj, item) => {
        return {
            ...obj,
            [item[key]]: item,
        };
    }, initialValue) as { [propkey in A[number][K]]: A[number]; };
}

Notes:
• TypeScript 4.0+ (tested on 4.0.x and 4.1.x)

• for the 2nd arg TS will allow and suggest only those keys for which all objects hold a value of type number or string (only those types by which any object props can be indexed).
So if all those objects had a prop BD with a string value for all the people, except John's BD was a Date object, TS won't suggest nor allow you to choose 'BD' for the 2nd arg.

• that single usage of any here is intentional. As a result, it's going to be collapsed, whereas using unknown would require several additional checks.

• usage of as operator to assert the type of return value is intentional. May be unnecessary in future TypeScript versions.

• works perfectly for variable data, but only almost perfect for constant data.
Here in this example if you denote that test array as const, TS will be able to tell you that the object has 5 properties 111, 112, 122, 123, 125, but won't be able to tell you which object exactly under each key and think its a union type of all objects for each key.
So, given the passed array had as const annotation and the returned object from the example is called fooobj:

const bar = fooobj[111];
// it thinks bar is of type of one of the objects from the original array

const name = bar.name;
// TS suggests name is of type 'John' | 'Sarah' | 'Kate' | 'Tom' | 'Emma'

if (bar.id === 111){
    // only here types collapse, asserting the exact value for bar props
    const name = bar.name;
    // TS is convinced that name === 'John'
}

[Abshir]

Thank you, this was very handy.
I remixed it a little:

const convertArrayToObject = (array) => {
    return _.reduce(array, (accumulator, currentValue) => {
        accumulator = Object.assign(accumulator, currentValue);
        return accumulator;
    }, {});
};

[Abshir]

I should have mentioned that I've used the lodash reduce method here instead of the javascript reduce

[nicole-ge]

I have another example to wotk with your Code, but I didn't reach it.

My Array:

[
{"name":"Product 1","series":{"name":"Area 1","value":3}},
{"name":"Product 2","series":{"name":"Area 2","value":3}},
{"name":"Product 1","series":{"name":"Area 2","value":1}},
]

Need to convert it to an object like this:

[ {"name": "Product 1",
"series":[{"name":"Area 1","value":3},{"name":"Area 2","value":1}] },
{"name":"Product 2",
"series":[{"name":"Area 2","value":3}] }
];

It would be nice if you could give my any advice how to reach this by adopting your Code?

[fetishlace]

I would do something like this below to get the result (not nice or fast but should do the job :D):

const arr = [
{"name":"Product 1","series":{"name":"Area 1","value":3}},
{"name":"Product 2","series":{"name":"Area 2","value":3}},
{"name":"Product 1","series":{"name":"Area 2","value":1}},
];

const result = [... new Set(arr.map(x=>x.name))].map(x=>({"name":x,"series":[]}));

arr.forEach(x=>result.find(y=>y.name===x.name).series.push(x.series));

JSON.stringify(result,null,3);

[nicole-ge]

Thank you this works very good.

I have one question:
what does arr.forEach(x=>result.find(y=>y.name===x.name).series.push(x.series));

I don't understand this function. I left it out and works. What does it?

[fetishlace]

It cannot work without it, it makes no sense. I've just generated array with unique names and empty series array at first (result), this one is pushing all the series objects data into these empty series arrays in result array.

It's just going through original array (arr) to pick series object and push it in result series array - names properties has to match hence y.name===x.name

[nicole-ge]

to understand the code a bit better, how I have to adjust it to get the following result:

[
{"name":"Area 1","value":3},
{"name":"Area 2","value":3},
{"name":"Area 2","value":1}
];

[fetishlace]

From that?
let arr = [
{"name":"Product 1","series":{"name":"Area 1","value":3}},
{"name":"Product 2","series":{"name":"Area 2","value":3}},
{"name":"Product 1","series":{"name":"Area 2","value":1}},
]
U can just remap it
arr.map(x=>x.series)

[Igor Irianto]

One Vuejs style guide recommends to normalize array into object. This will come in super handy!! Thanks!

[Aaron]

Good to hear! Also check out the comments as some awesome developers have taken the time to suggest some small improvements as well!

[Marco Antonio Dominguez]

This is good, anyhow is an antipattern, it should be something like:

const convertArrayToObject = (array, key) => 
  array.reduce((acc, curr) => {
    acc[curr[key]] = curr;
    return acc;
  }, {});

// Even concise
const convertArrayToObject = (array, key) => 
  array.reduce((acc, curr) =>(acc[curr[key]] = curr, acc), {});
// Basically everything inside parentheses will be evaluated, only the last value used will be only returned.

In this way, you avoid the spread op which is a way expensive than a single assignment

[Konrad]

I was a bit surprised by your claim regarding the spread operator being anti-pattern so I asked around and it seems that there are different opinions regarding that. Would you care to throw in some info to shed more light on the topic?

[Vinícius Hoyer]

Hey, man, I like code golfing too, but just because something can be written more concisely, it doesn't mean we should. Your second solution is kinda difficult to read because it's so cluttered and it also obligates people to know what the comma operator is and how it works. I just don't see how this is better than your first solution once the bundler will already minify the code for us, we don't need to write and read minified code, just be aware everyone, if you're gonna copy this, copy the first one, the people reading your code in the future will be happier :D

[Frank Lemanschik]

there is no item var around it should be curr did you even write that code alone?

[Aaron]

Thanks for taking the time to help me and others out here! I am going to read up on the Comma operator in JavaScript developer.mozilla.org/en-US/docs/W... do you have any other resources you recomend here?

[Ansagan Islamgali]

There is no comma operator here

[Vinícius Hoyer]

yes, there is, it's right above the line of the "Basically everything..."

[ygorbunkov]

What about Object.fromEntries()?

const convertArrayToObject = (arr,propName) => Object.fromEntries(arr.map(({[propName]:prop,...rest}) => [prop, {...rest}]));

[Aaron]

Thanks so much ygorbunkov, I have done some reading on this feature a few months back but hadn't seen how I could apply this to my problem, I will have a reference point if I need to use something similar again soon!

[ygorbunkov]

You shall bear in mind, though, it still works noticeably slower than reduce(). Hope, it will get optimised as this feature becomes more 'mainstream'.

[Aaron]

Thank you, I will keep this in mind. Have a great day