JS getting union of all object keys from an array of objects

Question

I have below array of objects:

var all = [
{f1: "v1", f2: "v2"},
{f1: "v1", f2: "v2", f3: "v3"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4", fn: "vn"}
];

Desired Output:

[f1, f2, f3, f4, f5, ...., fn];

Currently Using:

all.reduce((sum, item) => ([...new Set([...sum, ...Object.keys(item)])]), []);

Working Example: codepen

Any suggestions using es6 or any new js feature, for better performance.

It's not 100% clear what you're trying to achieve your example has 5 results from an array of 4 objects. Are you trying to get the final key of the object. Is order guaranteed? — Damian Green
– Damian Green, Commented Oct 13, 2021 at 20:32
Yet another option: new Set(all.map(o => Object.keys(o)).flat()) — Felix Kling
– Felix Kling, Commented Oct 14, 2021 at 8:07
@DamianGreen, desired output is "array of keys of heighest key:value pair object from the given array". — Priyesh Singhai
– Priyesh Singhai, Commented Oct 15, 2021 at 7:43
@FelixKling will it be havening better performance, as again we are having 4 operations. — Priyesh Singhai
– Priyesh Singhai, Commented Oct 15, 2021 at 7:45
"desired output is "array of keys of heighest key:value pair object from the given array" That's not the same as getting the union of all keys. What is it now? "will it be havening better performance, as again we are having 4 operations" It doesn't create intermediate objects like your solution. But in the end you'll have to actually measure the performance of different solutions. — Felix Kling
– Felix Kling, Commented Oct 15, 2021 at 7:59

CertainPerformance · Accepted Answer · 2021-10-13 20:27:58Z

6

You can flatten them into a single object, then take the keys.

var all = [
{f1: "v1", f2: "v2"},
{f1: "v1", f2: "v2", f3: "v3"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4", fn: "vn"}
];

const output = Object.keys(Object.assign({}, ...all));
console.log(output);

answered Oct 13, 2021 at 20:27

CertainPerformance

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Comments

rivercity · Accepted Answer · 2021-10-13 20:41:21Z

0

var all = [
{f1: "v1", f2: "v2"},
{f1: "v1", f2: "v2", f3: "v3"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4", fn: "vn"}
];

console.log(Object.keys(all[all.length-1]))

answered Oct 13, 2021 at 20:41

rivercity

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3 Comments

jfriend00 Over a year ago

This makes an assumption that the last element in the array contains all the keys and all the other elements in the array can be ignored. It's not clear that's a safe assumption.

Priyesh Singhai Over a year ago

No, any object of the array can have heighest number of key:value pairs, not the last one always.

rivercity Over a year ago

I know, but for the given input it shows the right results, why care about conercases if not asked?

Jason Goemaat · Accepted Answer · 2021-10-15 19:55:31Z

0

Using Array.flatMap and a Set object to get a Set of unique keys and then use the spread operator to convert the Set back into an array.

var all = [
{f1: "v1", f2: "v2"},
{f1: "v1", f2: "v2", f3: "v3"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4"},
{f1: "v1", f2: "v2", f3: "v3", f4: "v4", fn: "vn"}
];

const res = [...new Set(all.flatMap(Object.keys))];
console.log(res);

edited Oct 15, 2021 at 19:55

Jason Goemaat

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answered Oct 15, 2021 at 19:37

Siva Kondapi Venkata

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