Flatten javascript objects into a single-depth object

Object Flatten

var flattenObject = function(ob) {
	var toReturn = {};
	
	for (var i in ob) {
		if (!ob.hasOwnProperty(i)) continue;
		
		if ((typeof ob[i]) == 'object') {
			var flatObject = flattenObject(ob[i]);
			for (var x in flatObject) {
				if (!flatObject.hasOwnProperty(x)) continue;
				
				toReturn[i + '.' + x] = flatObject[x];
			}
		} else {
			toReturn[i] = ob[i];
		}
	}
	return toReturn;
};

very nicely done! Thank you for this!

anyone can revert this, i want to create object from single-depth object

@edwardEvans094 I found code to reverse this at https://stackoverflow.com/questions/42694980/how-to-unflatten-a-javascript-object-in-a-daisy-chain-dot-notation-into-an-objec

Really useful, thank you!

anybody has an opposite function? to unflatten it?

A Typescript 3.7+ version:

export function flatten<T extends Record<string, any>>(
  object: T,
  path: string | null = null,
  separator = '.'
): T {
  return Object.keys(object).reduce((acc: T, key: string): T => {
    const newPath = [path, key].filter(Boolean).join(separator);
    return typeof object?.[key] === 'object'
      ? { ...acc, ...flatten(object[key], newPath, separator) }
      : { ...acc, [newPath]: object[key] };
  }, {} as T);
}

@guillim
How do I remove the numbers from the keys in an array of objects? I have this fiddle. This data:

var data = {
  "ticker": "AAPL",
  "name": "Apple",
  "data": [
    1,2,3,4
  ],
  "obj": [
    {"revenue": 1, "income": 11},
    {"revenue": 2, "income": 22},
  ],
}

Results in:
["ticker", "name", "data.0", "data.1", "data.2", "data.3", "obj.0.revenue", "obj.0.income", "obj.1.revenue", "obj.1.income"]

I'd like:
["ticker", "name", "data", "data", "data", "data", "obj.revenue", "obj.income", "obj.revenue", "obj.income"]

What an awesome thread. Thanks to everyone here!

I made an improvement from @x47188 version to handle null values and empty [] arrays. Also added the Date and Regex object for the hell of it.

export function flatten<T extends Record<string, any>>(
  object: T,
  path: string | null = null,
  separator = '.'
): T {
  return Object.keys(object).reduce((acc: T, key: string): T => {
    const value = object[key];

    const newPath = [path, key].filter(Boolean).join(separator);

    const isObject = [
      typeof value === 'object',
      value !== null,
      !(value instanceof Date),
      !(value instanceof RegExp),
      !(Array.isArray(value) && value.length === 0),
    ].every(Boolean);

    return isObject
      ? { ...acc, ...flatten(value, newPath, separator) }
      : { ...acc, [newPath]: value };
  }, {} as T);
}

Example: https://stackblitz.com/edit/typescript-shdjoq

I'm wondering how to do it similarly to what @gitty-git-git asks.
But i want to keep arrays as they are unless they have objects, in that case flatten the objects.
For example this:

flatten({a:{b:2},c:[{e:{f:4}}]})

{
  "a.b": 2,
  "c": [
    {
      "e.f": 4
    }
  ]
}

Anyone has an idea? @gitty-git-git did you find a way to do yours?

slighlty modified from @codeBelt to output array indices as [0] instead of as property .0

export function flatten<T extends Record<string, any>>(object: T, path: string | null = null, separator = '.'): T {
  return Object.keys(object).reduce((acc: T, key: string): T => {
    const value = object[key];
    const newPath = Array.isArray(object)
      ? `${path ? path : ''}[${key}]`
      : [path, key].filter(Boolean).join(separator);
    const isObject = [
      typeof value === 'object',
      value !== null,
      !(value instanceof Date),
      !(value instanceof RegExp),
      !(Array.isArray(value) && value.length === 0),
    ].every(Boolean);

    return isObject
      ? { ...acc, ...flatten(value, newPath, separator) }
      : { ...acc, [newPath]: value };
  }, {} as T);
}

Forked example: https://stackblitz.com/edit/typescript-pwsl83

I have put together a simple module, Flatify-obj based on this original gist with some additional tweaks and tests.

Usage

   const flattenObject = require('flatify-obj');

   flattenObject({foo: {bar: {unicorn: '🦄'}}})
   //=> { 'foo.bar.unicorn': '🦄' }

   flattenObject({foo: {unicorn: '🦄'}, bar: 'unicorn'}, {onlyLeaves: true});
   //=> {unicorn: '🦄', bar: 'unicorn'}

For additional features PRs are welcome 🦄

Another one written in TypeScript: tree-to-flat-map.

you are the best!

@danzelbel like a charm!

Jewel!

This is my implementation, it just flattens every object in the parent object into one, without keying them with object.key.
Working example
Code:

function flatten(obj = {}) {
  const doneObject = {}
  for (const [k, v] of Object.entries(obj)) {
    if (typeof v == "object" && !(v instanceof  Date) && !Array.isArray(v) && !(v instanceof regExp)) {
       Object.assign(doneObject, flatten(v))
    } else {
      doneObject[k] = v
    }
  }
  return doneObject
}

Thank you for the original implementation!

A version that leaves undefined behind, array as is and don't try to flatten primitive objects (Number, Boolean, BigInt and String essentially) :

const flatten = <T extends Record<string, any>>(object: T, path?: string): Record<string, any> =>
  Object.entries(object).reduce((acc, [key, val]) => {
    if (val === undefined) return acc;
    if (path) key = `${path}.${key}`;
    if (typeof val === 'object' && val !== null && !(val instanceof Date) && !(val instanceof RegExp) && !Array.isArray(val)) {
      if (val !== val.valueOf()) {
        return { ...acc, [key]: val.valueOf() };
      }
      return { ...acc, ...flatten(val, key) };
    }
    return { ...acc, [key]: val };
  }, {});