Recursively return array of object and nested object keys - javascript

I am learning about recursion at the moment and have moved on from numbers, string and arrays into using it on objects... I'm trying to work out the best method for taking an object as an argument and collecting the keys of the object and all nested objects into an array
I can return the object keys of a single object not using recursion. So i was trying to create a variable as an empty array then iterate over the object using a for loop and if "i" is an object then push object keys into the array variable and return it. This wouldnt work unfortunate.
I would like the following:
{lamp: 2, candle: 2, pillow: {big: 2, small: 4}, bathroom: {toilet: 1, shower: {shampoo: 1, conditioner: 2}}}
To return:
[lamp, candle, pillow, big, small, bathroom, toilet, shower, shampoo, conditioner]
Hope this explains enough, let me know if not :)
I tried the following:
function(obj) {
let keysArray = [];
for (let i = 0, i < obj.length, i++)
if (obj[i] === typeOf object) {
keysArray.push(obj[i].keys);
}
return keysArray
}

You can write a recursive function as follows
let obj = {lamp: 2, candle: 2, pillow: {big: 2, small: 4}, bathroom: {toilet: 1, shower: {shampoo: 1, conditioner: 2}}};
function getKeys(o) {
let result = [];
for (let key in o) {
result.push(key);
if(o[key] && typeof o[key] === "object") result.push(...getKeys(o[key]));
}
return result;
}
console.log(getKeys(obj));

You need to loop through the object using for...in. The for loop is for arrays
obj[i] === typeOf object is not correct. It should be typeof obj[key] === "object".
If the nested property is an object, you need to recursively call the function and push keys to keysArray
function getKeys(obj) {
let keysArray = [];
for (let key in obj) {
keysArray.push(key);
if (typeof obj[key] === "object")
keysArray.push(...getKeys(obj[key]))
}
return keysArray
}
const input={lamp:2,candle:2,pillow:{big:2,small:4},bathroom:{toilet:1,shower:{shampoo:1,conditioner:2}}}
console.log(getKeys(input))
FYI: typeof null is "object". So, the above code will throw an error if any of the properties are null. So, Object(obj[k]) === obj[k] can be used. This is true for all objects EXCEPT for null
Also, if flatMap is supported, you could do something like this
const input={lamp:2,candle:2,pillow:{big:2,small:4},bathroom:{toilet:1,shower:{shampoo:1,conditioner:2}}};
const getKeys = obj =>
Object.keys(obj).flatMap(key => Object(obj[key]) === obj[key]
? [key, ...getKeys(obj[key])]
: key)
console.log(getKeys(input))

How about:
const keys = obj => Object.keys(obj).reduce((acc, key) => {
acc.push(key);
return (obj[key] !== null && typeof obj[key] === 'object') // Avoid evaluating null as an object
? acc.concat(keys(obj[key]))
: acc;
}, []);
Usage:
keys({foo: 1, bar: {foobar: 2}}); // Outputs ['foo', 'bar', 'foobar']

A very good use-case for generators. Here's a demonstration -
const data =
{lamp: 2, candle: 2, pillow: {big: 2, small: 4}, bathroom: {toilet: 1, shower: {shampoo: 1, conditioner: 2}}}
const keys = function* (o = {}) {
if (Object(o) === o)
for (const [k, v] of Object.entries(o)) {
yield k
yield* keys(v)
}
}
console.log(Array.from(keys(data)))
// [lamp, candle, pillow, big, small, bathroom, toilet, shower, shampoo, conditioner]
Another solution is to use Array.prototype.flatMap -
const data =
{lamp: 2, candle: 2, pillow: {big: 2, small: 4}, bathroom: {toilet: 1, shower: {shampoo: 1, conditioner: 2}}}
const keys = (o = {}) =>
Object(o) === o
? Object.entries(o).flatMap(([ k, v ]) =>
[ k, ...keys(v) ]
)
: []
console.log(keys(data))
// [lamp, candle, pillow, big, small, bathroom, toilet, shower, shampoo, conditioner]

Related

Comparing items in ojects in two arrays js [duplicate]

I want to compare 2 arrays of objects in JavaScript code. The objects have 8 total properties, but each object will not have a value for each, and the arrays are never going to be any larger than 8 items each, so maybe the brute force method of traversing each and then looking at the values of the 8 properties is the easiest way to do what I want to do, but before implementing, I wanted to see if anyone had a more elegant solution. Any thoughts?
As serialization doesn't work generally (only when the order of properties matches: JSON.stringify({a:1,b:2}) !== JSON.stringify({b:2,a:1})) you have to check the count of properties and compare each property as well:
const objectsEqual = (o1, o2) =>
Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => o1[p] === o2[p]);
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
If you need a deep comparison, you can call the function recursively:
const obj1 = { name: 'John', age: 33, info: { married: true, hobbies: ['sport', 'art'] } };
const obj2 = { age: 33, name: 'John', info: { hobbies: ['sport', 'art'], married: true } };
const obj3 = { name: 'John', age: 33 };
const objectsEqual = (o1, o2) =>
typeof o1 === 'object' && Object.keys(o1).length > 0
? Object.keys(o1).length === Object.keys(o2).length
&& Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: o1 === o2;
console.log(objectsEqual(obj1, obj2)); // true
console.log(objectsEqual(obj1, obj3)); // false
Then it's easy to use this function to compare objects in arrays:
const arr1 = [obj1, obj1];
const arr2 = [obj1, obj2];
const arr3 = [obj1, obj3];
const arraysEqual = (a1, a2) =>
a1.length === a2.length && a1.every((o, idx) => objectsEqual(o, a2[idx]));
console.log(arraysEqual(arr1, arr2)); // true
console.log(arraysEqual(arr1, arr3)); // false
EDIT: You cannot overload operators in current, common browser-based implementations of JavaScript interpreters.
To answer the original question, one way you could do this, and mind you, this is a bit of a hack, simply serialize the two arrays to JSON and then compare the two JSON strings. That would simply tell you if the arrays are different, obviously you could do this to each of the objects within the arrays as well to see which ones were different.
Another option is to use a library which has some nice facilities for comparing objects - I use and recommend MochiKit.
EDIT: The answer kamens gave deserves consideration as well, since a single function to compare two given objects would be much smaller than any library to do what I suggest (although my suggestion would certainly work well enough).
Here is a naïve implemenation that may do just enough for you - be aware that there are potential problems with this implementation:
function objectsAreSame(x, y) {
var objectsAreSame = true;
for(var propertyName in x) {
if(x[propertyName] !== y[propertyName]) {
objectsAreSame = false;
break;
}
}
return objectsAreSame;
}
The assumption is that both objects have the same exact list of properties.
Oh, and it is probably obvious that, for better or worse, I belong to the only-one-return-point camp. :)
Honestly, with 8 objects max and 8 properties max per object, your best bet is to just traverse each object and make the comparisons directly. It'll be fast and it'll be easy.
If you're going to be using these types of comparisons often, then I agree with Jason about JSON serialization...but otherwise there's no need to slow down your app with a new library or JSON serialization code.
I know this is an old question and the answers provided work fine ... but this is a bit shorter and doesn't require any additional libraries ( i.e. JSON ):
function arraysAreEqual(ary1,ary2){
return (ary1.join('') == ary2.join(''));
}
I have worked a bit on a simple algorithm to compare contents of two objects and return an intelligible list of difference. Thought I would share. It borrows some ideas for jQuery, namely the map function implementation and the object and array type checking.
It returns a list of "diff objects", which are arrays with the diff info. It's very simple.
Here it is:
// compare contents of two objects and return a list of differences
// returns an array where each element is also an array in the form:
// [accessor, diffType, leftValue, rightValue ]
//
// diffType is one of the following:
// value: when primitive values at that index are different
// undefined: when values in that index exist in one object but don't in
// another; one of the values is always undefined
// null: when a value in that index is null or undefined; values are
// expressed as boolean values, indicated wheter they were nulls
// type: when values in that index are of different types; values are
// expressed as types
// length: when arrays in that index are of different length; values are
// the lengths of the arrays
//
function DiffObjects(o1, o2) {
// choose a map() impl.
// you may use $.map from jQuery if you wish
var map = Array.prototype.map?
function(a) { return Array.prototype.map.apply(a, Array.prototype.slice.call(arguments, 1)); } :
function(a, f) {
var ret = new Array(a.length), value;
for ( var i = 0, length = a.length; i < length; i++ )
ret[i] = f(a[i], i);
return ret.concat();
};
// shorthand for push impl.
var push = Array.prototype.push;
// check for null/undefined values
if ((o1 == null) || (o2 == null)) {
if (o1 != o2)
return [["", "null", o1!=null, o2!=null]];
return undefined; // both null
}
// compare types
if ((o1.constructor != o2.constructor) ||
(typeof o1 != typeof o2)) {
return [["", "type", Object.prototype.toString.call(o1), Object.prototype.toString.call(o2) ]]; // different type
}
// compare arrays
if (Object.prototype.toString.call(o1) == "[object Array]") {
if (o1.length != o2.length) {
return [["", "length", o1.length, o2.length]]; // different length
}
var diff =[];
for (var i=0; i<o1.length; i++) {
// per element nested diff
var innerDiff = DiffObjects(o1[i], o2[i]);
if (innerDiff) { // o1[i] != o2[i]
// merge diff array into parent's while including parent object name ([i])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + i + "]" + o[0]; return o; }));
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if arrays equal
return undefined;
}
// compare object trees
if (Object.prototype.toString.call(o1) == "[object Object]") {
var diff =[];
// check all props in o1
for (var prop in o1) {
// the double check in o1 is because in V8 objects remember keys set to undefined
if ((typeof o2[prop] == "undefined") && (typeof o1[prop] != "undefined")) {
// prop exists in o1 but not in o2
diff.push(["[" + prop + "]", "undefined", o1[prop], undefined]); // prop exists in o1 but not in o2
}
else {
// per element nested diff
var innerDiff = DiffObjects(o1[prop], o2[prop]);
if (innerDiff) { // o1[prop] != o2[prop]
// merge diff array into parent's while including parent object name ([prop])
push.apply(diff, map(innerDiff, function(o, j) { o[0]="[" + prop + "]" + o[0]; return o; }));
}
}
}
for (var prop in o2) {
// the double check in o2 is because in V8 objects remember keys set to undefined
if ((typeof o1[prop] == "undefined") && (typeof o2[prop] != "undefined")) {
// prop exists in o2 but not in o1
diff.push(["[" + prop + "]", "undefined", undefined, o2[prop]]); // prop exists in o2 but not in o1
}
}
// if any differences were found, return them
if (diff.length)
return diff;
// return nothing if objects equal
return undefined;
}
// if same type and not null or objects or arrays
// perform primitive value comparison
if (o1 != o2)
return [["", "value", o1, o2]];
// return nothing if values are equal
return undefined;
}
I tried JSON.stringify() and worked for me.
let array1 = [1,2,{value:'alpha'}] , array2 = [{value:'alpha'},'music',3,4];
JSON.stringify(array1) // "[1,2,{"value":"alpha"}]"
JSON.stringify(array2) // "[{"value":"alpha"},"music",3,4]"
JSON.stringify(array1) === JSON.stringify(array2); // false
There is a optimized code for case when function needs to equals to empty arrays (and returning false in that case)
const objectsEqual = (o1, o2) => {
if (o2 === null && o1 !== null) return false;
return o1 !== null && typeof o1 === 'object' && Object.keys(o1).length > 0 ?
Object.keys(o1).length === Object.keys(o2).length &&
Object.keys(o1).every(p => objectsEqual(o1[p], o2[p]))
: (o1 !== null && Array.isArray(o1) && Array.isArray(o2) && !o1.length &&
!o2.length) ? true : o1 === o2;
}
Here is my attempt, using Node's assert module + npm package object-hash.
I suppose that you would like to check if two arrays contain the same objects, even if those objects are ordered differently between the two arrays.
var assert = require('assert');
var hash = require('object-hash');
var obj1 = {a: 1, b: 2, c: 333},
obj2 = {b: 2, a: 1, c: 444},
obj3 = {b: "AAA", c: 555},
obj4 = {c: 555, b: "AAA"};
var array1 = [obj1, obj2, obj3, obj4];
var array2 = [obj3, obj2, obj4, obj1]; // [obj3, obj3, obj2, obj1] should work as well
// calling assert.deepEquals(array1, array2) at this point FAILS (throws an AssertionError)
// even if array1 and array2 contain the same objects in different order,
// because array1[0].c !== array2[0].c
// sort objects in arrays by their hashes, so that if the arrays are identical,
// their objects can be compared in the same order, one by one
var array1 = sortArrayOnHash(array1);
var array2 = sortArrayOnHash(array2);
// then, this should output "PASS"
try {
assert.deepEqual(array1, array2);
console.log("PASS");
} catch (e) {
console.log("FAIL");
console.log(e);
}
// You could define as well something like Array.prototype.sortOnHash()...
function sortArrayOnHash(array) {
return array.sort(function(a, b) {
return hash(a) > hash(b);
});
}
My practice implementation with sorting, tested and working.
const obj1 = { name: 'John', age: 33};
const obj2 = { age: 33, name: 'John' };
const obj3 = { name: 'John', age: 45 };
const equalObjs = ( obj1, obj2 ) => {
let keyExist = false;
for ( const [key, value] of Object.entries(obj1) ) {
// Search each key in reference object and attach a callback function to
// compare the two object keys
if( Object.keys(obj2).some( ( e ) => e == key ) ) {
keyExist = true;
}
}
return keyExist;
}
console.info( equalObjs( obj1, obj2 ) );
Compare your arrays
// Sort Arrays
var arr1 = arr1.sort(( a, b ) => {
var fa = Object.keys(a);
var fb = Object.keys(b);
if (fa < fb) {
return -1;
}
if (fa > fb) {
return 1;
}
return 0;
});
var arr2 = arr2.sort(( a, b ) => {
var fa = Object.keys(a);
var fb = Object.keys(b);
if (fa < fb) {
return -1;
}
if (fa > fb) {
return 1;
}
return 0;
});
const equalArrays = ( arr1, arr2 ) => {
// If the arrays are different length we an eliminate immediately
if( arr1.length !== arr2.length ) {
return false;
} else if ( arr1.every(( obj, index ) => equalObjs( obj, arr2[index] ) ) ) {
return true;
} else {
return false;
}
}
console.info( equalArrays( arr1, arr2 ) );
I am sharing my compare function implementation as it might be helpful for others:
/*
null AND null // true
undefined AND undefined // true
null AND undefined // false
[] AND [] // true
[1, 2, 'test'] AND ['test', 2, 1] // true
[1, 2, 'test'] AND ['test', 2, 3] // false
[undefined, 2, 'test'] AND ['test', 2, 1] // false
[undefined, 2, 'test'] AND ['test', 2, undefined] // true
[[1, 2], 'test'] AND ['test', [2, 1]] // true
[1, 'test'] AND ['test', [2, 1]] // false
[[2, 1], 'test'] AND ['test', [2, 1]] // true
[[2, 1], 'test'] AND ['test', [2, 3]] // false
[[[3, 4], 2], 'test'] AND ['test', [2, [3, 4]]] // true
[[[3, 4], 2], 'test'] AND ['test', [2, [5, 4]]] // false
[{x: 1, y: 2}, 'test'] AND ['test', {x: 1, y: 2}] // true
1 AND 1 // true
{test: 1} AND ['test', 2, 1] // false
{test: 1} AND {test: 1} // true
{test: 1} AND {test: 2} // false
{test: [1, 2]} AND {test: [1, 2]} // true
{test: [1, 2]} AND {test: [1]} // false
{test: [1, 2], x: 1} AND {test: [1, 2], x: 2} // false
{test: [1, { z: 5 }], x: 1} AND {x: 1, test: [1, { z: 5}]} // true
{test: [1, { z: 5 }], x: 1} AND {x: 1, test: [1, { z: 6}]} // false
*/
function is_equal(x, y) {
const
arr1 = x,
arr2 = y,
is_objects_equal = function (obj_x, obj_y) {
if (!(
typeof obj_x === 'object' &&
Object.keys(obj_x).length > 0
))
return obj_x === obj_y;
return Object.keys(obj_x).length === Object.keys(obj_y).length &&
Object.keys(obj_x).every(p => is_objects_equal(obj_x[p], obj_y[p]));
}
;
if (!( Array.isArray(arr1) && Array.isArray(arr2) ))
return (
arr1 && typeof arr1 === 'object' &&
arr2 && typeof arr2 === 'object'
)
? is_objects_equal(arr1, arr2)
: arr1 === arr2;
if (arr1.length !== arr2.length)
return false;
for (const idx_1 of arr1.keys())
for (const idx_2 of arr2.keys())
if (
(
Array.isArray(arr1[idx_1]) &&
this.is_equal(arr1[idx_1], arr2[idx_2])
) ||
is_objects_equal(arr1[idx_1], arr2[idx_2])
)
{
arr2.splice(idx_2, 1);
break;
}
return !arr2.length;
}
Please try this one:
function used_to_compare_two_arrays(a, b)
{
// This block will make the array of indexed that array b contains a elements
var c = a.filter(function(value, index, obj) {
return b.indexOf(value) > -1;
});
// This is used for making comparison that both have same length if no condition go wrong
if (c.length !== a.length) {
return 0;
} else{
return 1;
}
}
The objectsAreSame function mentioned in #JasonBunting's answer works fine for me. However, there's a little problem: If x[propertyName] and y[propertyName] are objects (typeof x[propertyName] == 'object'), you'll need to call the function recursively in order to compare them.
not sure about the performance ... will have to test on big objects .. however, this works great for me.. the advantage it has compared to the other solutions is, the objects/array do not have to be in the same order ....
it practically takes the first object in the first array, and scans the second array for every objects .. if it's a match, it will proceed to another
there is absolutely a way for optimization but it's working :)
thx to #ttulka I got inspired by his work ... just worked on it a little bit
const objectsEqual = (o1, o2) => {
let match = false
if(typeof o1 === 'object' && Object.keys(o1).length > 0) {
match = (Object.keys(o1).length === Object.keys(o2).length && Object.keys(o1).every(p => objectsEqual(o1[p], o2[p])))
}else {
match = (o1 === o2)
}
return match
}
const arraysEqual = (a1, a2) => {
let finalMatch = []
let itemFound = []
if(a1.length === a2.length) {
finalMatch = []
a1.forEach( i1 => {
itemFound = []
a2.forEach( i2 => {
itemFound.push(objectsEqual(i1, i2))
})
finalMatch.push(itemFound.some( i => i === true))
})
}
return finalMatch.every(i => i === true)
}
const ar1 = [
{ id: 1, name: "Johnny", data: { body: "Some text"}},
{ id: 2, name: "Jimmy"}
]
const ar2 = [
{name: "Jimmy", id: 2},
{name: "Johnny", data: { body: "Some text"}, id: 1}
]
console.log("Match:",arraysEqual(ar1, ar2))
jsfiddle: https://jsfiddle.net/x1pubs6q/
or just use lodash :))))
const _ = require('lodash')
const isArrayEqual = (x, y) => {
return _.isEmpty(_.xorWith(x, y, _.isEqual));
};
using _.some from lodash: https://lodash.com/docs/4.17.11#some
const array1AndArray2NotEqual =
_.some(array1, (a1, idx) => a1.key1 !== array2[idx].key1
|| a1.key2 !== array2[idx].key2
|| a1.key3 !== array2[idx].key3);
There`s my solution. It will compare arrays which also have objects and arrays. Elements can be stay in any positions.
Example:
const array1 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const array2 = [{a: 1}, {b: 2}, { c: 0, d: { e: 1, f: 2, } }, [1,2,3,54]];
const arraysCompare = (a1, a2) => {
if (a1.length !== a2.length) return false;
const objectIteration = (object) => {
const result = [];
const objectReduce = (obj) => {
for (let i in obj) {
if (typeof obj[i] !== 'object') {
result.push(`${i}${obj[i]}`);
} else {
objectReduce(obj[i]);
}
}
};
objectReduce(object);
return result;
};
const reduceArray1 = a1.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const reduceArray2 = a2.map(item => {
if (typeof item !== 'object') return item;
return objectIteration(item).join('');
});
const compare = reduceArray1.map(item => reduceArray2.includes(item));
return compare.reduce((acc, item) => acc + Number(item)) === a1.length;
};
console.log(arraysCompare(array1, array2));
This is work for me to compare two array of objects without taking into consideration the order of the items
const collection1 = [
{ id: "1", name: "item 1", subtitle: "This is a subtitle", parentId: "1" },
{ id: "2", name: "item 2", parentId: "1" },
{ id: "3", name: "item 3", parentId: "1" },
]
const collection2 = [
{ id: "3", name: "item 3", parentId: "1" },
{ id: "2", name: "item 2", parentId: "1" },
{ id: "1", name: "item 1", subtitle: "This is a subtitle", parentId: "1" },
]
const contains = (arr, obj) => {
let i = arr.length;
while (i--) {
if (JSON.stringify(arr[i]) === JSON.stringify(obj)) {
return true;
}
}
return false;
}
const isEqual = (obj1, obj2) => {
let n = 0
if (obj1.length !== obj2.length) {
return false;
}
for (let i = 0; i < obj1.length; i++) {
if (contains(obj2, obj1[i])) {
n++
}
}
return n === obj1.length
}
console.log(isEqual(collection1,collection2))
if you take into consideration the order of the items use built in function in lodash isEqual
comparing with json is pretty bad. try this package to compare nested arrays and get the difference.
https://www.npmjs.com/package/deep-object-diff
If you stringify them...
type AB = {
nome: string;
}
const a: AB[] = [{ nome: 'Célio' }];
const b: AB[] = [{ nome: 'Célio' }];
console.log(a === b); // false
console.log(JSON.stringify(a) === JSON.stringify(b)); // true

How do I create a third object from two objects using the key values

Object 1:
{positive: ['happy', 'excited', 'joyful'], negative: ['depressed', 'sad', 'unhappy']}
Object 2:
{happy: 6, excited: 1, unhappy: 3}
What I want -
Result:
{positive: 7, negative: 3}
How do I achieve this?
I have used the following to get the category object 2 falls into but I'm not sure how to dynamically create the third object as the size of these objects grow. I cannot use for...of or for...in due to eslint airbnb rules
function getKeyByValue(value) {
return Object.keys(object).find((key) => object[key] === value);
}
This can be done using Array#reduce over Object.entries to create the new object. For each entry, destructured as [key, feelings], we calculate the sum using a nested reduce: for each feeling in feelings, we use the value from object2, if exists, or 0 otherwise using short circuit evaluation (object2[feeling] || 0), like so:
let result = Object.entries(object1).reduce((result, [key, feelings]) => {
result[key] = feelings.reduce((sum, feeling) => sum + (object2[feeling] || 0), 0);
return result;
}, {});
Demo:
let object1 = { positive: ["happy", "excited", "joyful"], negative: ["depressed", "sad", "unhappy"] };
let object2 = { happy: 6, excited: 1, unhappy: 3 };
let result = Object.entries(object1).reduce((result, [key, feelings]) => {
result[key] = feelings.reduce((sum, feeling) => sum + (object2[feeling] || 0), 0);
return result;
}, {});
console.log(result);
For example, you can use reduce:
obj1 = {positive: ['happy', 'excited', 'joyful'], negative: ['depressed', 'sad', 'unhappy']}
obj2 = {happy: 6, excited: 1, unhappy: 3}
const result ={}
for (let key of Object.keys(obj1)) {
result[key] = obj1[key].reduce(
(acc, value) => { return acc + (obj2[value] || 0) }, 0)
}
console.log(result)
Notes:
for ... of Object.keys() is more preferable in most situations than for ... in because for ... in iterate not only over the object's own keys, but also over its prototype chains.
Instead Object.keys() you can also use Reflect.ownKeys()

apply function to all instances of an attribute in an object tree

Let's say I have an object of unknown structure and depth, where I want to modify an attribute. For example, let's imagine that the attribute I want to change is id and the structure might be something like this:
"{
"a": "foo",
"b": "bar",
"c": {
"id": {
"a": "foo"
},
"b": "whatever"
},
"id": {
"a": "foo"
}
}"
Is there any stablished pattern or way to take any instance of a wanted key in the whole tree and apply a generic function to it? Ideally, checking array elements too if they existed.
In terms of functional programming, we need our structure to be a Functor, that is, supply it with a function fmap so that fmap(fn, someTree) will yield the same structure with fn applied to all values inside it. There are quite a few ways to implement that, for example:
let isObject = x => x && typeof x === 'object';
let map = fn => obj =>
Array.isArray(obj)
? obj.map((v, k) => fn([null, v]).pop())
: Object.fromEntries(Object.entries(obj).map(fn));
let _fmap = fn => ([key, val]) =>
isObject(val)
? [
key,
map(_fmap(fn))(val)
]
: fn([key, val]);
let fmap = fn => map(_fmap(fn));
//
const renameKey = (o, n) => ([key, val]) => (key === o ? [n, val] : [key, val]);
test = {
a: 1,
b: 2,
c: [
{a: 1}, {a: 2}, {b: 3}
]
};
newTest = fmap(renameKey('a', 'blah'))(test);
console.log(newTest);
Here, we treat a tree "node" as a [key, value] pair. The callback receives a pair and is supposed to return it back, with or without a modification.

(_.merge in ES6/ES7)Object.assign without overriding undefined values

There is _.merge functionality in lodash. I want to achieve the same thing in ES6 or ES7.
Having this snippet:
Object.assign({}, {key: 2}, {key: undefined})
I want to receive {key: 2}. Currently I receive {key: undefined}
This is NOT a deep merge.
Is it possible? If yes then how to achieve that?
You can't achieve that with a straight usage of Object.assign, because each next object will rewrite the same keys for prev merge. The only way, to filter your incoming objects with some hand-crafted function.
function filterObject(obj) {
const ret = {};
Object.keys(obj)
.filter((key) => obj[key] !== undefined)
.forEach((key) => ret[key] = obj[key]);
return ret;
}
You can simply filter out the keys with undefined values before passing them to Object.assign():
const assign = (target, ...sources) =>
Object.assign(target, ...sources.map(x =>
Object.entries(x)
.filter(([key, value]) => value !== undefined)
.reduce((obj, [key, value]) => (obj[key] = value, obj), {})
))
console.log(assign({}, {key: 2}, {key: undefined}))
Write a little utility to remove undefined values:
function removeUndefined(obj) {
for (let k in obj) if (obj[k] === undefined) delete obj[k];
return obj;
}
Then
Object.assign({}, {key: 2}, removeUndefined({key: undefined}))
This seems preferable to writing your own assign with wired-in behavior to remove undefined values.
use lodash to omit nil values and then combine the two objects into one via spread
{ ...(omitBy({key: 2}, isNil)), ...(omitBy({key: undefined}, isNil))}
See more info on lodash here https://lodash.com/docs/4.17.15
With ES2019/ES10's new object method, Object.fromEntries(), Michał's answer can be updated:
const assign = (target, ...sources) =>
Object.assign(target, ...sources.map(x =>
Object.fromEntries(
Object.entries(x)
.filter(([key, value]) => value !== undefined)
)
))
console.log(assign({}, {key: 2}, {key: undefined}))
If you just need the values and don't need an object, you could also use object destructuring:
const input = { a: 0, b: "", c: false, d: null, e: undefined };
const { a = 1, b = 2, c = 3, d = 4, e = 5, f = 6 } = input;
console.log(a, b, c, d, e, f);
// => 0, "", false, null, 5, 6
This will only override absent or undefined values.
I often use this for function argument default values like this:
function f(options = {}) {
const { foo = 42, bar } = options;
console.log(foo, bar);
}
f();
// => 42, undefined
f({})
// => 42, undefined
f({ foo: 123 })
// => 123, undefined
f({ bar: 567 })
// => 42, 567
f({ foo: 123, bar: 567 })
// => 123, 567

Return all properties in JS object to "empty" state

I wonder if there's a utility function in any framework like jQuery to truncate a JS object (there is more than one possible result, not sure what is the most useful yet):
{ // original object
foo: "bar",
arr: [1, 2, 3],
sub: {a: 4, b: 5, subsub: {c: 6}} }
{ // truncated, variant 1
foo: "",
arr: [],
sub: {} }
{ // truncated, variant 2
foo: "",
arr: [],
sub: {subsub: {}} }
{ // truncated, variant 3
foo: "",
arr: [],
sub: {a: 0, b: 0, subsub: {c: 0}} }
If no, is there a solution more clever than to recursively iterate over all properties, check types and remove / reset if necessary?
Call the constructor for each property to create an empty object of the same type.
for (var k in obj) {
obj[k] = obj[k] && obj[k].constructor();
}
This will also return numbers to 0, booleans to false, dates to now, and regexps to empty (but NaNs remain NaN).
To do this recursively, retaining object-valued properties but emptying them out too (variant 3):
function truncate(obj) {
for (var k in obj) {
var ctor = obj[k] && obj[k].constructor;
if (ctor === Object) truncate(obj[k]);
else if (ctor) obj[k] = ctor();
}
}
To eliminate numbers (variant 2), add a check:
function truncate(obj) {
for (var k in obj) {
var ctor = obj[k] && obj[k].constructor;
if (ctor === Object) truncate(obj[k]);
else if (ctor === Number) delete obj[k];
else if (ctor) obj[k] = ctor();
}
}
You may be looking for OMIT, part of underscore.js.
omit_.omit(object, *keys) Return a copy of the object, filtered to
omit the blacklisted keys (or array of keys). Alternatively accepts a
predicate indicating which keys to omit.
_.omit({name: 'moe', age: 50, userid: 'moe1'}, 'userid');
=> {name: 'moe', age: 50}
_.omit({name: 'moe', age: 50, userid: 'moe1'}, function(value, key, object) { return _.isNumber(value); });
=> {name: 'moe', userid: 'moe1'}
Object.prototype.clear=function(){
for (prop in this){
if (typeof this[prop]==='string') this[prop]='';
else if (this[prop] instanceof Array) this[prop]=[];
else{
this[prop]={}//variant 1
}
}
return this;
}
.
var obj={
foo:"bar",
arr:[1,2,3],
sub:{b:4,c:5,subsub:{e:6}}
}
obj.clear()//returns {foo:"",arr:[],sub:{}}

Categories