I have two object literals like so:
var firstObject =
{
x: 0,
y: 1,
z: 2,
a: 10,
b: 20,
e: 30
}
var secondObject =
{
x: 0,
y: 1,
z: 2,
a: 10,
c: 20,
d: 30
}
I want to get the intersection of the keys these two object literals have like so:
var intersectionKeys = ['x', 'y', 'z', 'a']
I can obviously do a loop and see if a key with the same name exists in the other object, but I am wondering if this would be a good case for some functional programming and map / filter / reduce usage? I myself have not done that much functional programming, but I have a feeling, that there could exist a clean and clever solution for this problem.
A solution without indexOf.
var firstObject = { x: 0, y: 1, z: 2, a: 10, b: 20, e: 30 },
secondObject = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30 };
function intersection(o1, o2) {
return Object.keys(o1).concat(Object.keys(o2)).sort().reduce(function (r, a, i, aa) {
if (i && aa[i - 1] === a) {
r.push(a);
}
return r;
}, []);
}
document.write('<pre>' + JSON.stringify(intersection(firstObject, secondObject), 0, 4) + '</pre>');
Second attempt with O(n).
var firstObject = { x: 0, y: 1, z: 2, a: 10, b: 20, e: 30 },
secondObject = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30 };
function intersection(o1, o2) {
return Object.keys(o1).filter({}.hasOwnProperty.bind(o2));
}
document.write('<pre>' + JSON.stringify(intersection(firstObject, secondObject), 0, 4) + '</pre>');
The given answers are nice and astonishing but there could be a problem in void's answer and that is:
"What if one of property values intentionally set to undefined."
Nina's answer is good (really fantastic) but as we are in era of fun JavaScript I think mine wont be too bad:
var a = { x: undefined, y: 1, z: 2, a: 10, b: 20, e: 30 }
var b = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30 }
function intersect(o1, o2){
return Object.keys(o1).filter(k => Object.hasOwn(o2, k))
}
document.write('<pre>' + JSON.stringify(intersect(a, b)) + '</pre>');
Update
onalbi mentioned some performance issue in comments which is rational and therefore the code bellow seems to be a better way to handle the problem:
var a = { x: undefined, y: 1, z: 2, a: 10, b: 20, e: 30};
var b = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30};
function intersect(o1, o2) {
const [k1, k2] = [Object.keys(o1), Object.keys(o2)];
const [first, next] = k1.length > k2.length ? [k2, o1] : [k1, o2];
return first.filter(k => k in next);
}
document.write('<pre>' + JSON.stringify(intersect(a, b)) + '</pre>');
The procedure i will suggest is:
Get the array of keys using Object.keys() for one of the objects.
Find the intersection the array using .filter and checking if the second object contains a key matching the first array.
var firstObject = {
x: 0,
y: 1,
z: 2,
a: 10,
b: 20,
e: 30
}
var secondObject = {
x: 0,
y: 1,
z: 2,
a: 10,
c: 20,
d: 30
}
function getIntKeys(obj1, obj2){
var k1 = Object.keys(obj1);
return k1.filter(function(x){
return obj2[x] !== undefined;
});
}
alert(getIntKeys(firstObject, secondObject));
Recursive function
This is other solution, maybe help you. I used a recursive function to intercept two objects. The advantage of this solution is that you not need worry about attributes that are objects at same time.
In this case the function intercept attributes that exist in both objects and asign the value of 'objSource' like final value of attribute intercepeted.
{
function interceptObjects(objSource, objInterface) {
let newObj = {};
for (const key in objSource) {
if (objInterface.hasOwnProperty(key)) {
// in javascript an array is a object too.
if (objSource[key] instanceof Object && !Array.isArray(objSource[key]) && objInterface[key] instanceof Object && !Array.isArray(objInterface[key])) {
newObj[key] = {};
newObj[key] = interceptObjects(objSource[key], objInterface[key])
} else {
newObj[key] = objSource[key];
}
}
}
return newObj;
}
// FOR TESTING
let objSource = {
attr1: '',
attr2: 2,
attr3: [],
attr4: {
attr41: 'lol',
attr42: 12,
attr43: 15,
attr45: [1, 4],
},
attr5: [2, 3, 4],
};
let objInterface = {
attr1: null,
attr4: {
attr41: null,
attr42: 12,
attr45: [1],
},
attr5: [],
attr6: null,
};
console.log(this.interceptObjects(objSource, objInterface));
}
Here is a simple entry, very functional, handles any number of objects, and returns the values of the matching keys from the first object passed.
This behavior is similar to that of array_intersect_key() in PHP in case anyone is searching for that.
function intersectKeys(first, ...rest) {
const restKeys = rest.map(o => Object.keys(o));
return Object.fromEntries(Object.entries(first).filter(entry => restKeys.every(rk => rk.includes(entry[0]))));
}
Expanded here for better explanation and commenting
function intersectKeys(first, ...rest) {
// extract the keys of the other objects first so that won't be done again for each check
const restKeys = rest.map(o => Object.keys(o));
// In my version I am returning the first objects values under the intersect keys
return Object.fromEntries(
// extract [key, value] sets for each key and filter them, Object.fromEntries() reverses this back into an object of the remaining fields after the filter
Object.entries(first).filter(
// make sure each of the other object key sets includes the current key, or filter it out
entry => restKeys.every(
rk => rk.includes(entry[0])
)
)
);
// to get JUST the keys as OP requested the second line would simplify down to this
return Object.keys(first).filter(key => restKeys.every(rk => rk.includes(key)));
}
It's important to note that this solution only works on string keys, Symbol keys will be ignored and the final object will not contain any. Though a similar function could be written to compare Symbol intersect as well.
I know this is an old post, however, I want to share a solution I wrote today that I believe is efficient and clean.
function intersectingKeys(...objects) {
return objects
.map((object) => Object.keys(object))
.sort((a, b) => a.length - b.length)
.reduce((a, b) => a.filter((key) => b.includes(key)));
}
This function can take in n number of objects, and find the intersecting keys.
This is how it works.
Map the objects, creating an array of key arrays.
Sort the array by length, this puts the smallest key arrays first.
Finally, reduce our key arrays, by filtering each list of keys against the next list.
I think the clever part of this algorithm is the pre sorting of the key arrays. By starting with the smallest list of keys, we have less work to do comparing keys.
Here is the usuage:
var firstObject = {
x: 0,
y: 1,
z: 2,
a: 10,
b: 20,
e: 30,
};
var secondObject = {
x: 0,
y: 1,
z: 2,
a: 10,
c: 20,
d: 30,
};
intersectingKeys(firstObject, secondObject);
// [ 'x', 'y', 'z', 'a' ]
Related
let obj = { T: 1, h: 1, i: 3, s: 3, t: 1, r: 1, n: 1, g: 1 };
In the given object we have 2 keys(i & s both have same value 3) contains same value. I need to keep one key/value pair and another wants to remove.
How we can achieve this?
Make a copy and leave one field out?
let obj2 = { T: obj.T, h: obj.h, i: obj.i, t: obj.t, r: obj.r, n: obj.n, g: obj.g };
;-)
It's not super clear if you want to omit all keys with duplicate values or just some keys. In case you want the latter, here's a reusable function for that:
function copyWithout(source, ...keysToOmit) {
return Object.entries(source).reduce(
(accumulator, [key, value]) => {
return keysToOmit.includes(key)
? accumulator
: Object.assign(accumulator, { [key]: value });
},
Object.create(null) // or just {}
);
}
And use it like so:
copyWithout(obj, 'i');
Can be used to omit multiple keys as well:
copyWithout(obj, 'i', 't', 'r');
You can loop over object keys and add values in some temp array. Verify from that temp array and if value already exist in it then delete that key from object. Try like below.
let obj = { T: 1, h: 1, i: 3, s: 3, t: 1, r: 1, n: 1, g: 1 };
let val = [];
// loop over keys and verify that if value is repeating then delete that key from object.
Object.keys(obj).forEach(k => {
if (val.includes(obj[k])) {
delete obj[k];
} else {
val.push(obj[k])
}
});
console.log(obj);
Alternatively if you wish to keep your object as it is and want output in another object then try like below.
let obj = { T: 1, h: 1, i: 3, s: 3, t: 1, r: 1, n: 1, g: 1 };
let val = [];
let newObj = {};
// loop over keys and verify that if value is repeating then delete that key from object.
Object.keys(obj).forEach(k => {
if (!Object.values(newObj).includes(obj[k])) {
newObj[k] = obj[k];
}
});
console.log('newObj', newObj);
console.log('obj', obj);
You can use the following and swap key values twice,
let obj = { T: 1, h: 1, i: 3, s: 3, t: 1, r: 1, n: 1, g: 1 };
function swapKV(ob) {
return Object.entries(ob).reduce((p,[key,value]) => ({...p,[value]: key}),{})
}
swapKV(swapKV(obj))
You could save all unique values in set. This is the most optimal solution since it takes O(n) in time.
let obj = { T: 1, h: 1, i: 3, s: 3, t: 1, r: 1, n: 1, g: 1 };
function removeDuplicateValues(obj) {
const uniqueValues = new Set();
const uniquePairs = {};
Object.keys(obj).forEach(key => {
const curValue = obj[key];
if (!uniqueValues.has(curValue)) {
uniqueValues.add(curValue);
uniquePairs[key] = curValue;
}
});
return uniquePairs;
}
console.log(removeDuplicateValues(obj)) // {T: 1, i: 3}
const object = { a: 1, b: 2, c: 3, d: 2, e: 3 };
const removeUniqueValuesFromObject = (object) => {
const map = new Map();
// loop through all the attributes on an object
for (const [key, value] of Object.entries(object)) {
// check if the value already exists in map, if yes delete the attribute
if (map.has(value)) {
delete object[`${key}`];
continue;
}
// if value not found, add it to the map
map.set(value, key);
}
// return the updated object
return object;
};
const result = removeUniqueValuesFromObject(object);
console.log(result);
Implement the sumObjects function, which takes an unlimited number of objects and returns an object that combines all the objects entered.
Notes:
All object properties will have numeric values only
If the object keys match, the values of the corresponding keys are summed
The function always returns an object
The numbers in the object can be positive or negative
You can use the rest operator to collect arguments passed to a function in one array
Examples:
const obj = {};
sumObjects() === {}
sumObjects(obj) === {}
and
const first = {a: 2, b: 4};
const second = {a: 2, b: 10};
const third = {d: -5};
sumObjects(first) === {a: 2, b: 4}
sumObjects(first, third) === {a: 2, b: 4, d: -5}
sumObjects(first, second, third) === {a: 4, b: 14, d: -5}
This is my code.
function sumObjects(...params) {
let C = Object.fromEntries(
Object.keys(params[0])
.concat(Object.keys(params[1]))
.map(k => [k,
(params[0][k] || 0) + (params[1][k])
])
)
return C
}
I don't know how to add all these objects into one.
The bottom line is that I need to combine all the objects, but I just don't know how to do this, I can't find anything.
You can iterate over the param objects using .reduce and for each object, set/update its properties in acc:
function sumObjects(...params) {
return params.reduce((acc,item) => {
Object.entries(item).forEach(([property,value]) => {
const prev = acc[property];
acc[property] = prev ? prev+value : value;
});
return acc;
}, {});
}
const first = { a: 2, b: 4 };
const second = { a: 2, b: 10 };
const third = { d: -5 };
console.log( sumObjects(first) );
console.log( sumObjects(first, third) );
console.log( sumObjects(first, second, third) );
You could reduce the array of objects and iterate the entries of a single object.
const
sumObjects = (...objects) => objects.reduce((r, o) => {
Object.entries(o).forEach(([k, v]) => r[k] = (r[k] ||0) + v);
return r;
}, {}),
first = { a: 2, b: 4 },
second = { a: 2, b: 10 },
third = { d: -5 };
console.log(sumObjects(first)); // {a: 2, b: 4}
console.log(sumObjects(first, third)); // {a: 2, b: 4, d: -5}
console.log(sumObjects(first, second, third)); // {a: 4, b: 14, d: -5}
.as-console-wrapper { max-height: 100% !important; top: 0; }
I have the following array structure:
const array = [array1, array2, array3];
Each one of the three arrays consists of objects of form:
array1 = [{x: 0, y: 1}, {x: 5, y: 9}, {x: 1, y: 8}, {x: 3, y: 2}, etc]
I am trying to find the most efficient way to go through arrays of array and return the array to which a particular (unique) object belongs. For example I have object
{x:9, y:5}
which can be uniquely found in array2, so I want to return array2.
here's what I've tried:
const array = [array1, array2, array3];
for (let x = 0; x < array.length; x++) {
for (let y = 0; y < array[x].length; y++) {
array[x].find(e => e === array[x][y])
return array[x];
}
}
You'll need two loops, but you can use methods that do the iteration for you:
let array1 = [{x: 0, y: 1}, {x: 5, y: 9}, {x: 1, y: 8}, {x: 3, y: 2}];
let array2 = [{x: 5, y: 4}, {x: 4, y: 5}, {x: 8, y: 8}, {x: 3, y: 2}];
let array3 = [{x: 4, y: 3}, {x: 0, y: 6}, {x: 7, y: 8}, {x: 5, y: 2}];
const array = [array1, array2, array3];
let obj = array2[2]; // let's find this one...
let result = array.find(arr => arr.includes(obj));
console.log(result);
Here use find
data = [
[{x:1, y:2}, {x:2, y:3}],
[{x:3, y:2}, {x:4, y:3}],
[{x:5, y:2}, {x:6, y:3}],
[{x:7, y:2}, {x:8, y:3}]
];
const getArray = ({x, y}) => data.find(a => a.some(o => o.x === x && o.y === y));
console.log(getArray({x:3, y:2}));
TLDR; There is a working example in this fiddle
This can be accomplished using the following 3 things:
a library such as lodash to check for object equality (https://lodash.com/docs/4.17.15#isEqual)
The reason for this is that the behaviour of directly comparing two objects is different than you might think more info here
array.findIndex to find the index of the outer array (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/findIndex)
array.find to find the element in an inner array (https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/find)
The following method findObjectInNestedArray will do what you'd like.
const findObjectArray = (obj, arr) => {
index = arr.findIndex(a => a.find(e => _.isEqual(e, obj)))
return arr[index] // will return `undefined` if not found
}
// Example code below
const array1 = [{x: 0, y: 1}, {x: 5, y: 9}, {x: 1, y: 8}, {x: 3, y: 2}];
const array2 = [{x: 1, y: 1}, {x: 2, y: 2}, {x: 3, y: 3}, {x: 4, y: 4}, {x:9, y:5}];
const array3 = [{x: 5, y: 5}];
const arrays = [array1, array2, array3];
const inArray2 = {x:9, y:5};
const notInAnyArray = {x:0, y:0};
console.log('array2', findObjectArray(inArray2, arrays));
console.log('not in array', findObjectArray(notInAnyArray, arrays));
I know I said a single iteration was impossible before, but I devised a possible method that could work under specific circumstances. Essentially, you can sort the properties then stringify the objects for instant lookups. The sort is necessary to ensure you always get a consistent stringified output regardless of the object's properties preexisting order. There are three caveats to this method:
the objects CANNOT contain functions. Properties with functions are dropped in the stringification process.
NaN and infinity are converted to null, which can cause unexpected "matches" in the cache
If the depth of the object is not known (i.e. the target objects can contain references to arrays and other objects), then you'll need to deeply traverse through every level before stringifying.
It's a trade-off that's only improves performance when comparing deeply nested or extremely large objects. It's scalable, though, I guess.. Here's an example of how it could be done:
// sort's an array's values, handling subarrays and objects with recursion
const sortArr = arr => arr.sort().map(el => typeof el === 'object' ? (Array.isArray(el) ? sortArr(el) : sortObj(el)) : el)
// sorts a key's objects, then recreates the object in a consistent order
const sortObj = obj => Object.keys(obj).sort().reduce((final, prop) => {
final[prop] = (
// if it's an object, we'll need to sort that...
typeof obj[prop] === 'object'
? (
Array.isArray(obj[prop])
? sortArr(obj[prop])//<-- recursively sort subarray
: sortObj(obj[prop])//<-- recursively sort subobject
)
// otherwise, just retrun the value
: obj[prop]
)
return final
}, {})
// for every element in the given array, deeply sort then stringify it
const deepSortObjectArray = (arr) => arr.map(el => JSON.stringify(sortObj(el)))
// from those strings, create an object with the strings as values and an associated 'true' boolean
const obejctCache = (obj) => deepSortObjectArray(obj).reduce((acc, el) => ({[el]: true, ...acc}), {})
// create an object string cache for every object in the array:
const cacheObjectArrays = arr => arr.map(obj => obejctCache(obj))
// perform an O(1) lookup in each of the caches for a matching value:
const findArrayContainer = (obj, caches) => {
const stringLookupObj = JSON.stringify(sortObj(obj))
return caches.findIndex(cache => cache[stringLookupObj])
}
const array = [
{y: 1, x: 0},
{x: 5, y: 9},
{x: 1, y: 8},
{x: 3, y: {z: 3, x: 1, y: 2}}
]
const arrayArray = [[], [], array]
const cachesArrays = cacheObjectArrays(arrayArray)
console.log(cachesArrays)
/* output: [
{},
{},
{ '{"x":3,"y":{"x":1,"y":2,"z":3}}': true,'{"x":1, "y":8}': true, '{"x":5,"y":9}': true,'{"x":0,"y":1}': true }
]
*/
console.log(findArrayContainer({y: 1, x: 0}, cachesArrays))
// output: 2; working normally!
console.log(findArrayContainer({x: 0, y: 1}, cachesArrays))
// output: 2; working regardless of order!
console.log(findArrayContainer({y: 1, x: 0, q: 0}, cachesArrays))
// output: -1; working as expected with non-found objects!
As you can see, it's pretty complicated. Unless you're 100% this is actually the performance bottleneck, these performance gains may not translate to making interaction smoother.
Let me know if you have any questions about it!
I've got this array of objects:
[
{x: 615, y: 293, a: 1},
{x: 340, y: 439, a: 0},
{x: 292, y: 505, a: 0}
]
Basically im trying to write a collider. I'd love to return indexes of objects that values of x and y that are equal to each other, how do i approach this?
You can write a function that iterates, via map, through the array and returns the index or a null depending on this condition (x === y) and than filter it, returning only those that are different from null
const collider = array =>
array.map( (item, index) => item.x === item.y ? index : null )
.filter( item => item !== null)
Working fiddle:
https://jsfiddle.net/c3qqmh3b/
var indexes = myArray.reduce((idxs, el, i) => {
if (el.x === el.y) {
return idxs.concat(i);
} else {
return idxs;
}
}, []);
If your myArray would be, for example:
myArray = [
{x: 615, y: 293, a: 1},
{x: 340, y: 340, a: 0},
{x: 292, y: 505, a: 0}
]
then you'll get [1] as a result (because element of index 1 has x===y)
You could just use a hashtable to check for duplicates:
const hash = {};
for(const {x, y} of array) {
if(hash[x + "|" + y])
alert("collides");
hash[x + "|" + y] = true;
}
I have an array of objects, e.g.:
[
{ a: 3, b: 2, c: 5, d: 6, e: 8 },
{ a: 1, b: 5, c: 3, d: 1, e: 2 }
]
Now I want to transform this to an array that contains only the values of specific properties, but without the objects themselves. E.g., if I am interested in a, b, c, and d, the result should look like this:
[ 3, 2, 5, 6, 1, 5, 3, 1 ]
My current approach looks like this:
const result = _.flatten(data.map(item => [ item.a, item.b, item.c, item.d ]));
Is there a better (i.e., more efficient, and maybe even more readable) way to get the result?
What you have seems plenty readable to me, and is likely efficient enough. But you can make it more efficient by avoiding all those temporary arrays and not looping twice:
const result = [];
data.forEach(item => result.push(item.a, item.b, item.c, item.d));
Example:
const data = [
{ a: 3, b: 2, c: 5, d: 6, e: 8 },
{ a: 1, b: 5, c: 3, d: 1, e: 2 }
];
const result = [];
data.forEach(item => result.push(item.a, item.b, item.c, item.d));
console.log(result);
Some engines do push really efficiently, others not so much. If efficiency were a critical requirement, you'd want to experiment comparing that with this on your target environments:
const result = [];
let index = 0;
let n, l, item;
for (n = 0, l = data.length; n < l; ++n) {
item = data[n];
result[index++] = item.a;
result[index++] = item.b;
result[index++] = item.c;
result[index++] = item.d;
}
Three things to note there:
Pushing directly onto the array instead of using push.
Using a simple for loop instead of forEach. The absolute overhead of using forEach is so small as to be virtually non-existant from a human perspective, but the relative overhead of it in a tight loop is quite large.
Declaring i, l, and item outside the for loop. If we declared them within the for, they'd be recreated on each loop iteration, adding overhead. (ES2015's semantics for let declarations are powerful and useful, but in this particular case, we don't want the overhead.)
Example:
const data = [
{ a: 3, b: 2, c: 5, d: 6, e: 8 },
{ a: 1, b: 5, c: 3, d: 1, e: 2 }
];
const result = [];
let index = 0;
let n, l, item;
for (n = 0, l = data.length; n < l; ++n) {
item = data[n];
result[index++] = item.a;
result[index++] = item.b;
result[index++] = item.c;
result[index++] = item.d;
}
console.log(result);
You could reduce with mapped keys.
var array = [{ a: 3, b: 2, c: 5, d: 6, e: 8 }, { a: 1, b: 5, c: 3, d: 1, e: 2 }],
keys = ['a', 'b', 'c', 'd'],
result = array.reduce((r, a) => r.concat(keys.map(k => a[k])), []);
console.log(result);
You can Array.prototype.map() the desired keys from the objects in the array and merge all them with Function.prototype.apply() and Array.prototype.concat():
let data = [{ a: 3, b: 2, c: 5, d: 6, e: 8 }, { a: 1, b: 5, c: 3, d: 1, e: 2 }],
result = [].concat.apply([], data.map(i => [i.a, i.b, i.c, i.d]));
console.log(result);
In ES6, and making no claims for efficiency, you could write
const input = [
{ a: 3, b: 2, c: 5, d: 6, e: 8 },
{ a: 1, b: 5, c: 3, d: 1, e: 2 }
];
console.log([].concat(...input.map(({a, b, c, d}) => [a, b, c, d])));
If you're really interested in efficiency, nothing would beat
var result = [];
for (let i = 0; i < input.length; i++) {
const o = input[i];
result.push(o.a, o.b, o.c, o.d);
}
although as another answer points out, it would be worth benchmarking push against result[cnt++] = val.
You can use 2 nested forEach loops to push targeted values of each element of your input array. I propose here an function receiving as parameters initial array and properties targeted as an other array :
var inflate = (data, options) => {
var results = [];
data.forEach(x => options.forEach(y => results.push(x[y])));
return results;
}
console.log(inflate([{ a: 3, b: 2, c: 5, d: 6, e: 8 }, { a: 1, b: 5, c: 3, d: 1, e: 2 }] ,["a","b","c","d"]));