How to improve the performance of 'for loop' to integrate some objects in JS

Question

I have three for loop as below to integrate their objects.
The problem is the length of 'cars' array is 20,000.
So it should runs every 20,000 times for finding same id between company.user and cars.
But the id of cars is unique.
Can I reduce this repeat number in JS? I want to reduce the taking time.
Thank you for reading it.

p.s. I uploaded same question adding the concrete logic inside of for loop.

for (let i = 0; i < company.length; i += 1) {
  for (let j = 0; j < company[i].user.length; j += 1) {
    for (let k = 0; k < cars.length; k += 1) {
      if (company[i].user[j].id === cars[k].id) {
        company[i].user[j] = {
          ...company[i].user[j],
          ...cars[k] 
        }
      }
    }
  }
}

is it possible to have multiple `company.user` and `cars` or is there only 1 match? — bill.gates, May 29 '20 at 11:50
if its possible, you could sort the "cars" array and then use some search algorithms to reduce the search time. Everyhing you need to know about search algorithms you can find here https://www.geeksforgeeks.org/searching-algorithms/ — ziga1337, May 29 '20 at 11:51
Don't check the length in the for loop, store the length and check against the stored value instead. https://stackoverflow.com/questions/8452317/do-loops-check-the-array-length-every-time-when-comparing-i-against-array-length — Trevor, May 29 '20 at 11:57
@Ifaruki Nope. I think it breaks just after matching once. I try to use binary search now. You gived me a inspiration. — Thomas Jason, May 29 '20 at 12:07
@ThomasJason you sure that it breaks? because there is no break — bill.gates, May 29 '20 at 12:08
for a given user.id, can the number of number of cars with the same id be **greater than 1** ? — Mister Jojo, May 29 '20 at 12:31
Make indexes out of plain objects or Map instances so that you can directly look up a car or user by id. — Pointy, May 29 '20 at 13:14
Also it would probably be better to avoid creating new objects for each user, and instead just `Object.assign()` the car properties to the existing user object. — Pointy, May 29 '20 at 13:15

score 1 · Accepted Answer · answered May 29 '20 at 11:53

If there is only 1 match then use break after you found that item. Imagine you find that item at index 1 for example, then the loop would still continue and do 19998 loops. Because of the information that you know there is only 1 possible match you can break it there.

    if (company[i].user[j].id === cars[k].id) {
      company[i].user[j] = {
        ...company[i].user[j],
        ...cars[k] 
      }
      break;
    }

score 0 · Answer 2 · answered May 29 '20 at 12:20

 for (let i = 0, leni = company.length;; i < leni ; i += 1) {
    for (let j = 0, lenj = company[i].user.length; j < lenj; j += 1) {
      for (let k = 0, lenk = cars.length; k < lenk; k += 1) {
        if (company[i].user[j].id === cars[k].id) {
          company[i].user[j] = {
            ...company[i].user[j],
            ...cars[k] 
          }
          break; // match only the first one, then stop searching for cars
        }
      }
    }
  }

Answer based on test results from https://jsperf.com/caching-array-length/4

Spread left in base on

https://thecodebarbarian.com/object-assign-vs-object-spread.html

`throw in an empty object` but `company[i].user[j]` is not empty... — Mister Jojo, May 29 '20 at 13:38

Mister Jojo · Answer 3 · 2020-05-29T15:27:25.923

this will optimize your code a little more, but ziga1337 is right, the only effective way is to sort your two objects

// company: [ { user: [ { id: '?1' 
// cars: [ { id: '?2' 

for (let iCompagny of compagny) {
  for (let iUser of iCompagny.user) {
    let fCar = cars.find(xCar.id===iUser.id)
    if (fCar) Object.assign(iUser, fCar)
  }
}

in case there is always a car.id unique to each user.id:

let indexCars = cars.reduce((a,c)=>{ a[c.id]=c; return a },{})

compagny.forEach(iCompagny=>{
  iCompagny.user.forEach(iUser=>{ Object.assign(iUser, indexCars[iUser.id]) })
})

How to improve the performance of 'for loop' to integrate some objects in JS

3 Answers3