A better way to handle grabbing an object from an array of objects?

Question

I have a JSON object with, say, 500 objects in an array. At the moment, to find the one I want, I'm using jQuery's "each" and comparing each object's id with the id I'm looking for, like so:

var desiredID = 500;
$.each(myObj.arrayOfObjects, function(k, oneObject){
    if((oneObject.lineid * 1) === (desiredID * 1)){
        // hooray! I have found my object
    }
});

Is there a better, more efficient way? The array of objects could potentially get impressively large.

(In this instance, I'm using jQuery, but it could be vanilla javascript, too.)

developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… , if you don't need IE support. And if you do I'm sure there's a polyfill some where. — Taplar
– Taplar, Commented Apr 7, 2020 at 22:12
You can achieve similar result using vanilla javascript as well: myObj.arrayOfObjects.find(oneObject => (oneObject.lineid * 1) === (desiredID *1)) — Ravi Nain
– Ravi Nain, Commented Apr 7, 2020 at 22:13
Though, from a data model standpoint, one could argue if the element have an "id", they might could be better represented as a map. Then id lookups are super easy — Taplar
– Taplar, Commented Apr 7, 2020 at 22:13
Side note; also not entirely sure why you are multiplying the desiredID by 1... — Taplar
– Taplar, Commented Apr 7, 2020 at 22:14

Chris · Accepted Answer · 2020-04-07 22:22:32Z

1

with vanilla js:

const obj = arrayOfObjects.find(({ lineid }) => lineid === desiredID)

if you want to convert string to number, use this triks:

         +'1' === 1 // true
        ~~'1' === 1 // true
  Number('1') === 1 // true
parseInt('1') === 1 // true

and for the desiredID you don't have to cast, because you know it's number

edited Apr 7, 2020 at 22:22

answered Apr 7, 2020 at 22:15

Chris

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6 Comments

WillD Over a year ago

for the sake of my own education: is Array.find() that much more efficient than just writing your own loop and checking against each? If so, why?

Chris Over a year ago

it depends on what size it is, but for the most part yes

WillD Over a year ago

Interesting... so up to a certain size array it is more efficient and then it stops being the faster way as the array grows? Or the opposite?

Chris Over a year ago

now it also depends on which search algorithm you write, if it is just 'for' is the same thing

Taplar Over a year ago

Also keep in mind, if you do not break out of your for loop once you find what you are looking for, an each will evaluate every element even once it is found, where as a find() will stop on the first match immediately

|

Taplar · Accepted Answer · 2020-04-07 22:23:10Z

0

myObj.mappedObjects = myObj.arrayOfObjects.reduce((r, e) => {
    r[+e.lineid] = e;
    return r;
}, {});

var desiredID = 500;
myObj.mappedObjects[desiredID];

Given that you mentioned possibly having 500ish elements, if there is the potential of having to look up elements repeatedly, I would suggest converting the array into a map, for quick id lookups going forward.

If the endpoint, or where ever, you are getting the original data could put it in that format to begin with, even better.

answered Apr 7, 2020 at 22:23

Taplar

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3 Comments

Chris Over a year ago

If you are looking for a single time in object, then 'reduce' it will consume much more than 'find'

Taplar Over a year ago

@Chris perhaps you missed my comment about repetitions?

Chris Over a year ago

I'm sorry, I wasn't paying attention

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