Javascript. Check if the Object contains object

Question

I have following two arrays of objects

//references
var refs=[
         {id:1,   name:'John',    state:'A'},
         {id:2,   name:'Obama',   state:'P'},
         {id:3,   name:'Lincoln', state:'P'}
];

//items
var items=[
          {ref:refs[0],    detailed:true },
          {ref:refs[1],    detailed:false}
];

Now I want to check if items contains refs[0] . How can I do It? Of course we can do something like this

for(item:items){
 if(item.ref==refs[0]){
  console.log('contains');
  break;
 }
}

but I might have 2000 refs. Any ideas. Thank you

Can you clarify why that approach won't work if you have 2000 refs? — JJJ
– JJJ, Commented Jul 6, 2014 at 12:29
How's having more refs change anything, you're only interested in refs[0] anyway according to your specs? I would not go through a for in loop though because both items and refs are arrays so you can do while(++i<len) since items[?].refs is a reference you can compare with like this === — HMR
– HMR, Commented Jul 6, 2014 at 12:38
@JumabekAlihanov - See my comment to Juhana under my answer. — Jared Farrish
– Jared Farrish, Commented Jul 6, 2014 at 13:07
Still not sure what the requirements are, if you only need to look up one item once then a loop is the way to go but if you have a lot of refs and you need to know if they are in items and if so what the values are then you need to redefine items. I've updated my answer with a possible optimization. — HMR
– HMR, Commented Jul 6, 2014 at 13:22

Jared Farrish · Accepted Answer · 2014-07-06 12:49:45Z

2

You can use Array.filter():

console.log(get_ref_item(refs[0], items));

function get_ref_item(ref, items) {
    return items.filter(function fi(item){
        return ref.id == item.ref.id;
    });
}

http://jsfiddle.net/5MfrT/

answered Jul 6, 2014 at 12:49

Jared Farrish

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3 Comments

JJJ Over a year ago

Note that this is practically identical to the original code performance-wise. There's just no way to avoid looping through the entire array when searching for elements unless you already know roughly where to start looking for.

Jared Farrish Over a year ago

@Juhana - From a performance point of view, the for with break is going to be faster in many cases, depending on where the item being searched for is in the array. So, in actuality, the OPs code is better in this regard. See: jsperf.com/for-vs-array-filter

Jared Farrish Over a year ago

Note, the only reason it could be equivalent (or maybe better, would have to test) is if the entire array had to be searched each time, for instance if the ref.id could exist more than once in the array.

HMR · Accepted Answer · 2014-07-06 13:20:52Z

0

Not clear what the problem is here but assuming you'd like to know if a ref is in items and you need to know this for multiple refs (look it up multiple times) then you better re define your items:

//references
var refs=[
         {id:1,   name:'John',    state:'A'},
         {id:2,   name:'Obama',   state:'P'},
         {id:3,   name:'Lincoln', state:'P'}
];

//items
var items=[
          {ref:refs[0],    detailed:true },
          {ref:refs[1],    detailed:false}
];

//if you need to continiously look up a ref in items
// it's better to convert items to an object
// {"refid":{ref:..,detailed:...}}
// assuming every item has a ref and has that ref only once
// and every ref has a unique id
var i = -1,len=items.length,tmp={};
while(++i<len){
  tmp[items[i].ref.id]=items[i];
}
items=tmp;

console.log("item for refs 0:",items[refs[0].id]);
console.log("item for refs 1:",items[refs[1].id]);
console.log("item for refs 2:",items[refs[2].id]);//undefined, that ref is not it items

edited Jul 6, 2014 at 13:20

answered Jul 6, 2014 at 13:02

HMR

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2 Comments

Jared Farrish Over a year ago

You can test your optimization against the for and Array.filter techniques here: jsperf.com/for-vs-array-filter Setting up performance tests can be tricky, so be sure to think it through.

HMR Over a year ago

@JaredFarrish I'm not using for ... in nor Array.filter or while(++i<len) because it's not using an array at all. But I can tell you right now that if you'd had 500+ items and 2000+ refs making the items an object with ref.id as keys is going to out perform whatever array lookup you're planning to do 200 times. (if op was planning to match every item with a ref)

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