JavaScript -- find matching object in array of objects

Question

I am trying to search for an object in an array of objects.

Note, vals and recs objects will be DYNAMIC.

var vals = {ID: "4", LOC: "LA", SEQ: "1"};

var recs = [
  {ID:"4", LOC:"LA", SEQ:"1"},
  {ID:"4", LOC:"NY", SEQ:"1"},
  {ID:"4", LOC:"CHI",SEQ:"1"}

];

Now I need to check if all key:value pairs in vals already exist in recs . In this case, recs[0] is an exact match of vals.

Heres my attempt:

var vals =  {ID: "4", LOC: "LA", SEQ: "1"};


var recs = [
  {ID:"4", LOC:"LA", SEQ:"1"},
  {ID:"3", LOC:"NY", SEQ:"2"},
  {ID:"2", LOC:"CHI",SEQ:"3"}

];


for(var i = 0; i<recs.length; i++){  
    if(recs[i]["ID"] == vals["ID"] && recs[i]["LOC"] == vals["LOC"] && recs[i]["SEQ"] == vals["SEQ"]){
      console.log(true);
    }
    else{
      console.log(false);
    }  
}

The above works only because I have hardcoded the keys from the vals object. In reality, the VALS object (and recs) will be DYNAMIC with X number of key:value pairs.

So how can I modify my for loop for a dynamic vals object?

thanks!

from someone who has done this a lot: Object.keys() is very fast, so comparing the length of each object's key array would be a fast way of eliminating misfits. sorting and joining those keys into a string provides a fast membership comparison. if both those quick tests pass, iterate and compare values. of course, if perf doesn't matter, just do two unfiltered nested loops and KISS. — dandavis
– dandavis, Commented Mar 17, 2016 at 0:51
the question is vuage. does {a:1} describe {a:1,b:2}? should it return the first result, all matches or a boolean? — dandavis
– dandavis, Commented Mar 17, 2016 at 0:56

YSharp · Accepted Answer · 2016-03-18 01:41:18Z

1

Try this:

for (var i = 0; i < recs.length; i++) {  
  var found = true;
  for (var p in vals) {
    if (vals.hasOwnProperty(p)) {
      if (recs[i][p] !== vals[p]) {
        found = false;
        break;
      }
    }
  }
  console.log(found);
}

edited Mar 18, 2016 at 1:41

answered Mar 17, 2016 at 0:49

YSharp

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1 Comment

dandavis Over a year ago

gimme a break for perf

George Chen · Accepted Answer · 2016-03-17 00:44:36Z

1

for(var i = 0; i<recs.length; i++) {  
    for (var prop in object) {
       if (recs[i][prop] != vals[prop]) {
           console.log(false);
           return;
       }

    }
    //check from both sides
    for (var prop in vals) {
       if (recs[i][prop] != vals[prop]) {
           console.log(false);
           return;
       }

    }
    console.log(true);


}

answered Mar 17, 2016 at 0:44

George Chen

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1 Comment

dandavis Over a year ago

shape mismatches will throw with this pattern. be careful what you iterate into when fetching properties

blurfus · Accepted Answer · 2016-03-17 05:31:09Z

1

You could iterate over the keys; something along the lines of:

var vals = {  ID: "4",  LOC: "LA",  SEQ: "1", REGION: "USA" };


var recs = [{    ID: 4,    LOC: "LA",    SEQ: "1",    REGION: "USA"  },
            {    ID: 3,    LOC: "NY",    SEQ: "2",    REGION: "USA"  },
            {    ID: 2,    LOC: "CHI",    SEQ: "3",    REGION: "USA" }

];

var isSame = true;

for (var i = 0; i < recs.length; i++) {
  console.log( i + '----------------' );
  var isSame = true;
  
  // get the keys of the record
  var keys = Object.keys(recs[i]);

  for (var j = 0; j < keys.length; j++) {
    
    var key = keys[j];
    var record = recs[i] 
    
    console.log( key + ": " + record[key] + '=' + vals[key] );   
    
    if (record[key] != vals[key] ) {                                
        isSame = false;// not equal
        break;
    }
  }

  console.log('isSame: ' + isSame );
  console.log('------------------' );
}

edited Mar 17, 2016 at 5:31

answered Mar 17, 2016 at 0:49

blurfus

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2 Comments

dandavis Over a year ago

might want to memorize keys[j] and or recs[i] when you use them so often to shorten and hasten.

blurfus Over a year ago

Good point; I'll update the answer when I get back to my desk... Thx @dandavis

Jonas Meinerz · Accepted Answer · 2016-03-18 00:20:41Z

1

You need to break it into two loops, one for each object of the array and one for each key of the object:

for(var i = 0; i<recs.length; i++){  
  var found = false
  for(var key in recs[i]) {
    if(recs[i].hasOwnProperty(key)){
      if(recs[i][key] != vals[key]){
        found = true
      }
    }
  console.log(found)
}

the hasOwnProperty call will make sure it doesn't break if the object does not have that key.

edited Mar 18, 2016 at 0:20

answered Mar 17, 2016 at 0:40

Jonas Meinerz

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1 Comment

dandavis Over a year ago

where does the vals.key come from? you need vals[key] et al

score 0 · Accepted Answer · 2016-03-17 03:55:05Z

0

You can try this:

function myFind(recs, vals) {
    return recs.some(function(obj) {
        for (var x in obj)
            if (x in vals && obj[x] != vals[x])
                return false;
        return true;
    });
}

var recs = [
    {ID:4, LOC:"LA", SEQ:"1", USA:"USA"},
    {ID:3, LOC:"NY", SEQ:"2", USA:"USA"},
    {ID:2, LOC:"CHI",SEQ:"3", USA:"USA"}
];

var vals =  {ID: "4", LOC: "LA", SEQ: "1"};

if (myFind(recs, vals)) {
    alert('found');
} else {
    alert('not found');
}

Hope it helps.

edited Mar 17, 2016 at 3:55

answered Mar 17, 2016 at 0:35

user4227915

6 Comments

Evan Trimboli Over a year ago

This won't find the item. The object references are different.

user4227915 Over a year ago

@EvanTrimboli. I didn't see the second declaration of recs sorry for that, I've updated now

Evan Trimboli Over a year ago

That edit won't find the value either. The references still won't be the same, with the added downside of mutating the values in recs.

user4227915 Over a year ago

@EvanTrimboli. Strange, It works here, I will add an snippet

dandavis Over a year ago

why are you using delete ?

|

Jairo Malanay · Accepted Answer · 2016-03-18 02:47:31Z

0

you can use underscorejs isEqual for this kind of problem

answered Mar 18, 2016 at 2:47

Jairo Malanay

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