javascript find objects based on search terms

Question

I need to search an array of objects with an object of search terms and get the index of results in an array.

Let's say I have an array like this:

[
  {
    name: "Mary",
    gender: "female",
    country: "USA",
    orientation: "straight",
    colorChoice: "red",
    shoeSize: 7
  },
  {
    name: "Henry",
    gender: "male",
    country: "USA",
    orientation: "straight",
    colorChoice: "red",
  },
  {
    name: "Bob",
    colorChoice: "yellow",
    shoeSize: 10
  },
  {
    name: "Jenny",
    gender: "female",
    orientation: "gay",
    colorChoice: "red",
  }
]

Now I need to search the array for:

{
  gender: "female"
}

and get result:

[ 0, 3 ]

The search object can be any length:

{
  gender: "female",
  colorChoice: "red"
}

What's the cleanest and most performant way to search the array?

Thanks.

Related: stackoverflow.com/questions/3624741/…

Marcel Korpel
– Marcel Korpel

2011-04-21 08:01:57 +00:00
Commented Apr 21, 2011 at 8:01 — Marcel Korpel
– Marcel Korpel, Commented Apr 21, 2011 at 8:01
You should get [0, 3] from the array you present.

KooiInc
– KooiInc

2011-04-21 08:20:16 +00:00
Commented Apr 21, 2011 at 8:20 — KooiInc
– KooiInc, Commented Apr 21, 2011 at 8:20

KooiInc · Accepted Answer · 2011-04-21 08:20:57Z

2

Here's the idea:

function getFemales(myArr){
 var i = myArr.length, ret = [];
 while (i--){
  if ('gender' in myArr[i] && myArr[i].gender === 'female') {
    ret.push(i);
  }
 }
 return ret.sort();
}

see jsfiddle

And more generic:

function findInElements(elArray, label, val){
 var i = elArray.length, ret = [];
 while (i--){
  if (label in elArray[i] && elArray[i][label] === val) {
    ret.push(i);
  }
 }
 return ret.sort();
}

see jsfiddle

answered Apr 21, 2011 at 8:20

KooiInc

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4 Comments

TJHeuvel Over a year ago

Any reason you traverse the array in reverse?

KooiInc Over a year ago

It's quicker.See devpro.it/examples/loopsbench or codeutopia.net/blog/2009/04/30/…

TJHeuvel Over a year ago

Thats pretty damn clever, since the type juggeling makes sure the while stops when i <= 0! You'd have to change your i-- to --i though :)

KooiInc Over a year ago

using --i I would miss element 0. In that case I'd have to use (--i>=0). It's not significantly faster, see jsperf.com/loop-descendening

TJHeuvel · Accepted Answer · 2011-04-21 08:08:52Z

2

This should do the trick:

function searchArray(fields, arr)
{
    var result = [];            //Store the results

    for(var i in arr)           //Go through every item in the array
    {
        var item = arr[i];
        var matches = true;     //Does this meet our criterium?

        for(var f in fields)    //Match all the requirements
        {
            if(item[f] != fields[f])    //It doesnt match, note it and stop the loop.
            {
                matches = false;
                break;
            }
        }

        if(matches)
            result.push(item);  //Add the item to the result
    }

    return result;
}

For example:

console.log(searchArray({
  gender: "female",
  colorChoice: "red"
},[
  {
    name: "Mary",
    gender: "female",
    country: "USA",
    orientation: "straight",
    colorChoice: "red",
    shoeSize: 7
  },
  {
    name: "Henry",
    gender: "male",
    country: "USA",
    orientation: "straight",
    colorChoice: "red",
  },
  {
    name: "Bob",
    colorChoice: "yellow",
    shoeSize: 10
  },
  {
    name: "Jenny",
    gender: "female",
    orientation: "gay",
    colorChoice: "red",
  }
]));

answered Apr 21, 2011 at 8:08

TJHeuvel

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1 Comment

TJHeuvel Over a year ago

You can edit the function to return the indexes (i) of the array too, just push that on the results array instead of the item.

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