Extracting attributes from array of objects based on other attributes

Question

I have array of objects , and would like to fetch some attributes based on another one

assume

var lngs = [{id : 3 , label : 'a'} , {id : 2 , label : 'b' , {id : 4 , label : 'c'}];

now I need to get array of labels based on ids

for example

getAttrs([4,3]);

function getAttrs(ids , arr)
{
   // as long as ids = [4,3] , it should return
   ["c","a"]
}

I tried this algorithm , but it did not work as expected

 langsFromIds(ids)
 {
   var rslt = [];
   var lngs = this.state.langs;
   for (var i = 0 ; i < lngs.length ; i++)
   {
     if (lngs[i].id == ids[i])
     {
       rslt.push(lngs[i].label);
     }
   }

   return rslt;
 }

because it runs in this way (for example)

first iteration (i==0) => lngs[i].id == 3 && ids[i] == 4 (will not match)

second iteration (i==1) => lngs[i].id == 2 && ids[i] == 3 (will not match)

and so on ..

so it will return empty array

Your lngs object is syntactically invalid. I have posted my answer assuming var lngs = [{id : 3 , label : 'a'} , {id : 2 , label : 'b'} , {id : 4 , label : 'c'}]; — Matt Ellen
– Matt Ellen, Commented Jul 10, 2019 at 14:03

Matt Ellen · Accepted Answer · 2019-07-10 14:02:29Z

1

First filter down to the corrects ids, then extract the desired labels:

var lngs = [{id : 3 , label : 'a'} , {id : 2 , label : 'b'} , {id : 4 , label : 'c'}];

function langsFromIds(ids, langs)
{
  return langs.filter(lng => ids.indexOf(lng.id) > -1).map(lng => lng.label);
}

console.log(langsFromIds([4, 3], lngs));

answered Jul 10, 2019 at 14:02

Matt Ellen

11.6k5 gold badges74 silver badges94 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

27mdmo7sn Over a year ago

It works well I think , thanks I'll accept your answer

freefaller · Accepted Answer · 2019-07-10 14:11:43Z

1

You could use the filter and find to get a list of individual lng objects...

var lngs = [{id:3,label:'a'},{id:2,label:'b'},{id:4,label:'c'}];

function lngsFromIds(ids) {
  return lngs.filter(x => ids.find(y => y == x.id));
}
console.log(lngsFromIds([3,4]));

And then use map to get the label as well...

var lngs = [{id:3,label:'a'},{id:2,label:'b'},{id:4,label:'c'}];

function labelsFromIds(ids) {
  return lngs.filter(x => ids.find(y => y == x.id)).map(z => z.label);
}
console.log(labelsFromIds([3,4]));

edited Jul 10, 2019 at 14:11

answered Jul 10, 2019 at 14:06

freefaller

20.1k7 gold badges62 silver badges96 bronze badges

Comments

Hichem CHTARA · Accepted Answer · 2019-07-10 14:08:48Z

0

Here is my proposition :

var lngs = [{id : 3 , label : 'a'} , {id : 2 , label : 'b' }, {id : 4 , label : 'c'}];

var npq = getAttrs([4,3]);
console.log(npq);

function getAttrs(ids)
{
   var rslt = [];
   //var lngs = this.state.langs;
    for (var k = 0; k < ids.length; k++)
    {
        for (var i = 0 ; i < lngs.length ; i++)
        {
          if (lngs[i].id == ids[k])
          {
             rslt.push(lngs[i].label);
          }
        }
    }
   return rslt;
}

answered Jul 10, 2019 at 14:08

Hichem CHTARA

251 silver badge7 bronze badges

Comments

mighty_dev · Accepted Answer · 2019-07-10 14:30:26Z

To answer your question, you are matching index to index and hence you won't get the desired result as the id could be found anywhere in ids.
A quick solution is to iterate inside again on ids and if a match is found, you add them in the result.

var lngs = [{id:3,label:'a'},{id:2,label:'b'},{id:4,label:'c'}];

function langsFromIds(ids)
 {
   var rslt = [];
   for (var i = 0 ; i < lngs.length ; i++)
   {
     for(var j=0; j < ids.length;j++)
     {
      	if (lngs[i].id == ids[j])
     	{
       		rslt.push(lngs[i].label);
     	}	 
     }
   }

   return rslt;
 }

console.log(langsFromIds([3,4]));

Although the above solution works, it isn't scalable, meaning above solution would take a lot of time to return a result if size of lngs goes huge along with ids. In other words, time complexity is O(n * m) where n is the size of lngs and m is the size of ids.
So, if you are aiming for an efficient approach, you could make a set of ids and check for a match in lngs.

var lngs = [{id:3,label:'a'},{id:2,label:'b'},{id:4,label:'c'}];

function lngsFromIds(ids) {
  var set = new Set(ids);
  return lngs.filter(x => set.has(x.id)).map(x => x.label);
}
console.log(lngsFromIds([3,4]));

Above approach can decide whether the current object's id is present in ids or not in on an average O(1) time, hence making the function faster. Time complexity would be O(n + m) time with an additional space trade off of O(m) where n and m are same as mentioned before.

Collectives™ on Stack Overflow

Extracting attributes from array of objects based on other attributes

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related