How to loop through Array of Objects to find ID then return a sibling key

Question

I have a complex Array of Objects below, and I have a term_id to search on. I'm trying to find the matching term_id, then return the associated ticker: name from the same Object from which I found the term_id.

container = [Object, Object];

// container:
[
    0: Object {
        tags: [
            0: {
                term: "tag_name_1",
                term_id: 1111
            },
            0: {
                term: "tag_name_2",
                term_id: 2222
            }
        ],
        ticker: {
            name: "ticker1"
        }
    },
    1: Object {
        tags: [
            0: {
                term: "tag_name_3",
                term_id: 3333
            }
        ],
        ticker: {
            name: "ticker2"
        }
    }
]

How would you accomplish this? Is there an easy way with _lodash?

ChadF · Accepted Answer · 2015-07-10 19:41:09Z

2

// You can do this with native JS:

var container = [{tags: [{term: "tag_name_1",term_id: 1111},{term: "tag_name_2",term_id: 2222}],ticker: {name: "ticker1"}},{tags: [{term: "tag_name_3",term_id: 3333}],ticker: {name: "ticker2"}}];

function search (container, id) {
  var contains = false;
  var result;

  container.forEach(function(obj){
    obj.tags.forEach(function(innerData){
      if (innerData.term_id === id) {
        contains = true;
      }
    })
    if (contains) {
      result = obj.ticker.name;
      contains = false;
    }
  });

  return result;
}

console.log(search(container, 1111));

edited Jul 10, 2015 at 19:41

answered Jul 10, 2015 at 19:26

ChadF

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3 Comments

StriplingWarrior Over a year ago

Shouldn't your search function do something with the result? Like return it?

Leon Gaban Over a year ago

Thanks yeah the forEach works :) I need to Gist this, 4 mins

ChadF Over a year ago

I had the return in the forEach which wasn't working. Fixed that and made it run as a code snippet. Check your console and hit run.

jabclab · Accepted Answer · 2015-07-10 19:40:10Z

2

You can use Array.prototype.some for this. For example:

function find(arr, t) {
    var ticker = null;

    arr.some(function (doc) {
        var tagMatch = doc.tags.some(function (tag) {
            return tag.term_id === t;
        });

        if (tagMatch) {
            ticker = doc.ticker.name;
        }

        return tagMatch;
    });

    return ticker;
}

Here's a JSFiddle.

answered Jul 10, 2015 at 19:40

jabclab

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1 Comment

Leon Gaban Over a year ago

Very nice and clean :)

Blauharley · Accepted Answer · 2015-07-10 19:36:13Z

Hope this helps you. It's a function that you can pass your objects into and a term_id you search for and it returns found ticker names:

var objs = [
    {
        tags: [
            {
                term: "tag_name_1",
                term_id: 1111
            },
            {
                term: "tag_name_2",
                term_id: 2222
            }
        ],
        ticker: {
            name: "ticker1"
        }
    },
    {
        tags: [
            {
                term: "tag_name_3",
                term_id: 3333
            }
        ],
        ticker: {
            name: "ticker2"
        }
    }
];

function getTickerNamesById(objs,id){
  var foundNames = [];
  objs.forEach(function(obj){
    obj.tags.forEach(function(term){
      if(term.term_id===id)foundNames.push(obj.ticker.name);
    });
  });
  return foundNames;
}

getTickerNamesById(objs,3333); // ["ticker2"]

Joel · Accepted Answer · 2015-07-10 20:09:57Z

A forEach() loop works, though there is no way to prevent it from cycling through the entire object once the id is matched. Assuming the id's are unique, a option with better performance would be the while loop:

function findId(id,container) {
  var i = 0,
      j;

  while (i < container.length) {
    j = 0;
    while (j < container[i].tags.length) {
      if (container[i].tags[j].term_id === id) {
       return container[i].ticker.name;
      }
      j += 1;
    }
    i += 1;
  }
  throw "item not found";
}

If your containers will be large you may want to consider this optimization. If you preferred a functional approach, you could accomplish a similar thing with some() or every(), both of which exit out given a specified condition.

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