Filter and Unique array of object in javascript with underscore.js based on two property

Question

In my application i have an array like this :

var arr=[
 {id:1,name:"mohammmad",deny:false},
 {id:1,name:"mohammmad",deny:false},
 {id:1,name:"mohammmad",deny:true},
 {id:2,name:"ali",deny:false},
 {id:3,name:"reza",deny:true},
 {id:2,name:"ali",deny:true}
];

now I want to return unique collection of this array based on id of object but if collection have true value for deny it must return the object that have deny=true. for example :

 {id:1,name:"mohammmad",deny:false},
 {id:1,name:"mohammmad",deny:false},
 {id:1,name:"mohammmad",deny:true},

it filter with id=1 but return the object that have deny=true. i also create a jsbin here .

What if there are matching ID's, but no objects with deny:true, only deny:false? What I'm really asking is, can we just remove anything that is deny:false and filter the rest ? — adeneo
– adeneo, Commented Aug 11, 2014 at 7:48
I don't think unique is going to help you, because the time you applied uniq it will filter the duplicates. — Mritunjay
– Mritunjay, Commented Aug 11, 2014 at 7:57

joews · Accepted Answer · 2014-08-11 08:15:27Z

2

Using underscore:

_.chain(arr)
  .sortBy(function(d) { return !d.deny })
  .uniq(function(d) { return d.id })
  .value()

Sort by deny (with true values first), then uniq on id. Underscore's implementation of uniq keeps the first matching value for each key. Therefore, if a given id had a value with deny: true, that value will be kept. If it didn't, it keeps first value with any value of deny.

JSFiddle

edited Aug 11, 2014 at 8:15

answered Aug 11, 2014 at 8:08

joews

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