Manipulating Javascript array of objects

Question

I have an array of objects and need to create a new array which contains the values of a specific key within the objects. Is there a more elegant way than the following code (note: without using anything more than JQuery). Expected result is: 455, 387, 495

var arr=[{sid:387,rank:2},{sid:455,rank:1},{sid:364,rank:4},{sid:495,rank:3}];
var topThreeTemp = arr.filter(function(a){return a.rank<4;}).sort(function(a,b){return a.rank>b.rank;});
var topThreeSIDs=[];
for(var i=0;i<topThreeTemp.length;i++){
    topThreeSIDs.push(topThreeTemp[i].sid);
}
console.log(topThreeSIDs.join(", "));

Array.map() so something like topThreeTemp.map(function( item ) { return item.sid; }); — Shilly
– Shilly, Commented Aug 24, 2018 at 10:04

Jordan Enev · Accepted Answer · 2018-08-24 10:10:17Z

4

Use ES6 map, instead of for and ES6 arrow functions (syntax sugars):

const arr = [
 { sid: 387,rank: 2 },
 { sid: 455,rank: 1 },
 { sid: 364,rank: 4 },
 { sid: 495,rank: 3 }
]

const topThreeSIDs = arr
  .filter(({ rank }) => rank < 4)
  .sort((a, b) => a.rank > b.rank)
  .map(({ sid }) => sid)
  

console.log(topThreeSIDs.join(', '))

edited Aug 24, 2018 at 10:10

answered Aug 24, 2018 at 10:04

Jordan Enev

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Comments

Nina Scholz · Accepted Answer · 2018-08-24 10:08:37Z

2

You could filter, sort with delta, slice and map sid as result array.

var array = [{ sid: 387, rank: 2 }, { sid: 455, rank: 1 }, { sid: 364, rank: 4 }, { sid: 495, rank: 3 }],
    top3 = array
        .filter(({ rank }) => rank < 4)
        .sort((a, b) => a.rank - b.rank)
        .slice(0, 3)
        .map(({ sid }) => sid);

console.log(top3.join(", "));

edited Aug 24, 2018 at 10:08

answered Aug 24, 2018 at 10:07

Nina Scholz

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4 Comments

Federico klez Culloca Over a year ago

Apparently there's no need for the slice, as it seems ranks are unique and they're already taking the ones smaller than 4

Nina Scholz Over a year ago

@FedericoklezCulloca, actually right, but if you have more data, you need to slice for top 3.

Nikhil Aggarwal Over a year ago

@NinaScholz - As you are filtering on based on rank ( < 4), and rank being unique, there will never be case when there will be more than 3 records left after filter. I think this is what Federico meant.

Abhijit Karnik Over a year ago

@NinaScholz : Yes, this is useful since the implicit assumption that ranks are unique is not necessarily true. Thanks.

Nikhil Aggarwal · Accepted Answer · 2018-08-24 10:06:16Z

1

Use Array.map

var arr=[{sid:387,rank:2},{sid:455,rank:1},{sid:364,rank:4},{sid:495,rank:3}];

var result = arr.filter(({rank}) => rank < 4).sort((a,b) => a.rank > b.rank).map(({sid}) => sid);
console.log(result.join(", "));

answered Aug 24, 2018 at 10:06

Nikhil Aggarwal

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Comments

Mohammad Mirzaee · Accepted Answer · 2018-08-24 10:14:44Z

0

function pluck(objs, name, condition) {
        var sol = [];
        for (var i in objs) {
            if (objs[i].hasOwnProperty(name) && condition && condition(objs[i])) {
                // console.log(objs[i][name]);
                sol.push(objs[i][name]);
            }
        }
        return sol;
    }

    var arr = [{sid: 387, rank: 2}, {sid: 455, rank: 1}, {sid: 364, rank: 4}, {sid: 495, rank: 3}];
    var pluckedArray = pluck(arr, sid, function (item) {
        return item.rand < 4; // your custom condition
    })

answered Aug 24, 2018 at 10:14

Mohammad Mirzaee

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Comments

Koushik Chatterjee · Accepted Answer · 2018-08-24 12:06:14Z

0

How about iterate the array only once, and build the resultant array?

arr.reduce((r, {rank, sid})=>(rank < 4 && (r[rank -1] = sid), r), [])

var arr = [{ sid: 387, rank: 2 }, { sid: 455, rank: 1 }, { sid: 364, rank: 4 }, { sid: 495, rank: 3 }],
    top3 = arr.reduce((r, {rank, sid})=>(rank < 4 && (r[rank -1] = sid), r), []);
    
console.log(top3);

Note: When you need top 3 within highest rank 4 and also know ranks are unique, you can use this. If ranks are not unique you can use r.splice(rank-1, 0, sid) instead of r[rank-1] = sid and slice top 3 at the end.

edited Aug 24, 2018 at 12:06

answered Aug 24, 2018 at 11:15

Koushik Chatterjee

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