How to merge multiple array of object by ID in javascript?

Question

I want to merge 4 array of object into one array

For example: 4 arrays like

var arr1 =[
  { memberID : "81fs", RatingCW:4.5},
  { memberID : "80fs", RatingCW:4},
  { memberID : "82fs", RatingCW:5 },
  { memberID : "83fs", RatingCW:3},
  { memberID : "84fs", RatingCW:4.7}
];
var arr2 =[
  { memberID : "80fs", ratingWW: 4},
  { memberID : "81fs", ratingWW: 4.5},
  { memberID : "83fs", ratingWW: 3},
  { memberID : "82fs", ratingWW: 5},
  { memberID : "84fs", ratingWW: 3.5}
];

var arr3 =  [
  { memberID : "80fs", incoCW:4},
  { memberID : "81fs", incoCW:4.5},
  { memberID : "82fs", incoCW:5},
  { memberID : "83fs", incoCW:3},
  { memberID : "84fs", incoCW:4.5}
  ];
var arr4 =  [
  { memberID : "80fs", incoWW:3},
  { memberID : "81fs", incoWW:2.5 },
  { memberID : "82fs", incoWW:5 },
  { memberID : "83fs", incoWW:3 },
  { memberID : "84fs", incoWW:6.5 }
];

and expected array like:

var finalArr = [
    { memberID : "80fs", RatingCW:4,ratingWW: 4, incoCW:4, incoWW:3},
    { memberID : "81fs", RatingCW:4.5,ratingWW: 4.5, incoCW:4.5, incoWW:2.5 },
    { memberID : "82fs", RatingCW:5,ratingWW: 5, incoCW:5, incoWW:5 },
    { memberID : "83fs", RatingCW:3,ratingWW: 3, incoCW:3, incoWW:3 },
    { memberID : "84fs", RatingCW:4.7,ratingWW: 3.5, incoCW:4.5, incoWW:6.5 }
  ];

What is the best way to merge using lodash or normal javascript?

order can be different @charlietfl

Shaishab Roy
– Shaishab Roy

2016-07-21 13:28:34 +00:00
Commented Jul 21, 2016 at 13:28 — Shaishab Roy
– Shaishab Roy, Commented Jul 21, 2016 at 13:28

castletheperson · Accepted Answer · 2016-07-21 19:25:01Z

18

With lodash, a lot more readable I think.

var arr1 = [{"memberID":"81fs","RatingCW":4.5},{"memberID":"80fs","RatingCW":4},{"memberID":"82fs","RatingCW":5},{"memberID":"83fs","RatingCW":3},{"memberID":"84fs","RatingCW":4.7}],
    arr2 = [{"memberID":"80fs","ratingWW":4},{"memberID":"81fs","ratingWW":4.5},{"memberID":"83fs","ratingWW":3},{"memberID":"82fs","ratingWW":5},{"memberID":"84fs","ratingWW":3.5}],
    arr3 = [{"memberID":"80fs","incoCW":4},{"memberID":"81fs","incoCW":4.5},{"memberID":"82fs","incoCW":5},{"memberID":"83fs","incoCW":3},{"memberID":"84fs","incoCW":4.5}],
    arr4 = [{"memberID":"80fs","incoWW":3},{"memberID":"81fs","incoWW":2.5},{"memberID":"82fs","incoWW":5},{"memberID":"83fs","incoWW":3},{"memberID":"84fs","incoWW":6.5}];

var merged = _(arr1)
  .concat(arr2, arr3, arr4)
  .groupBy("memberID")
  .map(_.spread(_.merge))
  .value();

console.log(merged);

<script src="https://cdn.jsdelivr.net/lodash/4.13.1/lodash.min.js"></script>

Here is the codepen: http://codepen.io/kuhnroyal/pen/Wxzdmw

edited Jul 21, 2016 at 19:25

castletheperson

33.6k11 gold badges74 silver badges111 bronze badges

answered Jul 21, 2016 at 14:10

kuhnroyal

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5 Comments

castletheperson Over a year ago

Just do .concat(arr2, arr3, arr4). Also, .groupBy('memberID') is shorthand for that grouping function.

ryeballar Over a year ago

You can use _.spread(). e.g. map(_.spread(_.merge)).

Jason Lunsford Over a year ago

Had a very similar issue and would have double looped if not for this solution. Beautiful - thank you for sharing.

KayakinKoder Over a year ago

Does anyone happen to know if this will maintain a specific order?

janhink Over a year ago

Could you please provide an equivalent using _.flow?

sbk201 · Accepted Answer · 2023-02-07 15:21:00Z

Vanilla JS way,it's a tough one,took hours to get it.

var arr1 = [{"memberID":"81fs","RatingCW":4.5},{"memberID":"80fs","RatingCW":4},{"memberID":"82fs","RatingCW":5},{"memberID":"83fs","RatingCW":3},{"memberID":"84fs","RatingCW":4.7}],
    arr2 = [{"memberID":"80fs","ratingWW":4},{"memberID":"81fs","ratingWW":4.5},{"memberID":"83fs","ratingWW":3},{"memberID":"82fs","ratingWW":5},{"memberID":"84fs","ratingWW":3.5}],
    arr3 = [{"memberID":"80fs","incoCW":4},{"memberID":"81fs","incoCW":4.5},{"memberID":"82fs","incoCW":5},{"memberID":"83fs","incoCW":3},{"memberID":"84fs","incoCW":4.5}],
    arr4 = [{"memberID":"80fs","incoWW":3},{"memberID":"81fs","incoWW":2.5},{"memberID":"82fs","incoWW":5},{"memberID":"83fs","incoWW":3},{"memberID":"84fs","incoWW":6.5}];

const arrs=[].concat(arr1,arr2,arr3,arr4);
const noDuplicate=arr=>[...new Set(arr)]
const allIds=arrs.map(ele=>ele.memberID);
const ids=noDuplicate(allIds);

const result=ids.map(id=>
    arrs.reduce((self,item)=>{
        return item.memberID===id?
        {...self,...item} : self
    },{})
)
console.log(result);

Leon Mak · Accepted Answer · 2016-07-21 14:40:33Z

Here's some steps using lodash:

Put all objects into an array (_.flatten then _.groupby)
Flatten 2d array into 1d array of 'user' obj (_.map then _.merge and apply to pass in array as arg)

var arr1 =[
  { memberID : "81fs", RatingCW:4.5},
  { memberID : "80fs", RatingCW:4},
  { memberID : "82fs", RatingCW:5 },
  { memberID : "83fs", RatingCW:3},
  { memberID : "84fs", RatingCW:4.7}
];
var arr2 =[
  { memberID : "80fs", ratingWW: 4},
  { memberID : "81fs", ratingWW: 4.5},
  { memberID : "83fs", ratingWW: 3},
  { memberID : "82fs", ratingWW: 5},
  { memberID : "84fs", ratingWW: 3.5}
];

var arr3 =  [
  { memberID : "80fs", incoCW:4},
  { memberID : "81fs", incoCW:4.5},
  { memberID : "82fs", incoCW:5},
  { memberID : "83fs", incoCW:3},
  { memberID : "84fs", incoCW:4.5}
  ];
var arr4 =  [
  { memberID : "80fs", incoWW:3},
  { memberID : "81fs", incoWW:2.5 },
  { memberID : "82fs", incoWW:5 },
  { memberID : "83fs", incoWW:3 },
  { memberID : "84fs", incoWW:6.5 }
];

var a  = _.groupBy(_.flatten([arr1,arr2,arr3,arr4]), 'memberID');
var b = _.map(a, function(val){ return _.merge.apply(_,val) });    
console.log(b);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/1.2.1/lodash.min.js"></script>

ifelse.codes · Accepted Answer · 2020-01-30 22:07:57Z

2

arr1.map(a =>
    Object.assign({},a,
        arr2.find(b => b.memberID === a.memberID),
        arr3.find(c => c.memberID === a.memberID),
        arr4.find(d => d.memberID === a.memberID)
    )
)

Here is the simplest code ... this is basically a oneliner solution. The performance may not be the best but works well for a small data sets.

answered Jan 30, 2020 at 22:07

ifelse.codes

2,38925 silver badges21 bronze badges

1 Comment

some_groceries Over a year ago

cool stuff, can you give more info about how this is bad at performance?

Nina Scholz · Accepted Answer · 2016-07-21 13:59:22Z

You could use a hash table for the right reference to the result array.

I works with unsorted arrays as well, because of the reference memberID in the object.

 var arr1 = [{ memberID: "80fs", RatingCW: 4 }, { memberID: "81fs", RatingCW: 4.5 }, { memberID: "82fs", RatingCW: 5 }, { memberID: "83fs", RatingCW: 3 }, { memberID: "84fs", RatingCW: 4.7 }],
    arr2 = [{ memberID: "80fs", ratingWW: 4 }, { memberID: "81fs", ratingWW: 4.5 }, { memberID: "82fs", ratingWW: 5 }, { memberID: "83fs", ratingWW: 3 }, { memberID: "84fs", ratingWW: 3.5 }],
    arr3 = [{ memberID: "80fs", incoCW: 4 }, { memberID: "81fs", incoCW: 4.5 }, { memberID: "82fs", incoCW: 5 }, { memberID: "83fs", incoCW: 3 }, { memberID: "84fs", incoCW: 4.5 }],
    arr4 = [{ memberID: "80fs", incoWW: 3 }, { memberID: "81fs", incoWW: 2.5 }, { memberID: "82fs", incoWW: 5 }, { memberID: "83fs", incoWW: 3 }, { memberID: "84fs", incoWW: 6.5 }],
    merged = [];

[arr1, arr2, arr3, arr4].forEach(function (a) {
    a.forEach(function (b) {
        if (!this[b.memberID]) {
            this[b.memberID] = {};
            merged.push(this[b.memberID]);
        }
        Object.keys(b).forEach(function (k) {
            this[b.memberID][k] = b[k];
        }, this);
    }, this);
}, Object.create(null));

console.log(merged);

ES6 for unsorted data.

 var arr1 = [{ memberID: "80fs", RatingCW: 4 }, { memberID: "81fs", RatingCW: 4.5 }, { memberID: "82fs", RatingCW: 5 }, { memberID: "83fs", RatingCW: 3 }, { memberID: "84fs", RatingCW: 4.7 }],
    arr2 = [{ memberID: "80fs", ratingWW: 4 }, { memberID: "81fs", ratingWW: 4.5 }, { memberID: "82fs", ratingWW: 5 }, { memberID: "83fs", ratingWW: 3 }, { memberID: "84fs", ratingWW: 3.5 }],
    arr3 = [{ memberID: "80fs", incoCW: 4 }, { memberID: "81fs", incoCW: 4.5 }, { memberID: "82fs", incoCW: 5 }, { memberID: "83fs", incoCW: 3 }, { memberID: "84fs", incoCW: 4.5 }],
    arr4 = [{ memberID: "80fs", incoWW: 3 }, { memberID: "81fs", incoWW: 2.5 }, { memberID: "82fs", incoWW: 5 }, { memberID: "83fs", incoWW: 3 }, { memberID: "84fs", incoWW: 6.5 }],
    merged = [];

[arr1, arr2, arr3, arr4].forEach((hash => a => a.forEach(b => {
    if (!hash[b.memberID]) {
        hash[b.memberID] = {};
        merged.push(hash[b.memberID]);
    }
    Object.assign(hash[b.memberID], b);
}))(Object.create(null)));


console.log(merged);

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