Sorting array of objects matching ids on a different set

Question

I have an array of objects

array = [
    {id: 5, name: "Helen", age: 20}, 
    {id: 15, name: "Lucy", age: 30}, 
    {id:7, name: "Carlos", age: 1}
]

Then I have a similar array sorted differently

arraySorted = [        
    {id: 15, name: "Lucy", age: 2}, 
    {id: 5, name: "Lara", age: 11}, 
    {id:7, name: "Carlos", age: 10}
]

The ids of the objects on both arrays will always match, the rest of the properties may or may not.

What I need is sorting the array in the same id order as arraySorted.

(it can be also done on plain JavaScript, lodash is not necessary but maybe it will be useful)

Arman Charan · Accepted Answer · 2018-03-28 07:26:04Z

3

See Map and Array.prototype.map() for more info.

// Raw.
const raw = [
    {id: 5, name: "Helen", age: 20}, 
    {id: 15, name: "Lucy", age: 30}, 
    {id:7, name: "Carlos", age: 1}
]

// Sorted.
const sorted = [        
    {id: 15, name: "Lucy", age: 2}, 
    {id: 5, name: "Lara", age: 11}, 
    {id:7, name: "Carlos", age: 10}
]

// Match.
const match = (raw, sorted) => (m => sorted.map(s => m.get(s.id)))(new Map(raw.map(r => [r.id, r])))

// Output.
const output = match(raw, sorted)

// Proof.
console.log(output)

edited Mar 28, 2018 at 7:26

answered Mar 28, 2018 at 7:20

Arman Charan

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4 Comments

trincot Over a year ago

This is a superior solution compared to those using find (+1).

user3808307 Over a year ago

@trincot why is it superior

Arman Charan Over a year ago

Array.prototype.find() executes a blind search every time it is called. Map.get() retrieves values using set keys. These attributes heavily influence time complexity.

trincot Over a year ago

Other solutions use a main loop with forEach or map like here, but then they have an implicit nested loop inside it by calling find which will start searching the array from left to right, meaning find runs with a O(n) time complexity. The overall time complexity thus becomes O(n²). Here, on the other hand there are three, not-nested loops (one loop runs after the other completes): two Array#map calls and an implicit iteration with the new Map call. m.get runs in constant time, so the overall time complexity is O(n+n+n) = O(n).

gurvinder372 · Accepted Answer · 2018-03-28 07:20:11Z

1

Rather than sort, use map, find and Object.assign instead

arraySorted.map( s => Object.assign( s, array.find( t => t.id == s.id ) ) );

Demo

var array = [{
    id: 5,
    name: "Helen",
    age: 20
  },
  {
    id: 15,
    name: "Lucy",
    age: 30
  },
  {
    id: 7,
    name: "Carlos",
    age: 1
  }
];
var arraySorted = [{
    id: 15,
    name: "Lucy",
    age: 2
  },
  {
    id: 5,
    name: "Lara",
    age: 11
  },
  {
    id: 7,
    name: "Carlos",
    age: 10
  }
];
array = arraySorted.map(s => Object.assign(s, array.find(t => t.id == s.id)));

console.log(array);

answered Mar 28, 2018 at 7:20

gurvinder372

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Comments

Jaye Renzo Montejo · Accepted Answer · 2018-03-28 07:26:29Z

1

The new array can be obtained via Array.prototype.map:

const newArray = arraySorted.map(sortedItem => 
  array.find(item => item.id === sortedItem.id)
)

console.log(newArray)

answered Mar 28, 2018 at 7:26

Jaye Renzo Montejo

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Comments

trincot · Accepted Answer · 2018-03-28 07:26:58Z

1

You can use a Map, and avoid using nested iteration methods (like find, which make the solution have a worse time complexity), assuming that indeed you have all the same id values in both arrays:

const array = [{id: 5, name: "Helen", age: 20}, {id: 15, name: "Lucy", age: 30}, {id:7, name: "Carlos", age: 1}],
    arraySorted = [{id: 15, name: "Lucy", age: 2},{id: 5, name: "Lara", age: 11},{id:7, name: "Carlos", age: 10}]

const result = arraySorted.map((map => row => array[map.get(row.id)])
                               (new Map(array.map((row, i) => [row.id, i]))));

console.log(result);

answered Mar 28, 2018 at 7:26

trincot

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1 Comment

user3808307 Over a year ago

Yes, I have the same ids all the time.

Deepshikha Chaudhary · Accepted Answer · 2018-03-28 07:35:14Z

1

Try this, hope this helps

var array=[{id:5,name:"Helen",age:20},{id:15,name:"Lucy",age:30},{id:7,name:"Carlos",age:1}];

var arraySorted=[{id:15,name:"Lucy",age:2},{id:5,name:"Lara",age:11},{id:7,name:"Carlos",age:10}];

array = arraySorted.map(s => array.find(t => t.id == s.id));

console.log(array);

answered Mar 28, 2018 at 7:35

Deepshikha Chaudhary

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Comments

Mohammad Usman · Accepted Answer · 2018-03-28 07:35:52Z

1

You can use Map to sort your array:

let a1 = [{id: 5, name: "Helen", age: 20}, {id: 15, name: "Lucy", age: 30}, {id:7, name: "Carlos", age: 1}],
    a2 = [{id: 15, name: "Lucy", age: 2}, {id: 5, name: "Lara", age: 11}, {id:7, name: "Carlos", age: 10}];

let map = ((m) => (
             a2.forEach(({id}) => m.set(id, a1.find(o => o.id === id))), m
          ))(new Map());

let result = [...map.values()];

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Mar 28, 2018 at 7:35

answered Mar 28, 2018 at 7:22

Mohammad Usman

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