Sorting an array based on property value of another array of objects

Question

Let's say I have an array of objects:

var list = [
       { name: "A", distance: 1},
       { name: "B", distance: 2},
       { name: "C", distance: 3},
       { name: "D", distance: 4},
       { name: "E", distance: 5},
       { name: "F", distance: 6},
       { name: "G", distance: 7},
       { name: "H", distance: 8} 
    ];

if I have another array like this one :

var disturbed = ["G", "B", "C", "F"];

how can I sort disturbed array based on distance property from the list array like this:

["B", "C", "F", "G"];

Edit: I have tried this code with no success:

items = [ 
       { name: "A", distance: 1},
       { name: "B", distance: 2},
       { name: "C", distance: 3},
       { name: "D", distance: 4},
       { name: "E", distance: 5},
       { name: "F", distance: 6},
       { name: "G", distance: 7},
       { name: "H", distance: 8}
]

sorting = [ 1, 2,3,4,5,6,7,8,9,10 ];
result = []
for (let i = 0; i < items.length; i++){
sorting.forEach(function(key) {
    var found = false;
    items = items.filter(function(item) {

        if(!found && items[i].distance == key) {
            result.push(item);
            found = true;
            return false;
        } else 
            return true;
    })
})

result.forEach(function(item) {
    document.writeln(item[i]) 
})
}

How can I Sort an array based on the property value of another array of objects

What is the expected output?

ellipsis
– ellipsis

2019-08-04 12:08:39 +00:00
Commented Aug 4, 2019 at 12:08 — ellipsis
– ellipsis, Commented Aug 4, 2019 at 12:08
stackoverflow.com/questions/50755037/…

user1425689
– user1425689

2019-08-04 12:09:30 +00:00
Commented Aug 4, 2019 at 12:09 — user1425689
– user1425689, Commented Aug 4, 2019 at 12:09

Cuong Le Ngoc · Accepted Answer · 2019-08-04 12:18:44Z

4

You can use .reduce() to change the list array to object and then sort based on this object.

Demo:

var list = [
       { name: "A", distance: 1},
       { name: "B", distance: 2},
       { name: "C", distance: 3},
       { name: "D", distance: 4},
       { name: "E", distance: 5},
       { name: "F", distance: 6},
       { name: "G", distance: 7},
       { name: "H", distance: 8} 
    ];
    
var disturbed = ["G", "B", "C", "F"];
    
var sort = list.reduce((acc, cur) => {
  acc[cur.name] = cur.distance;
  return acc;
}, {});

disturbed.sort((a, b) => sort[a] - sort[b]);

console.log(disturbed)

edited Aug 4, 2019 at 12:18

answered Aug 4, 2019 at 12:11

Cuong Le Ngoc

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Comments

Morphyish · Accepted Answer · 2019-08-04 12:11:30Z

3

You can use the sort method to do that

var list = [
       { name: "A", distance: 1},
       { name: "B", distance: 2},
       { name: "C", distance: 3},
       { name: "D", distance: 4},
       { name: "E", distance: 5},
       { name: "F", distance: 6},
       { name: "G", distance: 7},
       { name: "H", distance: 8} 
    ];

var disturbed = ["G", "B", "C", "F"];

disturbed.sort((a, b) => {
    var itemA = list.find(item => item.name === a);
    var itemB = list.find(item => item.name === b);
    return itemA.distance - itemB.distance;
});

disturbed.forEach(function(item) {
    document.writeln(item[i]) 
})

answered Aug 4, 2019 at 12:11

Morphyish

4,07210 silver badges28 bronze badges

Comments

Nick Parsons · Accepted Answer · 2019-12-17 03:15:02Z

3

You can use .find() to find the object with the specified name property that matches your elements in distributed. Once you have got this you can then get the distance property and calculate the difference to sort accordingly:

const list = [
  { name: "A", distance: 1},
  { name: "B", distance: 2},
  { name: "C", distance: 3},
  { name: "D", distance: 4},
  { name: "E", distance: 5},
  { name: "F", distance: 6},
  { name: "G", distance: 7},
  { name: "H", distance: 8} 
];

const disturbed = ["G", "B", "C", "F"];

const res = disturbed.sort((a, b) => {
  const {distance: d_a} = list.find(({name}) => name === a);
  const {distance: d_b} = list.find(({name}) => name === b);
  return d_a - d_b;
});

console.log(res);

A more efficient approach would be to create a new Map using .map() and then use .sort() on the keys form the map:

const list = [
  { name: "A", distance: 1},
  { name: "B", distance: 2},
  { name: "C", distance: 3},
  { name: "D", distance: 4},
  { name: "E", distance: 5},
  { name: "F", distance: 6},
  { name: "G", distance: 7},
  { name: "H", distance: 8} 
];
const disturbed = ["G", "B", "C", "F"];

const lut = new Map(list.map(({name, distance}) => [name, distance]));
const res = disturbed.sort((a, b) => lut.get(a) - lut.get(b));
console.log(res);

edited Dec 17, 2019 at 3:15

answered Aug 4, 2019 at 12:11

Nick Parsons

51.7k6 gold badges61 silver badges80 bronze badges

2 Comments

Jonas Wilms Over a year ago

.find inside a sort is very inefficient though, something like O(log n² * n²)

Nick Parsons Over a year ago

@JonasWilms yeah, I agree it's not very efficient. I was going to suggest a better approach but Cuong Le Ngoc provided that

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