Sorting Array of Objects based on Index in other array

Question

I have the following array of unique IDs:

idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"]

And I have another array of objects:

stories = [{"title": Story2, id = "56f4cf96dd2ca7275feaf7b7"},
{"title": Story4, id = "56f4cf96dd2ca7275feaf7ac"},
{"title": Story1, id = "56f4cf96dd2ca7275feaf802"},
{"title": Story3, id = "56f4cf96dd2ca7275feaf805"}]

How can I sort the second array based on the index of the first array? Preferably using lodash, as the arrays can grow a bit larger.

So far, I have the following to get the indexes from the first array:

var sortArray = _.toPairs(idArray)

[ [ '0', 56f4cf96dd2ca7275feaf802 ],
[ '1', 56f4cf96dd2ca7275feaf7b7 ],
[ '2', 56f4cf96dd2ca7275feaf805 ],
[ '3', 56f4cf96dd2ca7275feaf7ac ] ]

Trying different combinations of _.map() and _.sortBy() I can't seem to properly get the result I want which is:

desiredResult = [{"title": Story1, id = "56f4cf96dd2ca7275feaf802"},
          {"title": Story2, id = "56f4cf96dd2ca7275feaf7b7"},
          {"title": Story3, id = "56f4cf96dd2ca7275feaf805"},
          {"title": Story4, id = "56f4cf96dd2ca7275feaf7ac"}]

Redu · Accepted Answer · 2016-05-31 04:01:13Z

4

I believe the sort solution is very ineffective especially since you expect the arrays to grow bigger later. Sort is "at best" an O(2n) operation while you have two indexOf operations per cycle another O(2n^2). I propose the following which will outperform the sort method in large arrays.

var stories = [{"title": 'Story2', id : "56f4cf96dd2ca7275feaf7b7"},
{"title": 'Story4', id : "56f4cf96dd2ca7275feaf7ac"},
{"title": 'Story1', id : "56f4cf96dd2ca7275feaf802"},
{"title": 'Story3', id : "56f4cf96dd2ca7275feaf805"}],

    idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"],

ordered = idArray.reduce((p,c) => p.concat(stories.find(f => f.id == c)) ,[]);

console.log(ordered);

Only O(n^2)

answered May 31, 2016 at 4:01

Redu

26.3k6 gold badges61 silver badges84 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Jose Hermosilla Rodrigo · Accepted Answer · 2016-05-30 23:04:28Z

3

This is possible to do without any library using Array.sort()

var stories = [{"title": 'Story2', id : "56f4cf96dd2ca7275feaf7b7"},
{"title": 'Story4', id : "56f4cf96dd2ca7275feaf7ac"},
{"title": 'Story1', id : "56f4cf96dd2ca7275feaf802"},
{"title": 'Story3', id : "56f4cf96dd2ca7275feaf805"}];

var idArray = ["56f4cf96dd2ca7275feaf802",
"56f4cf96dd2ca7275feaf7b7",
"56f4cf96dd2ca7275feaf805",
"56f4cf96dd2ca7275feaf7ac"];

var ordered = stories.sort(function(a, b){
	return idArray.indexOf(a.id) - idArray.indexOf(b.id);
});

ordered.forEach( element =>{ console.log(element) });

edited May 30, 2016 at 23:04

answered May 30, 2016 at 22:58

Jose Hermosilla Rodrigo

3,7136 gold badges25 silver badges38 bronze badges

3 Comments

Pevara Over a year ago

Exactly my thought. You beat me to it... Here, have a fiddle: jsfiddle.net/jx3we8q7 ;)

user1202888 Over a year ago

Thank you, great answer. Is it ok to do 'return idArray.indexOf(b.id) - dArray.indexOf(a.id)' to reverse the order? Or should I use reverse() on the resulting array?

Jose Hermosilla Rodrigo Over a year ago

I think there's nothing bad to do it.

Alastair Brown · Accepted Answer · 2016-05-30 22:56:28Z

1

Try this

var idsToIndexes = {};

for (var i = 0; i < idArray.length; i++)
    idsToIndexes[idArray[i]] = i;

stories.sort(function(a, b){return idsToIndexes[a.id] - idsToIndexes[b.id];});

answered May 30, 2016 at 22:56

Alastair Brown

1,6169 silver badges12 bronze badges

Collectives™ on Stack Overflow

Sorting Array of Objects based on Index in other array

3 Answers 3

Comments

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related