Sort array of objects based on another array and move items not in that array to the end

Question

var arrayObject = [
{key: 'g'},
{key: 'a'},
{key: 'b'},
{key: 'c'},
{key: 'd'},
{key: 'e'}
];

var array = ['a', 'c', 'd', 'e'];

Desired Output:

[
{key: 'a'},
{key: 'c'},
{key: 'd'},
{key: 'e'},
{key: 'g'}
{key: 'b'}
]

Say I have an array of object to be sorted based on an array. I would like to get the above output. I tried the following.

arrayObject.sort(function(a, b){
  return array.indexOf(a.key) - array.indexOf(b.key);
});

I got the following output:

[
{key: 'g'},
{key: 'b'},
{key: 'a'},
{key: 'c'},
{key: 'd'}
{key: 'e'}
]

danh · Accepted Answer · 2020-05-13 03:46:32Z

6

I'd start by building a complete array to perform the sort, including missing elements from the keys, in their original order...

var arrayObject = [
{key: 'g'},
{key: 'a'},
{key: 'b'},
{key: 'c'},
{key: 'd'},
{key: 'e'}
];

var array = ['a', 'c', 'd', 'e'];

arrayObject.forEach(o => {
  if (!array.includes(o.key)) array.push(o.key)
})

// now your code works fine
 let result = arrayObject.sort((a, b) => {
    return array.indexOf(a.key) - array.indexOf(b.key);
 });

console.log(result)

answered May 13, 2020 at 3:46

danh

62.7k10 gold badges100 silver badges139 bronze badges

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

1 Comment

Jhecht Over a year ago

It might be worth it to use some variety of "map" to make the lookup of the key indexes faster. The example given to us is very simple, but if he should get back quite a few rows then the constant indexOf calls could get expensive.

Joundill · Accepted Answer · 2020-05-13 03:46:48Z

2

Just check in your sorting function whether a or b exist in the array, then return -1 for b and 1 for a if they don't exist in the array.

var arrayObject = [
{key: 'g'},
{key: 'a'},
{key: 'b'},
{key: 'c'},
{key: 'd'},
{key: 'e'}
];

var array = ['a', 'c', 'd', 'e'];

var newArr = arrayObject.sort(function(a, b){
  if(array.indexOf(b.key) === -1) return -1;
  if(array.indexOf(a.key) === -1) return 1;
  return array.indexOf(a.key) - array.indexOf(b.key);
});

console.log(newArr);

answered May 13, 2020 at 3:46

Joundill

7,82113 gold badges40 silver badges54 bronze badges

6 Comments

danh Over a year ago

This is close, but it doesn't preserve the order of the keys missing keys.

Joundill Over a year ago

@danh Yes it does. If both A and B don't exist in the array, then B will be put second. Admittedly, it does depend on which sort the browser implements.

Joundill Over a year ago

@danh Slightly better time complexity than your answer as well :P

danh Over a year ago

Please review the output of your code for "g" and "b". Time complexity of both solutions is O(n log n)

Igor Sukharev Over a year ago

incorrect order of "b","g", swap you condition statements and it will be the correct "g","b"

|

Collectives™ on Stack Overflow

Sort array of objects based on another array and move items not in that array to the end

2 Answers 2

1 Comment

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related