Sort an array in the same order of another array and if no match, add to the last

Question

I found this answer to sort an array based on another array. However, In this answer, when the array is not matched, it's added to first. Instead, I want the unmatched to add to the last. For that what change I need to make?

https://stackoverflow.com/a/28377564/890082

var reference_array = ["bob", "dan", "steven", "corbin"];
var array = ["ryan", "corbin",  "steven", "dan", "bob"];
array.sort(function(a, b) {
  return reference_array.indexOf(a) - reference_array.indexOf(b);
});
console.log(array);

Current output

["ryan", "bob", "dan", "steven", "corbin"]
  ^^^^

I want the output to be

["bob", "dan", "steven", "corbin", "ryan"]
                                    ^^^^

How can I do that?

Please read all answers on a question, not just the accepted one. This answer, for instance, shows how to sort in either direction. — Heretic Monkey
– Heretic Monkey, Commented Jun 4, 2021 at 12:23
Do you understand the logic inside the sort, do you know how the indexOf part works? — luk2302
– luk2302, Commented Jun 4, 2021 at 12:24

georg · Accepted Answer · 2021-06-04 12:32:03Z

3

This should do the trick:

var reference_array = ["bob", "dan", "steven", "corbin"];
var array = ["ryan", "corbin",  "steven", "xyz", "dan", "bob", "abc"];

array.sort((a, b) => 
  (reference_array.indexOf(a) >>> 0) -
  (reference_array.indexOf(b) >>> 0));


console.log(array);

See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Unsigned_right_shift . Basically, it leaves positive numbers as is and converts -1 to some big value.

You can also do that without any sorting at all:

var reference_array = ["bob", "dan", "steven", "corbin"];

var array = ["ryan", "corbin",  "steven", "xyz", "dan", "bob", "abc"];


var result = [
  ...reference_array.filter(x => array.includes(x)),
  ...array.filter(x => !reference_array.includes(x))
]
  
console.log(result)

edited Jun 4, 2021 at 12:32

answered Jun 4, 2021 at 12:29

georg

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3 Comments

Surjith S M Over a year ago

Thank you. it did the trick. Can also explain about >>>? Bit confused.

georg Over a year ago

@SurjithSM: yep, just added the link

Surjith S M Over a year ago

Perfect. Thank you.

James · Accepted Answer · 2021-06-04 12:32:29Z

1

In cases where a/b is not found, you want to hard-code the return value of the sort function to force the element to the end of the array.

var reference_array = ["bob", "dan", "steven", "corbin"];
var array = ["ryan", "corbin",  "steven", "dan", "bob"];
array.sort(function(a, b) {
  const ai = reference_array.indexOf(a);
  const bi = reference_array.indexOf(b);
  if (bi === -1) return -1;
  if (ai === -1) return 1;
  return ai - bi;
});
console.log(array);

answered Jun 4, 2021 at 12:32

James

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Comments

Reinis · Accepted Answer · 2021-06-04 12:33:58Z

1

Please check the question

var reference_array = ["bob", "dan", "steven", "corbin", "test"];
var array = ["ryan", "abc", "corbin", "steven", "dan", "bob"];
array.sort(function(a, b) {
  const a_idx = reference_array.indexOf(a)
  const b_idx = reference_array.indexOf(b);
  if (a_idx >= 0 && b_idx >= 0) return a_idx - b_idx; // normal 
  else if (a_idx < 0 && b_idx >= 0) return 1; // a doesn't not exist reference array and b exists
  else if (a_idx >= 0 && b_idx < 0) return -1; // b doesn't exist and a exists
  else return a.localeCompare(b) // both a, b don't exsit  and compare the string 
});
console.log(array);

answered Jun 4, 2021 at 12:33

Reinis

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