Sort object keys with a sorted array

Question

I have the below code that sorts object in an alphabetic order, I want to get away from having an alphabetic order and sort on my own chosen order.

var obj = {
  c: "see",
  a: "ay",
  b: "bee",
  d: "dee"
};

console.log("Object order:");
var key, keys = [];
for (key in obj) {
  keys.push(key);
}
console.log(keys.map(function(key) { return key + ": " + obj[key];}).join(", "));

console.log("Sorted order:");
keys = Object.keys(obj).sort()
console.log(keys.map(function(key) { return key + ": " + obj[key];}).join(", "));

The above will output the object order of a, b, c, d - but lets say I want an array to have my own order e.g.

var sortedArray = ['b','c','d','a'];

So how do I know implement this sortedArray into my .sort?

Expected outcome:

var newObj = {
  b: "bee",
  c: "cee",
  d: "dee",
  a: "ay"
};

Thanks

So you build a custom sort function.... developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… — epascarello
– epascarello, Commented Sep 6, 2018 at 12:14

Calvin Nunes · Accepted Answer · 2018-09-06 13:19:05Z

1

It's a very weird pattern to a sort, but as you want, I made a function called CustomOrderBy which will do a sort based on an array of keys.

Something like this:

You have an array with the keys you want ordered from first to last, like you already have (sortedArray). Then you pass this array to the function.
The function then loop through the obj keys and get the first letter of it to check if it matches with the arra of keys, getting its index. Based on the index, it sorts.
After the sort, then it creates a new obj and set the key and the value in the order it created in the sort function used before

Please, see the code to undertand it better.

var sortedArray = ['b', 'c', 'd', 'a'];

var obj = {
  c: "see",
  a: "ay",
  b: "bee",
  d: "dee"
};

function CustomOrderBy(keyOrder){
  let keysSorted = Object.keys(obj).sort(function(keyA, keyB) {
    let idxA = keyOrder.indexOf(keyA[0]);  
    let idxB = keyOrder.indexOf(keyB[0]);
    if (idxA > idxB){
      return 1
    }else if(idxA < idxB){
      return -1
    }else{
      return 0;
    }  
  });

  let sorted = {};
  for (var key of keysSorted){
    if (obj[key] != null){
      sorted[key] = obj[key];
    }
  }
  return sorted;
}
console.log(CustomOrderBy(sortedArray))

answered Sep 6, 2018 at 13:19

Calvin Nunes

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1 Comment

Tom Rudge Over a year ago

Thank you so much, this was perfect. I had to adopt it slightly as my key changed to 2 characters. Awesome.

Calvin Nunes · Accepted Answer · 2018-09-06 12:28:00Z

1

You can pass your custom sort function into the sort() API. See the example below:

var items = [
  { name: 'Edward', value: 21 },
  { name: 'Sharpe', value: 37 },
  { name: 'And', value: 45 },
  { name: 'The', value: -12 },
  { name: 'Magnetic', value: 13 },
  { name: 'Zeros', value: 37 }
];
    
// sort by value
items.sort(function (a, b) {
  return a.value - b.value;
});

// sort by name
items.sort(function(a, b) {
  var nameA = a.name.toUpperCase(); // ignore upper and lowercase
  var nameB = b.name.toUpperCase(); // ignore upper and lowercase
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names must be equal
  return 0;
});

console.log(items)

edited Sep 6, 2018 at 12:28

Calvin Nunes

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answered Sep 6, 2018 at 12:18

vizsatiz

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Comments

charlietfl · Accepted Answer · 2018-09-06 12:27:19Z

0

Use custom sorter that uses indexOf()

var sortedArray = ['b', 'c', 'd', 'a'];

var obj = {
  c: "see",
  a: "ay",
  b: "bee",
  d: "dee"
};


var keys = Object.keys(obj).sort(function(a, b) {
  // TODO - cases where not in sortedArray
  return sortedArray.indexOf(a) - sortedArray.indexOf(b)
});


console.log(keys)

answered Sep 6, 2018 at 12:27

charlietfl

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8 Comments

Calvin Nunes Over a year ago

Isn't your code doing just the same as console.log(sortedArray) ?

charlietfl Over a year ago

@CalvinNunes it defintiely is. Not really sure what OP is trying to do to be honest

Calvin Nunes Over a year ago

I think that he wants to sort some array with a different order than the default (lexographical alphabetic order) but he not specified the criteria, so, it's really hard to know what he really wants

Tom Rudge Over a year ago

Yer this is great - it orders the keys, but it does not maintain the values in the keys array. The output is only outputing the keys in order. cee, ay, bee and dee are missing

charlietfl Over a year ago

Javascript objects don't have order. Whole thing seems like an XY problem.

|

Akrion · Accepted Answer · 2018-10-07 05:40:09Z

0

Pardon me asking but if the relation between the sort order and the object is one to one why don't we do something like this (starting from the sort order array):

var myObject = {c: "see", a: "ay", b: "bee", d: "dee"};
var sortOrder = ['b','c','d','a'];

const mySort = (obj, order) => order.reduce((r,c) => (r[c] = obj[c], r), {})

console.log(mySort(myObject, sortOrder))

It would seem that it would be quite more concise and simpler since it uses only one reduce.

answered Oct 7, 2018 at 5:40

Akrion

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Comments

Abdul Samad · Accepted Answer · 2018-09-06 12:23:35Z

-1

you can create a function like i did below. and pass your object it will return sorted object to you.

    var obj = {
        c: "see",
        a: "ay",
        b: "bee",
        d: "dee"
    };


    function sortObject(obj)
    {
        var keys = [];
        var outputObj = {};

        for (key in obj)
        {
            keys.push(key);
        }

        keys.sort();

        for(var i=0; i<keys.length; i++){
            outputObj[keys[i]] = obj[keys[i]];
        }

        return outputObj;
    }
    
    console.log(sortObject(obj));

answered Sep 6, 2018 at 12:23

Abdul Samad

71 bronze badge

1 Comment

Calvin Nunes Over a year ago

Your function does exactly what the asker don't want, that is a sort() by alphabetic order, he wants his own sort order, not the default.

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