使用排序数组对对象键进行排序

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 eg

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

Answer 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) 

Answer 2

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:

1. 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.

2. 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.

3. 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)) 

Answer 3

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) 

Answer 4

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 .

Answer 5

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));