Sort array by order according to another array

Question

I have an object that is being returned from a database like this: [{id:1},{id:2},{id:3}]. I have another array which specified the order the first array should be

Answer 1

If you want linear time, first build a hashtable from the first array and then pick items in order by looping the second one:

data = [{id:5},{id:2},{id:9}]
order = [9,5,2]

hash = {}
data.forEach(function(x) { hash[x.id] = x })

sorted = order.map(function(x) { return hash[x] })

document.write(JSON.stringify(sorted))

Answer 2

Why not just create new array and push the value from second array in?? Correct me if i wrong

array1 = [];
array2 = [2,3,1];

for ( var i = 0; i < array2 .length; i++ )
{
    array1.push({
         id : array2[i]
     })
}

Answer 3

    function sortArrayByOrderArray(arr, orderArray) {
        return arr.sort(function(e1, e2) {
            return orderArray.indexOf(e1.id) - orderArray.indexOf(e2.id);
        });
    }

    console.log(sortArrayByOrderArray([{id:1},{id:2},{id:3}], [2,3,1]));

Answer 4

To my understanding, sorting is not necessary; at least in your example, the desired resulting array can be generated in linear time as follows.

var Result;
for ( var i = 0; i < Input.length; i++ )
{
    Result[i] = Input[Order[i]-1];
}

Here Result is the desired output, Input is your first array and Order the array containing the desired positions.

Answer 5

In your example, the objects are initially sorted by id, which makes the task pretty easy. But if this is not true in general, you can still sort the objects in linear time according to your array of id values.

The idea is to first make an index that maps each id value to its position, and then to insert each object in the desired position by looking up its id value in the index. This requires iterating over two arrays of length n, resulting in an overall runtime of O(n), or linear time. There is no asymptotically faster runtime because it takes linear time just to read the input array.

function objectsSortedBy(objects, keyName, sortedKeys) {
  var n = objects.length,
      index = new Array(n);
  for (var i = 0; i < n; ++i) {  // Get the position of each sorted key.
    index[sortedKeys[i]] = i;
  }
  var sorted = new Array(n);
  for (var i = 0; i < n; ++i) {  // Look up each object key in the index.
    sorted[index[objects[i][keyName]]] = objects[i];
  }
  return sorted;
}

var objects = [{id: 'Tweety', animal: 'bird'},
               {id: 'Mickey', animal: 'mouse'},
               {id: 'Sylvester', animal: 'cat'}],
    sortedIds = ['Tweety', 'Mickey', 'Sylvester'];

var sortedObjects = objectsSortedBy(objects, 'id', sortedIds);

// Check the result.
for (var i = 0; i < sortedObjects.length; ++i) {
  document.write('id: '+sortedObjects[i].id+', animal: '+sortedObjects[i].animal+'<br />');
}

Answer 6

var objArray = [{id:1},{id:2},{id:3}];
var sortOrder = [2,3,1];

var newObjArray = [];
for (i in sortOrder) {
    newObjArray.push(objArray[(sortOrder[i]) - 1])
};