Sort an array based on another array in javascript while keeping not sorted elements at the end

Question

I have an array that i want to sort based on another array. If both array is the same length, the code below works fine. However, if my array called order is missing an element, i still want to display that at the end of the new sorted array. Here is a quick example:

//Source arrays
var order = [{_id: '123', type: '2'},{_id: '123', type: '1'}];
var elements = [{_id: '124', type: '1', name: '(should be last, not sorted)'},{_id: '123', type: '1', name: 'second(should be second)'},{_id: '123', type: '2', name: 'third(should be first)'}];

var sorted = [];
for (var i = 0, len = order.length; i < len; i++) {
    var element = elements.filter(function (el) {
		return el.type == order[i]['type'] &&
               el._id == order[i]['_id'];
    });
    sorted.push(element[0]);
}

//Just for testing
var list = document.createElement('ul');
for (var i = 0, len = sorted.length; i < len; i++) {
      var item = document.createElement('li');
      item.appendChild(document.createTextNode(sorted[i]['name']));
      list.appendChild(item);
}
document.getElementById('list').appendChild(list);

<div id="list"></div>

As you can see, now i only have two items in my sorted array. How can i add the missing item(s) from elements at the end?

Rajesh Dixit · Accepted Answer · 2016-09-30 09:45:58Z

1

You can use array.findIndex to get index and then sort accordingly

var order = [{_id: '123', type: '2'},{_id: '123', type: '1'}];
var elements = [{_id: '124', type: '1', name: '(should be last, not sorted)'},{_id: '123', type: '1', name: 'second(should be second)'},{_id: '123', type: '2', name: 'third(should be first)'}];


elements.sort(function(a, b) {
  var MAX_VALUE = 9999999;
  var index_a = order.findIndex(function(el) {
    return el._id === a._id && el.type === a.type;
  });
  var index_b = order.findIndex(function(el) {
    return el._id === b._id && el.type === b.type;
  });
  
  index_a = index_a < 0? MAX_VALUE : index_a;
  index_b = index_b < 0? MAX_VALUE : index_b;

  return index_a > index_b ? 1 : index_a < index_b ? -1 : 0
});

console.log(elements)

answered Sep 30, 2016 at 9:45

Rajesh Dixit

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Nina Scholz · Accepted Answer · 2016-09-30 10:07:26Z

1

You could use Array#sort and an object for the sort order. If no order is given, then the element is moved to the end.

function sortWithOrder(array) {
    function getOrder(o) { return (orderObj[o._id] || {})[o.type] || Infinity; }

    var orderObj = Object.create(null);
    array.forEach(function (a, i) {
        orderObj[a._id] = orderObj[a._id] || {};
        orderObj[a._id][a.type] = i + 1;
    });
    return function (a, b) {
        return getOrder(a) - getOrder(b);
    };
}

var order = [{ _id: '123', type: '2' }, { _id: '123', type: '1' }],
    elements = [{ _id: '124', type: '1', name: '(should be last, not sorted)' }, { _id: '123', type: '1', name: 'second(should be second)' }, { _id: '123', type: '2', name: 'third(should be first)' }];

elements.sort(sortWithOrder(order));
console.log(elements);

.as-console-wrapper { max-height: 100% !important; top: 0; }

edited Sep 30, 2016 at 10:07

answered Sep 30, 2016 at 9:46

Nina Scholz

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4 Comments

Rajesh Dixit Over a year ago

Any specific reason to define getOrder a local function inside Sort?

Nina Scholz Over a year ago

@Rajesh, because it is only locally used.

Rajesh Dixit Over a year ago

But will that not define/override same function for every iteration. Also does this have any performance benefit? Sorry for so many queries.

Nina Scholz Over a year ago

@Rajesh, i changed the style. but the original function does not change for every call. it stays the same.

sabithpocker · Accepted Answer · 2016-09-30 10:33:26Z

var sorted = elements.sort((a, b) => 
               (order.findIndex(i => a._id === i._id && a.type === i.type) + 1 || order.length + 1) 
             - (order.findIndex(i => b._id === i._id && b.type === i.type) + 1 || order.length + 1)
             );

'use strict';

var order = [{
  _id: '123',
  type: '2'
}, {
  _id: '123',
  type: '1'
}];
var elements = [{
  _id: '124',
  type: '1',
  name: '(should be last, not sorted)'
}, {
  _id: '123',
  type: '2',
  name: 'third(should be first)'
}, {
  _id: '124',
  type: '1',
  name: '(should be last, not sorted)'
}, {
  _id: '123',
  type: '1',
  name: 'second(should be second)'
}, {
  _id: '123',
  type: '1',
  name: 'second(should be second)'
}, {
  _id: '123',
  type: '2',
  name: 'third(should be first)'
}, {
  _id: '123',
  type: '2',
  name: 'third(should be first)'
}, {
  _id: '123',
  type: '1',
  name: 'second(should be second)'
}, {
  _id: '124',
  type: '1',
  name: '(should be last, not sorted)'
}];

var sorted = elements.sort(function (a, b) {
  return (order.findIndex(function (i) {
    return a._id === i._id && a.type === i.type;
  }) + 1 || order.length + 1) - (order.findIndex(function (i) {
    return b._id === i._id && b.type === i.type;
  }) + 1 || order.length + 1);
});

//Just for testing
var list = document.createElement('ul');
for (var i = 0, len = sorted.length; i < len; i++) {
      var item = document.createElement('li');
      item.appendChild(document.createTextNode(sorted[i]['name']));
      list.appendChild(item);
}
document.getElementById('list').appendChild(list);

<div id="list"></div>

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