Sort an object on indices of an array in JavaScript?

Question

OK so given this input (other properties have been stripped for brevity):

var names = [{
    name: 'Michael'
}, {
    name: 'Liam'
}, {
    name: 'Jake'
}, {
    name: 'Dave'
}, {
    name: 'Adam'
}];

I'd like to sort them by another array's indices, and if they aren't in that array, sort alphabetically.

var list = ['Jake', 'Michael', 'Liam'];

Giving me an output of:

Jake, Michael, Liam, Adam, Dave

I've tried using lo-dash but it's not quite right:

names = _.sortBy(names, 'name');
names = _.sortBy(names, function(name) {
    var index = _.indexOf(list, name.name);
    return (index === -1) ? -index : 0;
});

as the output is:

Jake, Liam, Michael, Adam, Dave

Any help would be really appreciated!

i hope you are aware that there is a function called sort() within javascript and it can sort your array alphabetically ! so the problem now is to move your elements into another array ? list.sort(); will output Adam, Dave,Jake, Michael, Liam — ProllyGeek
– ProllyGeek, Commented Nov 25, 2013 at 0:17
@elclanrs - "if they aren't in that array, sort alphabetically", so in other terms the whole array is sorted alphabetically but those three keys come first if that makes sense. Yes, I am aware of sort() although I was already using lodash in the project so it doesn't matter to me to use a native method over the (in my opinion, neater) lodash one. — Ben
– Ben, Commented Nov 25, 2013 at 18:41

go-oleg · Accepted Answer · 2013-11-25 00:26:37Z

3

You are close. return (index === -1) ? -index : 0; is the problem.

Following your approach, it should look like this:

names = _.sortBy(names, 'name')

var listLength = list.length;

_.sortBy(names, function(name) {
    var index = _.indexOf(list, name.name);
    // If the name is not in `list`, put it at the end
    // (`listLength` is greater than any index in the `list`).
    // Otherwise, return the `index` so the order matches the list.
    return (index === -1) ? listLength : index;
});

edited Nov 25, 2013 at 0:26

answered Nov 25, 2013 at 0:18

go-oleg

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