Javascript custom sort algorithm according to another array

Question

I have this two arrays

var refArray = ['India','Pakistan','Nepal','Bhutan','SreeLanka','Singapore','Thailand','China','Russia']
var beenThere = ['Russia','Bhutan','India']

I need to sort beenThere on the order of refArray for display purpose, so if I sort beenThere, result should be,

['India','Bhutan','Russia']

Where can I get one algorithm.

dandavis · Accepted Answer · 2014-09-15 06:03:58Z

3

a faster way is to use the pre-sorted array as a template, and limit the indexOf() work to a single indexOf() call on the sub-set items, instead of 2-indexOf() calls on all items.

var refArray = ['India','Pakistan','Nepal','Bhutan','SreeLanka','Singapore','Thailand','China','Russia']
var beenThere = ['Russia','Bhutan','India'];
function contains(a){return this.indexOf(a)!==-1; }
refArray.filter(contains, beenThere); // == ["India", "Bhutan", "Russia"]

answered Sep 15, 2014 at 6:03

dandavis

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4 Comments

RobG Over a year ago

"a faster way is"—than what? I think you need to back that up. ;-)

dandavis Over a year ago

indexOf() is expensive. if you run indexOf() on the big array twice per iteration, it does more work and thus takes more time than running one indexOf() on the smaller array each iteration. this will be especially true where indexOf() is polyfilled by user-land code.

RobG Over a year ago

Well, testing in various browsers shows filter is slightly faster in Safari, but slower in Firefox and Chrome. Tests with different size arrays may change that. Regardless of theory, it's the implementation that matters. ;-)

dfsq Over a year ago

@RobG And it actually makes sense, because filter has to check all the refArray items, while with sort+indexOf we need to visit only few items in beenArray.

dfsq · Accepted Answer · 2014-09-15 05:58:01Z

2

Try to use indexOf:

beenThere.sort(function(a, b) {
    return refArray.indexOf(a) - refArray.indexOf(b); 
}); // ['India','Bhutan','Russia']

answered Sep 15, 2014 at 5:58

dfsq

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Comments

BatScream · Accepted Answer · 2014-09-15 05:59:45Z

2

Just compare the indices of each element in the refArray using indexOf method.

beenThere.sort(function(a,b){
return refArray.indexOf(a)-refArray.indexOf(b);
})

answered Sep 15, 2014 at 5:59

BatScream

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Comments

dandavis · Accepted Answer · 2014-09-15 06:33:24Z

since the data are strings, and the strings don't have commas, you can avoid all user-land iteration with a dynamic RegExp:

var refArray = ['India','Pakistan','Nepal','Bhutan','SreeLanka','Singapore','Thailand','China','Russia']
var beenThere = ['Russia','Bhutan','India'];


(","+refArray+",").match(RegExp(",("+beenThere.join("|")+"),","g")).join(",").split(/,+/).slice(1,-1); 
// ==  ["India", "Bhutan", "Russia"]

that one is nice in that it doesn't need [].indexOf(), so it works in older browsers. you can use another delimeter besides comma if the data has commas, with some slightly uglier code...

or, using filter for the iteration, but with a native method instead of a user-land function:

var refArray = ['India','Pakistan','Nepal','Bhutan','SreeLanka','Singapore','Thailand','China','Russia']
var beenThere = ['Russia','Bhutan','India'];

refArray.filter(/./.test, RegExp("("+beenThere.join("|")+")","g"));
// == ["India", "Bhutan", "Russia"]

these would probably perform faster than indexOf(), but sorting is a weird operation, with lots of opportunity for behind-the-scenes optimization, so results may vary.

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