Question

142

Sort an array in the same order of another array

rated 0 times [ 148] [ 6] / answers: 1 / hits: 26041 / 9 Years ago, sat, february 7, 2015, 12:00:00

I have a few arrays of 50+ names like this.

[dan, ryan, bob, steven, corbin]

[bob, dan, steven, corbin]

I have another array that has the correct order. Note that the second array above does not include all of the names, but I still want it to follow the order of the following:

[ryan, corbin, dan, steven, bob]

There is no logical order to it, they are just in this order. What makes sense to me is to compare each array against the correctly ordered one. I think I saw some people doing this with PHP, but I was not able to find a JavaScript solution. Does anyone have any idea how to do this? I've been trying for a few hours and I'm stumped.

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domeniccolti

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answered 9 Years ago beatrizrheaq · Accepted Answer

Use indexOf() to get the position of each element in the reference array, and use that in your comparison function.

var reference_array = [ryan, corbin, dan, steven, bob];

var array = [bob, dan, steven, corbin];

array.sort(function(a, b) {

  return reference_array.indexOf(a) - reference_array.indexOf(b);

});

console.log(array); // [corbin, dan, steven, bob]

Searching the reference array every time will be inefficient for large arrays. If this is a problem, you can convert it into an object that maps names to positions:

var reference_array = [ryan, corbin, dan, steven, bob];

reference_object = {};

for (var i = 0; i < reference_array.length; i++) {

    reference_object[reference_array[i]] = i;

}

var array = [bob, dan, steven, corbin];

array.sort(function(a, b) {

  return reference_object[a] - reference_object[b];

});

console.log(array);