Sort an array with arrays in it by string

Question

I have an array that contains several arrays and I would like to order the arrays based on a certain string within those arrays.

var myArray = [
                [1, 'alfred', '...'],
                [23, 'berta', '...'],
                [2, 'zimmermann', '...'],
                [4, 'albert', '...'],
              ];

How can I sort it by the name so that albert comes first and zimmermann comes last?

I know how I would do it if I could use the integer for sorting but the string leaves me clueless.

Thank for your help! :)

stackoverflow.com/questions/2301856/… is not dissimilar (though not a duplicate). The accepted answer to that question should help you here, though. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Mar 25, 2011 at 16:29

dakab · Accepted Answer · 2016-06-16 05:26:19Z

105

This can be achieved by passing a supporting function as an argument to the Array.sort method call.

Something like this:

 function Comparator(a, b) {
   if (a[1] < b[1]) return -1;
   if (a[1] > b[1]) return 1;
   return 0;
 }

 var myArray = [
   [1, 'alfred', '...'],
   [23, 'berta', '...'],
   [2, 'zimmermann', '...'],
   [4, 'albert', '...'],
 ];

 myArray = myArray.sort(Comparator);
 console.log(myArray);

edited Jun 16, 2016 at 5:26

dakab

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answered Mar 25, 2011 at 16:34

Martin Milan

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3 Comments

dmnkhhn Over a year ago

Thanks Martin, that is working nicely. However, I have problems with names that are lowercase, those are sorted differently from names that start with the same uppercase letter.

Martin Milan Over a year ago

In that case, use the toUpperCase method of the strings (a[1].toUpperCase() and b[1].toUpperCase) in the Comparator method instead. That will make the sort case insensitive

SamGoody Over a year ago

OT, but in many languages (Perl, Ruby, PHP, etc) there is a "spaceship operator" which natively compares and returns -1, 0, or 1 (as per your Comparator method). At the rate ES copies ideas, I would expect it to land in Babel soon.

user456584 · Accepted Answer · 2014-08-21 19:27:56Z

42

You can still use array.sort() with a custom function. Inside the function, simply compare the element that you want to use as your key. For you example, you could use:

myArray.sort(function(a, b) { 
    return a[1] > b[1] ? 1 : -1;
});

edited Aug 21, 2014 at 19:27

user456584

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answered Mar 25, 2011 at 16:31

vhallac

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4 Comments

wong2 Over a year ago

I think you forgot the return

Aleadam Over a year ago

This does not handle equal strings (i.e., it does not return either -1, 1, or 0. @Martin Milan's reply is a better one

vhallac Over a year ago

@Aleadam In theory, you are correct. The comparator is not within the required specs. However, in practice, this works just well enough if you don't care about elements with equal keys being swapped during the sort algorithm. For large arrays, this solution may end up being slower - depending on the implementation of array.sort(). But I need to return 1 or -1, at least. Editing answer.

gordon Over a year ago

so if you need to leave elements in place if equal: return a[1] == b[1] ? 0 : (a[1] > b[1] ? 1 : -1);

andyrandy · Accepted Answer · 2015-10-07 07:22:04Z

15

There´s an easier way now:

myArray = myArray.sort(function(a, b) {
    return a[1].localeCompare(b[1]);
})

It is case insensitive too.

Source: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare

edited Oct 7, 2015 at 7:22

answered Sep 6, 2015 at 22:07

andyrandy

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3 Comments

user776942 Over a year ago

"cannot find function localeCompare in object"

andyrandy Over a year ago

your array elements must be strings, not objects

user776942 Over a year ago

yeah, I've tried converting numbers to strings and still get messed up results. stackoverflow.com/questions/33004116/…

Brian M. Hunt · Accepted Answer · 2017-08-18 12:10:37Z

9

In ES6 one might do the relatively pithy:

myArray.sort(([a], [b]) => a.localeCompare(b))

or

myArray.sort(([a], [b]) => a < b ? -1 : a > b ? 1 : 0)

The helpful/modern bits being the => lambda operator, and the [X] argument destructuring.

answered Aug 18, 2017 at 12:10

Brian M. Hunt

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Comments

gordon · Accepted Answer · 2013-08-01 15:33:02Z

7

Awesome! Compound sort on first element, second element and then third, all ascending in this example, would be

myArray=myArray.sort(function(a,b){
    retVal=0;
    if(a[0]!=b[0]) retVal=a[0]>b[0]?1:-1;
    else if(a[1]!=b[1]) retVal=a[1]>b[1]?1:-1;
    else if(a[2]!=b[2]) retVal=a[2]>b[2]?1:-1;
    return retVal
});

(multiple sort on more than one column)

answered Aug 1, 2013 at 15:33

gordon

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3 Comments

gjonte Over a year ago

can this be done dynamically based on array length

gordon Over a year ago

I haven't played with dynamic array length, but I'd head in direction of

retVal=0; var lenA=a.length; var lenB=b.length; var toLen=Math.min(lenA,lenB); for(i=0;i<toLen;i++){if(a[i]!=b[i]) retVal=a[i]>b[i]?1:-1;} return retVal;

(ps Sorry not to see for so long.)

gordon Over a year ago

This assumes the ordering of the elements of your arrays matches the hierarchy of comparing the elements. If you need to bounce comparisons around among the elements, I suppose you could try passing that order in as another argument.

Capagris · Accepted Answer · 2021-10-23 02:06:00Z

0

Kept getting issues with the solutions posted here. For a simple sorter of simple strings inside an array, I used this:

arr = ["aaa","abc","aba"]
arr.sort((a, b) => a.localeCompare(b));
console.log(arr);
//['aaa','aba','abc']

answered Oct 23, 2021 at 2:06

Capagris

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