Sorting an array based on alphabets?

Question

I need to sort an array based on Alphabets. I have tried sort() method of javascript, but it doesn't work since my array consists of numbers, lowercase letters and uppercase letters. Can anybody please help me out with this? Thanks

For e.g my array is:

[
    "@Basil",
    "@SuperAdmin",
    "@Supreme",
    "@Test10",
    "@Test3",
    "@Test4",
    "@Test5",
    "@Test6",
    "@Test7",
    "@Test8",
    "@Test9",
    "@a",
    "@aadfg",
    "@abc",
    "@abc1",
    "@abc2",
    "@abc5",
    "@abcd",
    "@abin",
    "@akrant",
    "@ankur",
    "@arsdarsd",
    "@asdd",
    "@ashaa",
    "@aviral",
    "@ayush.kansal",
    "@chris",
    "@dgji",
    "@dheeraj",
    "@dsfdsf",
    "@dualworld",
    "@errry",
    "@george",
    "@ggh",
    "@gjhggjghj"
]

What "Alphabets" would you like it to be sorted after if that doesn't contain letters and numbers? What is wrong with the default Alphabet, what result do you expect? — Bergi
– Bergi, Commented Jul 8, 2013 at 9:14
@Bergi: "@" is consistent in the array that I get from the backend,the next letter to it must start with an 'a' or'A'. Say in my example,the first result should be any instances of '@a' then '@aadc' '@abc1' '@abc2' '@chris' and so on — Sumodh Nair
– Sumodh Nair, Commented Jul 8, 2013 at 9:19

Riri · Accepted Answer · 2013-07-08 09:16:54Z

7

a.sort(function(a, b) {
    var textA = a.toUpperCase();
    var textB = b.toUpperCase();
    return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
});

This should work (jsFiddle)

answered Jul 8, 2013 at 9:16

Riri

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Comments

user2555388 · Accepted Answer · 2013-07-08 09:20:54Z

4

function alphabetical(a, b){
     var c = a.toLowerCase();
     var d = b.toLowerCase();
     if (c < d){
        return -1;
     }else if (c > d){
       return  1;
     }else{
       return 0;
     }
}


yourArray.sort(alphabetical);

answered Jul 8, 2013 at 9:20

user2555388

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Comments

georg · Accepted Answer · 2013-07-08 09:46:05Z

1

To sort an array by a key function applied to an element (toUpperCase here), Schwartzian transform, aka "decorate-sort-undecorate" is your technique of choice:

cmp = function(x, y) { return x > y ? 1 : x < y ? -1 : 0 }

sorted = array.map(function(x) {
    return [x.toUpperCase(), x]
}).sort(function(x, y) {
    return cmp(x[0], y[0])
}).map(function(x) {
    return x[1]
})

The major advantage of this approach is that the key function is called exactly once for each element, which can matter when the key is heavy or has side effects.

I realize that you're looking for a simple answer right now, but this might be something for you to consider learning in the future.

edited Jul 8, 2013 at 9:46

answered Jul 8, 2013 at 9:27

georg

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4 Comments

Bergi Over a year ago

No, it yields ["a!", "a"] while it should be ["a", "a!"]. You will need to sort by the decorated part of the array only.

georg Over a year ago

@Bergi: I don't see any ["a!", "a"] in the question.

Bergi Over a year ago

Of course it's not in the question, OP only posted an example array. How can you be sure that his actual data will contain none of !"#$%&'()*+?

Christoph Over a year ago

@thg435 well, the argument "It works on the 3 example values" is not much of a valid standpoint for arguing. Although I like the general approach of using a ST, therefore +1 for the general concept.

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