JavaScript: Sorting with exception

Question

Lets say I have an array of objects (for simpler display purposes I will just show it as an array)

[ 'TEST', 'NEW', 'ALPHA', 'ZOO', 'WHATEVER' ]

I need to sort it alphabetically (easy part), however, I need to sort it in a way, that a certain word, lets say NEW will end up at the very end.

[ 'ALPHA', 'TEST', 'WHATEVER', 'ZOO', 'NEW' ]

Here is the function that I sort with

var sortedWords = function(a, b) {
    return a.word > b.word ? 1 : -1 ;
};

So, I get the sortedWords array, and then go over it again, create a yet another array and push words into the new array, unless the word equals NEW. In that case I set it aside and append at the very end of this new array before returning it. There got to be a better and more efficient way of doing this.

Thank you.

@BrianHannay Remember that if a and b are both the target word, the result must be 0. — C. K. Young
– C. K. Young, Commented Jun 22, 2013 at 19:34

C. K. Young · Accepted Answer · 2013-06-22 19:32:59Z

7

Change your comparator to look for NEW first:

function (a, b) {
    if ((a.word === 'NEW') != (b.word === 'NEW')) {
        return a.word === 'NEW' ? 1 : -1;
    }
    return a.word > b.word ? 1 :
           a.word < b.word ? -1 : 0;
}

answered Jun 22, 2013 at 19:32

C. K. Young

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5 Comments

solefald Over a year ago

Thanks, Chris! This does exactly what I needed, but apparently I have to wait for 7 minutes before I can accept this as an answer :)

AntouanK Over a year ago

@solefald can you check this jsfiddle.net/blackjim/X6krQ ? Maybe I missed something.

solefald Over a year ago

@Antonis: yep. since you are sorting a flat array (i sort array of objects), you need to change a.word to a and b.word to b

AntouanK Over a year ago

@solefald this is not the code that was submitted though. The initial code didn't work.

solefald Over a year ago

@Antonis: In my original question? See the first line. I mentioned that for simple display purposes I will show it is a flat array. Chris's solution works great in my case, obviously with some adjustments to match my object element names and the words I match for.

AntouanK · Accepted Answer · 2024-02-28 14:29:16Z

4

Use sort with a custom compareFunction

var ar = [ 'ALPHA', 'WHATEVER', 'NEW', 'ZOO', 'TEST' ];

ar.sort(function(a, b){
    var wordToBeLast = 'NEW';  // set your word here
    
    if(a===wordToBeLast){
        return 1;
    } else if(b===wordToBeLast){
        return -1;
    } else {
        return a.localeCompare(b);
    }
});

edited Feb 28, 2024 at 14:29

answered Jun 22, 2013 at 19:47

AntouanK

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1 Comment

Malcolm Over a year ago

return a > b; is incorrect: the compare function needs to return a number, not a boolean, and the number should be 0 if the two items are equal.

Marius Schulz · Accepted Answer · 2013-06-22 19:44:23Z

2

You could define your comparator function as follows:

var sortedWords = function(a, b) {
    if (a.word === b.word) {
        // a and b are considered equal, the order is okay
        return 0;   
    }

    if (a.word === "NEW") {
        // "NEW" is considered larger than every other value, the order is wrong
        return 1;
    }

    // Correct order if b equals "NEW", otherwise lexicographical comparison
    return b.word === "NEW" ? -1 : (a.word > b.word ? 1 : -1);
};

edited Jun 22, 2013 at 19:44

answered Jun 22, 2013 at 19:33

Marius Schulz

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2 Comments

Blender Over a year ago

Can you explain how this is an answer and not a comment?

Marius Schulz Over a year ago

@Blender: I didn't paste my fiddle code at that point of time yet.

GarryOne · Accepted Answer · 2023-01-17 17:47:13Z

0

There is simpler way using lodash, which can be extended to a more complex scenario, while keeping the code simple.

const strings = [ 'TEST', 'NEW', 'ALPHA', 'ZOO', 'WHATEVER' ];

_.sortBy(strings, [
    (string) => string !== 'NEW',
    (string) => string
  ])

answered Jan 17, 2023 at 17:47

GarryOne

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