使用JavaScript对多维数组进行排序：整数

Question

I have a 2D array called "results." Each "row" array in results contains both string and integer values. I'm using this script to sort the array by any "column" on an onclick event:

function sort_array(results, column, direction) {
var sorted_results = results.sort(value);
function value(a,b) {
    a = a[column];
    b = b[column];
    return a == b ? 0 : (a < b ? -1*direction : 1*direction)
    }
}

This works fine for the columns with strings. But it treats the columns of integers like strings instead of numbers. For example, the values 15, 1000, 200, 97 would be sorted 1000, 15, 200, 97 if "ascending" or 97, 200, 15, 1000 "descending."

I've double-checked the typeof the integer values, and the script knows they're numbers. How can I get it to treat them as such?

Answer 1

Make the types of a and b be part of the comparison that decides what your value function returns. In the process you'll have to decide how to sort integers relative to strings.

Alternately you can have a comparison function that takes the values, and does a substitution to replace every string of digits with a string of digits of fixed length, with leading zeros, and then does string comparisons. A benefit of this approach is that you wind up sorting things like "a2", "a9", "a10", etc. Which people generally like.

Answer 2

Are you sure they're numbers? How did you check? Try this modification to force them to be numbers just in case:

if (isNumberColumn(column)) {
    a = +a;
    b = +b;
}

Answer 3

I think that you're going to have to make a preliminary pass over the values in a column before sorting. Why? Well you need to know in advance (before sorting, that is) whether all the values can be treated as numbers. If they can, then the "value" function can convert them. Otherwise, it should sort them as strings.

Answer 4

you could try to make your own comparison function, one that suit your needs. You could define your own way of comparing strings with ints. ex :

function getDirection(a,b){
    if(typeof(a) == typeof(b))
        return a == b ? 0 : (a>b?1:-1);
    a+="";b+="";
    if(a[0] == '-')
        {
            if(b[0] != '-')
                return 1;
        }
    else 
        if(b[0] != '-')
            return -1;
     return a[0] == b[0] ? 0 : (a[0]>b[0]?1:-1);
}

Hope it helps.

Answer 5

Firstly: sort is a mutator, which means sorting happens in place. So to avoid unintended consequeces, i'd change

var sorted_results = results.sort(value);

to

var sorted_results = results.slice(0).sort(value);

unless of course you wish it sorted in place, but then what do you need the sorted_results variable for?

As for the sorting itself - the integer sorting seems to work just fine for me, the problem is really the strings in the mixed integer and string scenario, as seen in his example: http://jsfiddle.net/QJ5fM/ , where it sort the following arrays quite differently:

[[16],[131],['aa'],['0hey'],[176],[100],['hey'],[1],[12]];
[['aa'],[16],[131],['0hey'],[176],[100],['hey'],[1],[12]];
[['aa'],['0hey'],[16],[131],[176],[100],['hey'],[1],[12]];

giving the following result in chrome 9:

1,12,0hey,aa,hey,16,100,131,176
1,12,0hey,16,100,131,176,aa,hey
1,12,16,100,131,176,0hey,aa,hey

the following in firefox 3.6:

1,12,0hey,aa,hey,16,100,131,176
1,12,0hey,hey,16,100,131,176,aa
1,12,hey,16,100,131,176,0hey,aa

and the following in ie8:

0hey,aa,hey,1,12,16,100,131,176
0hey,aa,hey,1,12,16,100,131,176
0hey,aa,hey,1,12,16,100,131,176

As a surprise here, it seems ie8 has the only sane, or at least consistent, implementation :P