How to sort Array of Arrays? [duplicate]

Question

I need to sort an array like the one below based on the difference between "over" "under" odds of a match, so the most balanced outcome goes to the first row of the main array!

var object = [
        [{"0.5": { "over": 1.2, "under": 3.8 }}], //difference 2.6
        [{"1.5": { "over": 1.5, "under": 2.8 }}], //difference 1.3
        [{"2.5": { "over": 1.9, "under": 2.0 }}], //difference 0.1
        [{"2.5": { "over": 2.2, "under": 1.8 }}], //difference 0.4
    ];

Like this

var object = [
    [{ "2.5": { "over": 1.9, "under": 2.0 } }], //difference 0.1
    [{ "2.5": { "over": 2.2, "under": 1.8 } }], //difference 0.4
    [{ "1.5": { "over": 1.5, "under": 2.8 } }], //difference 1.3
    [{ "0.5": { "over": 1.2, "under": 3.8 } }], //difference 2.6
];

I have read this post but it's not of any help in my case

Sort Complex Array of Arrays by value within

I just solved it. Thanks guys!

var obj = [{ value: 0.5, over: 1.2, under: 3.8 },
        { value: 1.5, over: 1.5, under: 3.2 },
        { value: 2.5, over: 1.8, under: 2.5 },
        { value: 3.5, over: 1.9, under: 1.9 },
        { value: 3.5, over: 2.5, under: 1.75 }];



obj.sort(function(a,b) { 
    return Math.min(Math.abs(parseFloat(b.over) - parseFloat(a.under)));
    });

How is it not of any help? It shows you how you can sort an array by passing a custom compare callback function. All you need to do is sort your main array, and compare two items in it in your callback. — GolezTrol
– GolezTrol, Commented May 17, 2014 at 23:32
You have too many arrays there, you don't need the nested arrays. If you just use a colleciton you can try Underscore's sortBy — elclanrs
– elclanrs, Commented May 17, 2014 at 23:33
function(a,b) {return Math.abs(a.over-a.under) - Math.abs(b.over-a.under);} — Niet the Dark Absol
– Niet the Dark Absol, Commented May 17, 2014 at 23:36
@NiettheDarkAbsol: That would work if OP didn't have nested everything. I would look into fixing the source where this is coming from, so I get a decent data-structure to work with. — elclanrs
– elclanrs, Commented May 17, 2014 at 23:37
@elclanrs Oh yeah. I guess a[0].over et al. but yeah, over-nested. — Niet the Dark Absol
– Niet the Dark Absol, Commented May 17, 2014 at 23:39

CrossUI · Accepted Answer · 2014-05-17 23:40:19Z

1

object.sort(function(x,y){
    var xx,yy,x1,y1;
    for(var i in x[0]){xx=x[0][i];break;}
    for(var i in y[0]){yy=y[0][i];break;}
    x1=Math.abs(xx.over-xx.under);
    y1=Math.abs(yy.over-yy.under);
    return x1===y1?0:x1>y1?1:-1;
});

answered May 17, 2014 at 23:40

CrossUI

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1 Comment

Olsi Over a year ago

Thank's CrossUI, it's perfect. A bit disappointed with the quickdownvoters here though. I don't think at all this could be solved with the example mentioned above. Besides I offered it myself, which means I had done the research, proved it and it didn't work. Thanks Again CrossUi!

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