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I'm sitting over the following array-problem:

  • I have four arrays 1,2,3,4
  • each can contain any number of elements
  • I'm looking for the array(s) with the highest number of elements (and its 2nd to last element)

I'm using this:

var arr = [],
$panels = $(':jqmData(hash="history")'),
$panels.each(function(index) {
     //data("stack") contains the array
     arr.push($(this).data("stack"));
     });
arr.sort(function(a,b){return b.length - a.length});
console.log( arr[0] );

So if the arrays look like this:
[a,b,c] [d,e] [f,g,h,i] [k,l] console says [ [f,g,h,i] ]

OK, but if two arrays have the same highest length, like so:
[a,b,c,d] [d,e] [f,g,h,i] [k,l] I don't get [ [a,b,c,d] [f,g,h,i] ]

because arr[0] will only return the first array again.

I'm clueless...

Is there a way to give me the array or arrays with the highest number of elements and their 2nd-to-last respective element?

Thanks!

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You could add a simple loop which iterates over the arrays and collects every array which has the same length as the first one. As your result is already sorted, you don't even have to iterate over all of them. – Felix Kling Oct 21 '11 at 10:27
so a loop looking for arr[0].length? – frequent Oct 21 '11 at 10:28
That's what I mean. The first array is already the longest one. Edit: yes, something like var result = [], mlength = arr[0].length; for(var i = 0, l = arr.length; i < l; i++) { if(arr[i].length < mlength) break; result.push(arr[i]); }. That said, there might be still other, better, ways. – Felix Kling Oct 21 '11 at 10:28
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3 Answers

up vote 3 down vote accepted

Use this code:

var longest = [];
var longestLen = 0;
$('div:jqmData(hash="history")').each(function(){
    var data = $(this).data("stack");
    if(data.length > longestLen){
        longest = [data];
        longestLen = data.length;
    }
    else if(data.length == longestLen) longest.push(data)
})

What happens?

  1. var data = ... The array is retrieved.
  2. if(data.length > longestLen) - Comparing the length of the current array to the known longest length.
    If the new array has a greater size, replace longest by a new array consisting of element data.
    Update longestLen to the length of the current longest array.
  3. else if(data.length == longestLen) longest.push(data) - If the current array has the same length as the longest array, push the current array in the existing array longest.
share|improve this answer
I think I like this. Trying (and trying to understand...) – frequent Oct 21 '11 at 10:32
@frequent Updated answer with explanation and a small fix. – Rob W Oct 21 '11 at 10:44
@frequent .pop removes and returns the last element from an array. If you want to get a copy of the 2nd-to-last element, use data = data.splice(1). – Rob W Oct 21 '11 at 13:23
Thanks. I was cutting off the last element and then grab the "new" last element. Using you way now though. – frequent Oct 21 '11 at 15:52
@frequent Hang on, are you looking for the "element before the last element" + "last element"? In that case, use element.splice(-2) (get the last two elements from an array). – Rob W Oct 21 '11 at 15:55
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up vote 1 down vote

You can map the arrays to their length, and only return the elements with the highest amount of elements (which is the amount of the first element in the sorted array):

[[1, 2], [7], [3, 4], [], [5]] // the array

.sort(function(a, b) {
    return b.length - a.length;
    // sort by highest to lowest amount of elements
})

.filter(function(value, index, array) {
    return value.length === array[0].length;
    // filter by highest length (length of first element)
});

// [[1, 2], [3, 4]]
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hm. also trying this. Thanks so far – frequent Oct 21 '11 at 10:38
@frequent: It was a little excessive. Tried to make it more readable. – pimvdb Oct 21 '11 at 10:40
I also got this to work, but I'm liking the above better. Still, thanks! – frequent Oct 21 '11 at 11:38
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up vote 0 down vote 
function arrayFunc(arr) {
    var c = [[]];
    $.each(arr, function(key, value) { 
        if(c[0].length == value.length) {
            c.push([value]);
        }
        else {
            if(c[0].length <= value.length) {
                c = [];
                c.push(value);
            }
        }
    });
    return c;
}

var a = [[0,1,2],[0,1,2,3],[0,1]];
var b = [[0,1,2],[0,1,2,3],[0,1],[0,1,3,4]];

console.log('result var a');
console.dir(arrayFunc(a));

console.log('result var b');
console.dir(arrayFunc(b));
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