Sort four (4) javascript arrays with 10 objects each by a value

Question

model w/ sample data => 
[{
    date: '13413413',
    name: 'asdfasdf',
    other: 'kjh'
}]

getJSON returns 4 arrays of 10 model objects each.

array1 = 10 of resultsObj sorted by date from newest to oldest
array2 = 10 of resultsObj sorted by date from newest to oldest
array3 = 10 of resultsObj sorted by date from newest to oldest
array4 = 10 of resultsObj sorted by date from newest to oldest

I need 1 array with all 40 objects sorted by date, newest to latest.

I think can do this with a lot of if/else checks, but would love to see some examples/ideas, thanks.

You're looking for a multi-merge algorithm

Bergi
– Bergi

2015-06-16 18:07:27 +00:00
Commented Jun 16, 2015 at 18:07 — Bergi
– Bergi, Commented Jun 16, 2015 at 18:07

Shankar Regmi · Accepted Answer · 2015-06-16 18:13:23Z

5

This will be easy to understand for you. first concatenate the array into one and chain the output to Array.sort

code

var result = a.concat(b,c,d).sort(function(a, b) {
    return b.date-a.date
});

answered Jun 16, 2015 at 18:13

Shankar Regmi

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1 Comment

atomicadam Over a year ago

wonderful! Thanks so much. Clean and works! good idea to concat all together, I wasn't thinking about that at all.

Denys Séguret · Accepted Answer · 2015-06-16 18:07:22Z

1

You can do it in one statement without intermediate variables:

var sorted = [].concat.call(array1,array2,array3,array4).sort(function(a,b){
    return b.date-a.date
})

You should have a second look at your design, though, and check you shouldn't have an array of arrays or another structure instead of four variables.

answered Jun 16, 2015 at 18:07

Denys Séguret

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Comments

Oriol · Accepted Answer · 2015-06-16 18:21:02Z

1

Since the 4 arrays are already sorted, the fastest way is having an index for each one, and find the minimum of the current four:

var indices = [0, 0, 0, 0],
    arrays = [array1, array2, array3, array4],
    array = Array(40);
for(var i=0; i<40; ++j) {
  var min = Infinity, argmin;
  for(var j=0; j<4; ++i) if(indices[j] < 10 && arrays[j][indices[j]] < min) {
    min = arrays[j][indices[j]].date;
    argmin = j;
  }
  array[i] = arrays[argmin][indices[argmin]++];
}

So, if the arrays have n elements, this algorithm is O(4n) = O(n).

A simpler approach would be concatenating all the arrays and sorting, but that would be O(4n log(4n)) = O(n log n).

edited Jun 16, 2015 at 18:21

answered Jun 16, 2015 at 18:13

Oriol

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