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Sort an array according to the order of another array

I have an array with the following dimensions:

var arr = [
  'small',
  'small',
  'small',
  'small',
  ...
  'medium',
  'medium',
  'medium',
  'medium',
  ...
  'big',
  'big',
  ...
];

I need to reorganize this array according to this order:

var order = ['small', 'small', 'medium', 'medium', 'big'];

Thus, the result ends approximately as follows:

var arr = [
  'small',
  'small',
  'medium',
  'medium',
  'big',

  'small',
  'small',
  'medium',
  'medium',
  'big'

  ...
];

I know about other similar issues in SO, but haven’t found anything yet. I do not know how to approach this. I thought I sortshould do, but what am I testing? It seems simple, but I'm stuck, I don't know where to start. Any clues?

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6 answers

Well, I finally finished this solution that works:

function orderBy(arr, order) {
  var result = [],
      i = 0, len = arr.length,
      index;

  while (result.length < len) {
    index = arr.indexOf(order[i]);
    result.push(arr[index]);
    arr.splice(index, 1);
    i = i >= order.length-1 ? 0 : ++i;
  }

  return result;
}

It modifies the original array, but that's fine.

Demo: http://jsbin.com/umizat/1/edit

+1
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. .

var init_arr = ['small', 'big', 'big', 'medium'];

var scorer = {
   small: 0,
   medium: 1,
   big: 2
}

// You can use the same array too. I am creating new one.
var final_arr = init_arr.sort(function(a,b) {
   return scorer[a]-scorer[b];
});

alert(final_arr); //small, medium, big, big

+3

, , .. 5, , .

, length % 5 == 0, :

function reOrder(array) {
  var result = [];
  var f = array.length/5; // this must be an integer
  var t;

  for (var i=0; i<f; i++) {
    t = i*2;
    result.push.call(result, array.slice(t, t+2));
    t = f*2 + i*2;
    result.push.call(result, array.slice(t, t+2));
    t = f*4 + i;
    result.push.call(result, array.slice(t, t+1));
  }
  return result; 
} 

var a = ['s','s','s','s','s','s','m','m','m','m','m','m','b','b','b'];
alert(reOrder(a)); // s,s,m,m,b,s,s,m,m,b,s,s,m,m,b
+1

3 , .

SMALL[];
MEDIUM[];
BIG[];

for(i,j,k : upto array lengths : i++,j++,k++){

   nwArray.add(SMALL[i]);
   nwArray.add(SMALL[++i]);
   nwArray.add(MEDIUM[j]);
   nwArray.add(MEDIUM[++j]);
   nwArray.add(BIG[k]);

}

return newArray;
0

order , .

var sorted = [];
var arr = [
  'small',
  'small',
  'small',
  'small',
  ...
  'medium',
  'medium',
  'medium',
  'medium',
  ...
  'big',
  'big',
  ...
];
var order = ['small', 'small', 'medium', 'medium', 'big'];

out:
while (true) {
  for (var i = 0; i < order.length; i++) {
    var index = arr.indexOf(order[i]);
    if ((index === -1) or (arr.length === 0)) {
      break out;
    } else {
      var elem = arr.splice(index, 1);
      sorted.append(elem[0]);
    }
}
0

, , , , JSFiddle , , .

, , . order, .

, :

function mySort(arr, order) {
    var newAr = [];
    var orIndex = 0;
    while(arr.length != 0) {
        var type = order[orIndex];
        var ind = arr.indexOf(type);
        if(ind != -1) {
            newAr.push(type);
            arr.splice(ind, 1);
        }
        orIndex++;
        if(orIndex >= order.length) {
            orIndex = 0;
        }
    }
    return newAr;
}

, :

var arrTest = [
  'small',
  'small',
  'small',
  'small',
  'small',
  'small',
  'small',
  'small',
  'medium',
  'medium',
  'medium',
  'medium',
  'medium',
  'medium',
  'medium',
  'medium',
  'big',
  'big',
  'big',
  'big',
];
var orderTest = ['small', 'small', 'medium', 'medium', 'big'];

, , , , , , , , . , , :

var sortedArray = mySort(arrTest, orderTest);

, :

alert(mySort(arrTest, orderTest).join('\n'));

And this should work for any situation that involves two, and the order can be easily changed.

Hope this helps!

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