How to sort an array sequencial in javascript?

Question

I'm not sure if the title of this question is correct or not and also not sure what the appropriate keyword to search on google.

I have an array look like:

var myArray = [1,1,2,2,2,3,4,4,4];

and I want to sort my array into:

var myArray = [1,2,3,4,1,2,4,2,4];

Please in to my expected result. the order is ascending but duplicate value will repeated on last sequence instead of put it together in adjacent keys. So the expected result grouped as 1,2,3,4 1,2,4 and 2,4.

Thank you for your help and sorry for my bad English.

@NinaScholz, the expected result should be [1, 2, 3, 7, 7, 7, 7] — Habibillah
– Habibillah, Commented Sep 23, 2015 at 9:38

Magus · Accepted Answer · 2015-09-23 07:38:15Z

3

This code works. But it may exist a better solution.

// We assume myArray is already sorted
var myArray = [1,1,2,2,2,3,4,4,4],
    result = [];

while (myArray.length) {
   var value = myArray.shift();

   // Find place
   var index = 0;
   while(result[index] && result[index][result[index].length - 1] == value) index++;

   if(!result[index]) {
      result[index] = [];
   }  

   result[index][result[index].length] = value;
}

result.reduce(function(current, sum) {
  return current.concat(sum);
});

console.log(result) // Display [1, 2, 3, 4, 1, 2, 4, 2, 4]

answered Sep 23, 2015 at 7:38

Magus

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Comments

Darren Willows · Accepted Answer · 2015-09-23 08:11:08Z

Here is my method using JQuery and it does not assume the array is already sorted.

It will iterate through the array and no duplicates to tempResultArray, once finished, it will then add them to the existing result and repeat the process again to find duplicates.

This is not the most efficient method, but it can be handled by one function and does not require the array to be sorted.

var myArray = [1,1,2,2,2,3,4,4,4],result = [];

while (myArray && myArray.length) {
    myArray = customSort(myArray);
}
console.log(result);

function customSort(myArray){
    var tempResultArray = [], tempMyArray = [];
    $.each(myArray, function(i, el){
        if($.inArray(el, tempResultArray ) === -1){ 
            tempResultArray.push(el);
        }else{
            tempMyArray.push(el); 
        }
   });  
    tempResultArray.sort(function(a, b){return a-b});
    $.merge( result,tempResultArray)
    return tempMyArray;
}

JSFiddle

Nina Scholz · Accepted Answer · 2015-09-23 14:34:10Z

2

This proposal features a straight forward approach with focus on array methods.

function sprout(array) {
    return array.reduce(function (r, a) {
        !r.some(function (b) {
            if (b[b.length - 1] < a) {
                b.push(a);
                return true;
            }
        }) && r.push([a]);
        return r;
    }, []).reduce(function (r, a) {
        return r.concat(a);
    });
}

document.write('<pre>' + JSON.stringify(sprout([1, 1, 2, 2, 2, 3, 4, 4, 4]), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(sprout([1, 2, 3, 7, 7, 7]), 0, 4) + '</pre>');
document.write('<pre>' + JSON.stringify(sprout([1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 6, 6, 6, 7]), 0, 4) + '</pre>');

edited Sep 23, 2015 at 14:34

answered Sep 23, 2015 at 11:48

Nina Scholz

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Comments

maioman · Accepted Answer · 2015-09-23 20:41:05Z

2

here's another solution:

var myArray = [1, 1, 2, 2, 2, 3, 4, 4, 4];


function mySequelArray(arr) {
    var res = arguments[1] || [];
    var nextVal;
    var min = Math.min.apply(null, arr);

    if (res.length > 0) {
        nextVal = arr.filter(function (x) {
            return x > res[res.length - 1]
        }).sort()[0] || min;
    } else {
        nextVal = min;
    }
    res.push(nextVal);
    arr.splice(arr.indexOf(nextVal), 1);

    return (arr.length > 0) ? mySequelArray(arr, res) : res;


}

console.log(mySequelArray(myArray))

fiddle

answered Sep 23, 2015 at 20:41

maioman

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Comments

Bardo · Accepted Answer · 2015-09-23 07:28:28Z

1

My best approach will be to split your array into separate arrays for each repeated value, then arrange each separate array and join altogether.

answered Sep 23, 2015 at 7:28

Bardo

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Comments

Sotiris Kiritsis · Accepted Answer · 2015-09-23 11:46:31Z

UPDATED: I wrote a quick code sample that should work if the same number in inputArray is not given more than twice. You could improve it by making it recursive thus creating new arrays for each new number and removing the limitation. Had some free time so i re-wrote a recursive function to sort any given array in sequence groups like you wanted. Works like a charm, inputArray does not need to be sorted and doesn't require any libraries. Jsfiddle here.

var inputArray = [3, 4, 1, 2, 3, 1, 2, 4, 1, 2, 5, 1];
var result = sortInSequence(inputArray);
console.log(result); //output: [1, 2, 3, 4, 5, 1, 2, 3, 4, 1, 2, 1]

function sortInSequence(inputArray){
    var inputArraySize = inputArray.length,
        tempArray = [], //holds new array we are populating
        sameValuesArray = [], //holds same values that we will pass as param in recursive call
        rSorted = []; //init sorted array in case we have no same values

    for(var i = inputArraySize; i > 0; i--){
        var value = inputArray.pop();
        tempArray.push(value);

        var counter = 0,
            tempArraySize = tempArray.length;
        for(var j = 0; j < tempArraySize; j++){
            if(tempArray[j] == value){
                counter++;
            }
        }

        if(counter == 2){
            //value found twice, so remove it from tempArray and add it in sameValuesArray
            var sameValue = tempArray.pop();
            sameValuesArray.push(sameValue);
        }
    }

    if(sameValuesArray.length > 0){
        rSorted = sortInSequence(sameValuesArray);
    }

    tempArray.sort();
    return tempArray.concat(rSorted);
}

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