[javascript] Javascript - sort array based on another array

Is it possible to sort and rearrange an array that looks like this:

itemsArray = [ 
    ['Anne', 'a'],
    ['Bob', 'b'],
    ['Henry', 'b'],
    ['Andrew', 'd'],
    ['Jason', 'c'],
    ['Thomas', 'b']
]

to match the arrangement of this array:

sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]

Unfortunately, I don’t have any IDs to keep track on. I would need to priority the items-array to match the sortingArr as close as possible.

Update:

Here is the output I’m looking for:

itemsArray = [    
    ['Bob', 'b'],
    ['Jason', 'c'],
    ['Henry', 'b'],
    ['Thomas', 'b']
    ['Anne', 'a'],
    ['Andrew', 'd'],
]

Any idea how this can be done?

You can do something like this:

function getSorted(itemsArray , sortingArr ) {
  var result = [];
  for(var i=0; i<arr.length; i++) {
    result[i] = arr[sortArr[i]];
  }
  return result;
}

You can test it out here.

Note: this assumes the arrays you pass in are equivalent in size, you'd need to add some additional checks if this may not be the case.

refer link

refer

let a = ['A', 'B', 'C' ]

let b = [3, 2, 1]

let c = [1.0, 5.0, 2.0]

// these array can be sorted by sorting order of b

const zip = rows => rows[0].map((_, c) => rows.map(row => row[c]))

const sortBy = (a, b, c) => {
  const zippedArray = zip([a, b, c])
  const sortedZipped = zippedArray.sort((x, y) => x[1] - y[1])

  return zip(sortedZipped)
}

sortBy(a, b, c)

Use the $.inArray() method from jQuery. You then could do something like this

var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
var newSortedArray = new Array();

for(var i=sortingArr.length; i--;) {
 var foundIn = $.inArray(sortingArr[i], itemsArray);
 newSortedArray.push(itemsArray[foundIn]);
}

this is probably too late but, you could also use some modified version of the code below in ES6 style. This code is for arrays like:

var arrayToBeSorted = [1,2,3,4,5];
var arrayWithReferenceOrder = [3,5,8,9];

The actual operation :

arrayToBeSorted = arrayWithReferenceOrder.filter(v => arrayToBeSorted.includes(v));

The actual operation in ES5 :

arrayToBeSorted = arrayWithReferenceOrder.filter(function(v) {
    return arrayToBeSorted.includes(v);
});

Should result in arrayToBeSorted = [3,5]

Does not destroy the reference array.

If you use the native array sort function, you can pass in a custom comparator to be used when sorting the array. The comparator should return a negative number if the first value is less than the second, zero if they're equal, and a positive number if the first value is greater.

So if I understand the example you're giving correctly, you could do something like:

function sortFunc(a, b) {
  var sortingArr = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
  return sortingArr.indexOf(a[1]) - sortingArr.indexOf(b[1]);
}

itemsArray.sort(sortFunc);

One Line answer.

itemsArray.sort(function(a, b){  
  return sortingArr.indexOf(a) - sortingArr.indexOf(b);
});

this should works:

var i,search, itemsArraySorted = [];
while(sortingArr.length) {
    search = sortingArr.shift();
    for(i = 0; i<itemsArray.length; i++) {
        if(itemsArray[i][1] == search) {
            itemsArraySorted.push(itemsArray[i]);
            break;
        }
    } 
}

itemsArray = itemsArraySorted;

You could try this method.

const sortListByRanking = (rankingList, listToSort) => {
  let result = []

  for (let id of rankingList) {
    for (let item of listToSort) {
      if (item && item[1] === id) {
        result.push(item)
      }
    }
  }

  return result
}

  const result = sortingArr.map((i) => {
    const pos = itemsArray.findIndex(j => j[1] === i);
    const item = itemsArray[pos];
    itemsArray.splice(pos, 1);
    return item;
  });

I had to do this for a JSON payload I receive from an API, but it wasn't in the order I wanted it.

Array to be the reference array, the one you want the second array sorted by:

var columns = [
    {last_name: "last_name"},
    {first_name: "first_name"},
    {book_description: "book_description"},
    {book_id: "book_id"},
    {book_number: "book_number"},
    {due_date: "due_date"},
    {loaned_out: "loaned_out"}
];

I did these as objects because these will have other properties eventually.

Created array:

 var referenceArray= [];
 for (var key in columns) {
     for (var j in columns[key]){
         referenceArray.push(j);
     }
  }

Used this with result set from database. I don't know how efficient it is but with the few number of columns I used, it worked fine.

result.forEach((element, index, array) => {                            
    var tr = document.createElement('tr');
    for (var i = 0; i < referenceArray.length - 1; i++) {
        var td = document.createElement('td');
        td.innerHTML = element[referenceArray[i]];
        tr.appendChild(td);

    }
    tableBody.appendChild(tr);
}); 

Use intersection of two arrays.

Ex:

var sortArray = ['a', 'b', 'c',  'd', 'e'];

var arrayToBeSort = ['z', 's', 'b',  'e', 'a'];

_.intersection(sortArray, arrayToBeSort) 

=> ['a', 'b', 'e']

if 'z and 's' are out of range of first array, append it at the end of result

with numerical sortingArr:

itemsArray.sort(function(a, b){  
  return sortingArr[itemsArray.indexOf(a)] - sortingArr[itemsArray.indexOf(b)];
});

let sortedOrder = [ 'b', 'c', 'b', 'b' ]
let itemsArray = [ 
    ['Anne', 'a'],
    ['Bob', 'b'],
    ['Henry', 'b'],
    ['Andrew', 'd'],
    ['Jason', 'c'],
    ['Thomas', 'b']
]
a.itemsArray(function (a, b) {
    let A = a[1]
    let B = b[1]

    if(A != undefined)
        A = A.toLowerCase()

    if(B != undefined)
        B = B.toLowerCase()

    let indA = sortedOrder.indexOf(A)
    let indB = sortedOrder.indexOf(B)

    if (indA == -1 )
        indA = sortedOrder.length-1
    if( indB == -1)
        indB = sortedOrder.length-1

    if (indA < indB ) {
        return -1;
    } else if (indA > indB) {
        return 1;
    }
    return 0;
})

This solution will append the objects at the end if the sorting key is not present in reference array

Why not something like

//array1: array of elements to be sorted
//array2: array with the indexes

array1 = array2.map((object, i) => array1[object]);

The map function may not be available on all versions of Javascript

this.arrToBeSorted =  this.arrToBeSorted.sort(function(a, b){  
  return uppthis.sorrtingByArray.findIndex(x => x.Id == a.ByPramaeterSorted) - uppthis.sorrtingByArray.findIndex(x => x.Id == b.ByPramaeterSorted);
});

I would use an intermediary object (itemsMap), thus avoiding quadratic complexity:

function createItemsMap(itemsArray) { // {"a": ["Anne"], "b": ["Bob", "Henry"], …}
  var itemsMap = {};
  for (var i = 0, item; (item = itemsArray[i]); ++i) {
    (itemsMap[item[1]] || (itemsMap[item[1]] = [])).push(item[0]);
  }
  return itemsMap;
}

function sortByKeys(itemsArray, sortingArr) {
  var itemsMap = createItemsMap(itemsArray), result = [];
  for (var i = 0; i < sortingArr.length; ++i) {
    var key = sortingArr[i];
    result.push([itemsMap[key].shift(), key]);
  }
  return result;
}

See http://jsfiddle.net/eUskE/

For getting a new ordered array, you could take a Map and collect all items with the wanted key in an array and map the wanted ordered keys by taking sifted element of the wanted group.

var itemsArray = [['Anne', 'a'], ['Bob', 'b'], ['Henry', 'b'], ['Andrew', 'd'], ['Jason', 'c'], ['Thomas', 'b']],_x000D_
    sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ],_x000D_
    map = itemsArray.reduce((m, a) => m.set(a[1], (m.get(a[1]) || []).concat([a])), new Map),_x000D_
    result = sortingArr.map(k => (map.get(k) || []).shift());_x000D_
_x000D_
console.log(result);

In case you get here needing to do this with an array of objects, here is an adaptation of @Durgpal Singh's awesome answer:

const itemsArray = [
  { name: 'Anne', id: 'a' },
  { name: 'Bob', id: 'b' },
  { name: 'Henry', id: 'b' },
  { name: 'Andrew', id: 'd' },
  { name: 'Jason', id: 'c' },
  { name: 'Thomas', id: 'b' }
]

const sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ]

Object.keys(itemsArray).sort((a, b) => {
  return sortingArr.indexOf(itemsArray[a].id) - sortingArr.indexOf(itemsArray[b].id);
})

Case 1: Original Question (No Libraries)

Plenty of other answers that work. :)

Case 2: Original Question (Lodash.js or Underscore.js)

var groups = _.groupBy(itemArray, 1);
var result = _.map(sortArray, function (i) { return groups[i].shift(); });

Case 3: Sort Array1 as if it were Array2

I'm guessing that most people came here looking for an equivalent to PHP's array_multisort (I did) so I thought I'd post that answer as well. There are a couple options:

1. There's an existing JS implementation of array_multisort(). Thanks to @Adnan for pointing it out in the comments. It is pretty large, though.

2. Write your own. (JSFiddle demo)

function refSort (targetData, refData) {
  // Create an array of indices [0, 1, 2, ...N].
  var indices = Object.keys(refData);

  // Sort array of indices according to the reference data.
  indices.sort(function(indexA, indexB) {
    if (refData[indexA] < refData[indexB]) {
      return -1;
    } else if (refData[indexA] > refData[indexB]) {
      return 1;
    }
    return 0;
  });

  // Map array of indices to corresponding values of the target array.
  return indices.map(function(index) {
    return targetData[index];
  });
}

3. Lodash.js or Underscore.js (both popular, smaller libraries that focus on performance) offer helper functions that allow you to do this:

    var result = _.chain(sortArray)
      .pairs()
      .sortBy(1)
      .map(function (i) { return itemArray[i[0]]; })
      .value();

...Which will (1) group the sortArray into [index, value] pairs, (2) sort them by the value (you can also provide a callback here), (3) replace each of the pairs with the item from the itemArray at the index the pair originated from.

This is what I was looking for and I did for sorting an Array of Arrays based on another Array:

It's On^3 and might not be the best practice(ES6)

function sortArray(arr, arr1){_x000D_
      return arr.map(item => {_x000D_
        let a = [];_x000D_
        for(let i=0; i< arr1.length; i++){_x000D_
          for (const el of item) {_x000D_
            if(el == arr1[i]){_x000D_
              a.push(el);_x000D_
            }   _x000D_
            }_x000D_
          }_x000D_
          return a;_x000D_
      });_x000D_
    }_x000D_
    _x000D_
    const arr1 = ['fname', 'city', 'name'];_x000D_
  const arr = [['fname', 'city', 'name'],_x000D_
  ['fname', 'city', 'name', 'name', 'city','fname']];_x000D_
  console.log(sortArray(arr,arr1));

var sortedArray = [];
for(var i=0; i < sortingArr.length; i++) {
    var found = false;
    for(var j=0; j < itemsArray.length && !found; j++) {
        if(itemsArray[j][1] == sortingArr[i]) {
            sortedArray.push(itemsArray[j]);
            itemsArray.splice(j,1);
            found = true;
        }
    }
}

http://jsfiddle.net/s7b2P/

Resulting order: Bob,Jason,Henry,Thomas,Anne,Andrew

ES6

const arrayMap = itemsArray.reduce(
  (accumulator, currentValue) => ({
    ...accumulator,
    [currentValue[1]]: currentValue,
  }),
  {}
);
const result = sortingArr.map(key => arrayMap[key]);

More examples with different input arrays