Javascript - sort array based on another array

Question

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?

Answer 1

One-Line answer.

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

Or even shorter:

itemsArray.sort((a, b) => sortingArr.indexOf(a) - sortingArr.indexOf(b));

Answer 2

Something like:

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

sorting = [ 'b', 'c', 'b', 'b', 'c', 'd' ];
result = []

sorting.forEach(function(key) {
    var found = false;
    items = items.filter(function(item) {
        if(!found && item[1] == key) {
            result.push(item);
            found = true;
            return false;
        } else 
            return true;
    })
})

result.forEach(function(item) {
    document.writeln(item[0]) /// Bob Jason Henry Thomas Andrew
})

Here's a shorter code, but it destroys the sorting array:

result = items.map(function(item) {
    var n = sorting.indexOf(item[1]);
    sorting[n] = '';
    return [n, item]
}).sort().map(function(j) { return j[1] })

Answer 3

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);

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

function sortFunc(a, b) {
  var sortingArr = ["A", "B", "C"];
  return sortingArr.indexOf(a.type) - sortingArr.indexOf(b.type);
}

const itemsArray = [
  {
    type: "A",
  },
  {
    type: "C",
  },
  {
    type: "B",
  },
];
console.log(itemsArray);
itemsArray.sort(sortFunc);
console.log(itemsArray);

Answer 8

ES6

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

More examples with different input arrays

Answer 9

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/

Answer 10

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);
})

Answer 11

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

Answer 12

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']],
    sortingArr = [ 'b', 'c', 'b', 'b', 'a', 'd' ],
    map = itemsArray.reduce((m, a) => m.set(a[1], (m.get(a[1]) || []).concat([a])), new Map),
    result = sortingArr.map(k => (map.get(k) || []).shift());

console.log(result);

Answer 13

I hope that I am helping someone, but if you are trying to sort an array of objects by another array on the first array's key, for example, you want to sort this array of objects:

const foo = [
  {name: 'currency-question', key: 'value'},
  {name: 'phone-question', key: 'value'},
  {name: 'date-question', key: 'value'},
  {name: 'text-question', key: 'value'}
];        

by this array:

const bar = ['text-question', 'phone-question', 'currency-question', 'date-question'];

you can do so by:

foo.sort((a, b) => bar.indexOf(a.name) - bar.indexOf(b.name));

Answer 14

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)

Answer 15

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

It might help someone

Answer 16

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

Answer 17

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);
}); 

Answer 18

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

Answer 19

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
}

Answer 20

with numerical sortingArr:

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

Answer 21

This seems to work for me:

var outputArray=['10','6','8','10','4','6','2','10','4','0','2','10','0'];
var template=['0','2','4','6','8','10'];
var temp=[];

for(i=0;i<template.length;i++) {
  for(x=0;x<outputArray.length;x++){
    if(template[i] == outputArray[x]) temp.push(outputArray[x])
  };
}

outputArray = temp;
alert(outputArray)

Answer 22

This approach features a sorting algoritm where for each item of the order array, the data is searched for a corresponding item and this item is swapped by the left neighbor until the item is at the right position.

 a   b   b   d   c   b   data
<b>  a   b   d   c   b   right order of <>
 b  <c>  a   b   d   b
 b   c  <b>  a   d   b
 b   c   b  <b>  a   d
 b   c   b   b  <a>  d
 b   c   b   b   a  <d>

const
    data = [['Anne', 'a'], ['Bob', 'b'], ['Henry', 'b'], ['Andrew', 'd'], ['Jason', 'c'], ['Thomas', 'b']],
    order = ['b', 'c', 'b', 'b', 'a', 'd'];
    
let i = 0;

for (const o of order) {
    let j = i;
    while (data[j][1] !== o) j++;
    while (i !== j) {
        [data[j], data[j - 1]] = [data[j - 1], data[j]];
        j--;
    }
    i++;
}

console.log(data);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 23

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;

Answer 24

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);
});

Answer 25

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]);
}

Answer 26

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

Answer 27

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