Ranking array elements

Question

I need an algorithm to rank elements of an array in Javascript.

Example : I have an array as follows:

[79, 5, 18, 5, 32, 1, 16, 1, 82, 13]

I need to rank the entries by value. So 82 should receive rank 1 , 79 rank 2 etc. If two entries have the same value they receive the same rank and the rank for a lower value is raised.

So for this array, the new ranking array would be:

[2, 7, 4, 7, 3, 9, 5, 9, 1, 6] 

How can I do this?

Answer 1

 var arr = [79, 5, 18, 5, 32, 1, 16, 1, 82, 13]; var sorted = arr.slice().sort(function(a,b){return ba}) var ranks = arr.map(function(v){ return sorted.indexOf(v)+1 }); console.log(ranks);

Result :

[2, 7, 4, 7, 3, 9, 5, 9, 1, 6]

If you want to be compatible with old browsers, you may have to define a shim for indexOf and for map (note that if you want to do this very fast for very big arrays, you'd better use for loops and use an object as map instead of indexOf ).

Answer 2

This won't work with older browsers because it uses ECMAScript 5 features , but it allows you to quickly and succinctly produce an array of rankings even for very large arrays. (It doesn't use indexOf which does a linear search and thus can be slow for large arrays.)

function cmp_rnum(a,b) {
    // comparison function: reverse numeric order
    return b-a;
}
function index_map(acc, item, index) {
    // reduction function to produce a map of array items to their index
    acc[item] = index;
    return acc;
}
function ranks(v) {
    var rankindex = v.slice().sort(cmp_rnum).reduceLeft(index_map, Object.create(null));
    // reduceLeft() is used so the lowest rank wins if there are duplicates
    // use reduce() if you want the highest rank
    return v.map(function(item){ return rankindex[item]+1; });
}

Example output:

> ranks([79, 5, 18, 5, 32, 1, 16, 1, 82, 13]);
  [2, 7, 4, 7, 3, 9, 5, 9, 1, 6]

Answer 3

function rank(arr, f) {
    return arr
    .map((x, i) => [x, i])
    .sort((a, b) => f(a[0], b[0]))
    .reduce((a, x, i, s) => (a[x[1]] =
        i > 0 && f(s[i - 1][0], x[0]) === 0 ? a[s[i - 1][1]] : i + 1, a), []);
}

Usage:

rank([79, 5, 18, 5, 32, 1, 16, 1, 82, 13], (a, b) => b - a);
// [2, 7, 4, 7, 3, 9, 5, 9, 1, 6] 

Looks a bit ugly, but it doesn't use indexOf() or an object/map, so not only does it run a little faster, but more importantly, it respects the meaning of "same rankedness" as defined by the comparison function . If one uses indexOf() or an object, "same rankedness" can only mean a === b or String(a) === String(b) .

Alternatively, use findIndex() :

function rank(arr, f) {
    const sorted = arr.slice().sort(f)
    return arr.map(x => sorted.findIndex(s => f(x, s) === 0) + 1)
}

Answer 4

JavaScript ES6 simple two lines solution.

 var arrayRankTransform = arr => { const sorted = [...arr].sort((a, b) => b - a); return arr.map((x) => sorted.indexOf(x) + 1); }; console.log(arrayRankTransform([79, 5, 18, 5, 32, 1, 16, 1, 82, 13]));

Answer 5

I am not good at Javascript but in PHP it can be done quite easily the following way. Somebody good at JavaScript can come up with the relevant code.

$marks = [79, 5, 18, 5, 32, 1, 16, 1, 82, 13];

public function getRank($marks) {
    $rank = 1; $count = 0; $ranks = [];
    //sort the marks in the descending order
    arsort($marks,1);
    foreach($marks as $mark) {
      //check if this mark is already ranked
      if(array_key_exists($mark, $ranks)) {
       //increase the count to keep how many times each value is repeated
       $count++;
       //no need to give rank - as it is already given
      } else {
        $ranks[$mark] = $i+$j;
        $i++;
      }
    return $ranks;
}

Answer 6

I needed the same piece of code for an operations scheduling script I was writing. I used objects and their properties/keys, which can have any value and can be accessed whenever needed. Also, as far as I read in some articles, the search of properties in objects can be faster than search in arrays.

The script below has three simple steps:

1. sort the values (ascending or descending doesn't matter for the rest of the script)

2. find the ranks and number of occurrences for each value

3. replace the given values with ranks using the data from step 2

Note! The below script will not output duplicate ranks, but instead increments ranks for duplicate values/elements.

function rankArrayElements( toBeRanked ) {

// STEP 1
var toBeRankedSorted = toBeRanked.slice().sort( function( a,b ) { return b-a; } ); // sort descending
//var toBeRankedSorted = toBeRanked.slice().sort( function( a,b ) { return a-b; } ); // sort ascending

var ranks = {}; // each value from the input array will become a key here and have a rank assigned
var ranksCount = {}; // each value from the input array will become a key here and will count number of same elements

// STEP 2
for (var i = 0; i < toBeRankedSorted.length; i++) { // here we populate ranks and ranksCount
    var currentValue = toBeRankedSorted[ i ].toString();

    if ( toBeRankedSorted[ i ] != toBeRankedSorted[ i-1 ] ) ranks[ currentValue ] = i; // if the current value is the same as the previous one, then do not overwrite the rank that was originally assigned (in this way each unique value will have the lowest rank)
    if ( ranksCount[ currentValue ] == undefined ) ranksCount[ currentValue ] = 1; // if this is the first time we iterate this value, then set count to 1
    else ranksCount[ currentValue ]++; // else increment by one
}

var ranked = [];

// STEP 3
for (var i = toBeRanked.length - 1; i >= 0; i--) { // we need to iterate backwards because ranksCount starts with maximum values and decreases
    var currentValue = toBeRanked[i].toString();

    ranksCount[ currentValue ]--;
    if ( ranksCount[ currentValue ] < 0 ) { // a check just in case but in theory it should never fail
        console.error( "Negative rank count has been found which means something went wrong :(" );
        return false;
    }
    ranked[ i ] = ranks[ currentValue ]; // start with the lowest rank for that value...
    ranked[ i ] += ranksCount[ currentValue ]; // ...and then add the remaining number of duplicate values
}

return ranked;}

I also needed to do something else for my script.

The above output has the following meaning:

• index - the ID of the element in the input array

• value - the rank of the element from the input array

And I needed to basically 'swap the index with the value', so that I have a list of element IDs, arranged in the order of their ranks:

function convertRanksToListOfElementIDs( ranked ) {  // elements with lower ranks will be first in the list

var list = [];

for (var rank = 0; rank < ranked.length; rank++) { // for each rank...
    var rankFound = false;
    for (var elementID = 0; elementID < ranked.length; elementID++) { // ...iterate the array...
        if ( ranked[ elementID ] == rank ) { // ...and find the rank
            if ( rankFound ) console.error( "Duplicate ranks found, rank = " + rank + ", elementID = " + elementID );
            list[ rank ] = elementID;
            rankFound = true;
        }
    }
    if ( !rankFound ) console.error( "No rank found in ranked, rank = " + rank );
}

return list;}

And some examples:

ToBeRanked:

[36, 33, 6, 26, 6, 9, 27, 26, 19, 9]

[12, 12, 19, 22, 13, 13, 7, 6, 13, 5]

[30, 23, 10, 26, 18, 17, 20, 23, 18, 10]

[7, 7, 7, 7, 7, 7, 7, 7, 7, 7]

[7, 7, 7, 7, 7, 2, 2, 2, 2, 2]

[2, 2, 2, 2, 2, 7, 7, 7, 7, 7]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

rankArrayElements( ToBeRanked ):

[0, 1, 8, 3, 9, 6, 2, 4, 5, 7]

[5, 6, 1, 0, 2, 3, 7, 8, 4, 9]

[0, 2, 8, 1, 5, 7, 4, 3, 6, 9]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[5, 6, 7, 8, 9, 0, 1, 2, 3, 4]

[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

convertRanksToListOfElementIDs( rankArrayElements( ToBeRanked ) ):

[0, 1, 6, 3, 7, 8, 5, 9, 2, 4]

[3, 2, 4, 5, 8, 0, 1, 6, 7, 9]

[0, 3, 1, 7, 6, 4, 8, 5, 2, 9]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

[5, 6, 7, 8, 9, 0, 1, 2, 3, 4]

[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Answer 7

IMHO several solutions here are incorrect as they do not correctly handle values that occur after repeated values. Such followers should get the next rank. The highest rank should equal the number of unique values in the array. This solution (in PHP) is, IMHO, correct. Basically @Suresh's solution with the bugs removed.

  function rank($marks){
    $rank = 1; $ranks = [];
    rsort($marks,SORT_NUMERIC);
    foreach($marks as $mark) {
      if(!isset($ranks[$mark])) {
        $ranks[$mark] = $rank++;
      }
    }
    return $ranks;
   }

Answer 8

This should work with duplicate keys in the array

function rank(arry) {
    let sorted = arry.slice().sort(function (a, b) {
        return b - a
    });


    let currentRank = sorted.length;
    let rankValue = null;
    let ranks = [];

    sorted.forEach(value => {
        if(value !== rankValue && rankValue !==null) {
            currentRank--;
        }

        ranks.push({value,currentRank});
        rankValue = value;
    });

    let mapRanksToArrayValues = arry.map(function (x) {
        let _rank = null;
        ranks.forEach( rank => {
            if(rank.value === x ) {
                _rank =  rank.currentRank;
                return;
            }
        });
        return _rank;
    });

    return mapRanksToArrayValues;
}

Answer 9

I created Rank_JS Pro.

<script>https://cdn.statically.io/gl/maurygta2/mquery/master/Rank Tools/rank.js</script>

Basics Methods:

var a = {
  b: 2,
  c: 7
}
Rank_Tools.rank(a,(pos,name,value) => {
  return pos + ". "+name+" "+value;
})
// result
// rank1 = 1. c 7
// rank 2 = 2. b 2

Answer 10

This alternate way doesn't require the input array to be sorted:

 // O(n^2) const rank = (arr) => { // Create a temporary array to keep metadata // regarding each entry of the original array const tmpArr = arr.map(v => ({ value: v, rank: 1, })); // Get rid of douplicate values const unique = new Set(arr); // Loops through the set for (let a of unique) { for (let b of tmpArr) { // increment the order of an element if a larger element is pressent if (b.value < a) { b.rank += 1; } } } // Strip out the unnecessary metadata return tmpArr.map(v => v.rank); }; console.log(rank([2600, 200, 36, 36, 400, 2, 0, 0])); // => [1, 3, 4, 4, 2, 5, 6, 6] console.log(rank([79, 5, 18, 5, 32, 1, 16, 1, 82, 13])); // => [2, 7, 4, 7, 3, 8, 5, 8, 1, 6]

Answer 11

I had the same homework and this one works well, also it's easier to understand if you are new to this.

function rankings(arr) {
    let rankingsArr = [];
    for (let i = 0; i < arr.length; i++) {
        var rank = 1;
        for (let j = 0; j < arr.length; j++) {
            if (arr[j] > arr[i]) rank++;
        }
        rankingsArr.push(rank);
    }
    return rankingsArr;
}