Order array by predefined rules

Question

I have an array of currencies ["GBP", "EUR", "NOK", "DKK", "SKE", "USD", "SEK", "BGN"]. I would like to order it by moving predefined list if the currency is present to the beginning of array. Predefined list is ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP']. So in this case it should return ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP', 'SEK', BGN'].

But in case unfiltered array does not contain all values in predifined list it should also order it correctly. For example : ["GBP", "EUR", "NOK", "LTU", "ZGN"] should look like ['EUR', 'NOK', 'GBP', 'LTU', 'ZGN'

I was trying to sort it using this function:

list.sort(c => ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP'].indexOf(c))

but it puts all predefined currencies at the end of the list, not at the from. Maybe there is a better way of doing that?

the callback for .sort() takes to arguments - the first and second comparator. you can do this same thing but with list.sort(( c, d ) => [...].indexOf(c) > [...].indexOf(d) ? 1 : [...].indexOf(c) < [...].indexOf(d) ? -1 : 0); but, adjust the > and < signs as needed. — Deryck
– Deryck, Commented Jun 7, 2016 at 6:12

Nina Scholz · Accepted Answer · 2016-06-07 15:17:39Z

2

You could use sorting with map and a hash table for the sort order. If the value is not in the hash table, the original order is taken.

var order = ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP'],
    orderObj = Object.create(null),
    data = ["GBP", "EUR", "NOK", "DKK", "SKE", "USD", "SEK", "BGN"];

// generate hash table
order.forEach((a, i) => orderObj[a] = i + 1);

// temporary array holds objects with position and sort-value
var mapped = data.map((el, i) => { return { index: i, value: orderObj[el] || Infinity }; });

// sorting the mapped array containing the reduced values
mapped.sort((a, b) => a.value - b.value || a.index - b.index);

// assigning the resulting order
var data = mapped.map(el => data[el.index]);

console.log(data);

edited Jun 7, 2016 at 15:17

answered Jun 7, 2016 at 6:26

Nina Scholz

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3 Comments

kuldarim Over a year ago

Good answer. But since i'm using ES6 the last code part can be written in one line

data         .map((el, i) => { return { index: i, value: order[el] || 1000 }; })         .sort((a, b) => a.value - b.value || a.index - b.index)         .map(el => d[el.index]);

Bergi Over a year ago

Instead of 1000, better use the proper value Infinity :-)

Nina Scholz Over a year ago

@Bergi, great idea :-)

Redu · Accepted Answer · 2016-06-07 15:28:37Z

2

I guess this can also be achieved like this

Array.prototype.intersect = function(a) {
  return this.filter(e => a.includes(e));
};
Array.prototype.excludes = function(a) {
  return this.filter(e => !a.includes(e));
};
var getCur = (p,c) => p.intersect(c).concat(c.excludes(p)),
      cur1 = ["GBP", "EUR", "NOK", "DKK", "SKE", "USD", "SEK", "BGN"],
      cur2 = ["GBP", "EUR", "NOK", "LTU", "ZGN"],
       pdl = ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP', 'SEK', 'BGN'];
console.log(getCur(pdl,cur1));
console.log(getCur(pdl,cur2));

edited Jun 7, 2016 at 15:28

answered Jun 7, 2016 at 7:10

Redu

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1 Comment

Bergi Over a year ago

Since you're using ES6 anyway, better go for a.includes(e)

Fabrice Mpenge · Accepted Answer · 2016-06-07 08:26:20Z

0

var tabCurrency = ['GBP', 'EUR', 'NOK', 'DKK', 'SKE', 'USD', 'SEK', 'BGN'];

var tabPredef = ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP'];

var newTabGood = [];

    tabPredef.forEach(function (itemPredef, indexPref) {

         var indexTemp;

         tabCurrency.forEach(function (itemCurrency, indexCurrency) {
               if(itemPredef == itemCurrency)
               {
                  newTabGood.push(itemPredef);
                  indexTemp = indexCurrency;
               }
         })

         tabCurrency.splice(indexTemp, 1)
   }) 


 var resultat = newTabGood.concat(tabCurrency);

 console.log(resultat)

answered Jun 7, 2016 at 8:26

Fabrice Mpenge

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Comments

Brajendra Swain · Accepted Answer · 2016-06-07 18:43:56Z

0

here is my solution. :-)

//custom index of
Array.prototype.customIndexOf = function(a){
  return this.indexOf(a) === -1 ? Infinity : this.indexOf(a);
}

let orderArr = ['EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP'];


/*test case 1*/
let urList = ["GBP", "EUR", "NOK", "DKK", "SKE", "USD", "SEK", "BGN"];
urList.sort((a, b) => { return orderArr.customIndexOf(a) - orderArr.customIndexOf(b); });
console.log(urList); //[ 'EUR', 'USD', 'DKK', 'SKE', 'NOK', 'GBP', 'SEK', 'BGN' ]

/*test case 2*/
let newList = ["GBP", "EUR", "NOK", "LTU", "ZGN"];

newList.sort((a, b) => { return orderArr.customIndexOf(a) - orderArr.customIndexOf(b); });
console.log(newList); //[ 'EUR', 'NOK', 'GBP', 'LTU', 'ZGN' ]

hope this is what u need :-)

edited Jun 7, 2016 at 18:43

answered Jun 7, 2016 at 7:14

Brajendra Swain

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3 Comments

Bergi Over a year ago

Why Number.MAX_VALUE? Just use Infinity.

Brajendra Swain Over a year ago

anything.. we need one big value.. for comparison.. :)

Bergi Over a year ago

But how big? Is 10000 enough? Is 2^31 enough? Better use the biggest value available - Infinity. Admittedly, as indexOf can't return anything larger than an array index, so anything above 2^32 will do, and Number.MAX_VALUE is way above that so it's fine.

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