Group array items based on variable javascript

Question

I have an array that is created dynamic from an xml document looking something like this:

myArray[0] = [1,The Melting Pot,A]
myArray[1] = [5,Mama\'s Mexican         


        

Answer 1

You have to create an empty JavaScript object and assign an array to it for each letter.

var object = {};

for ( var x = 0; x < myArray.length; x++ )
{
    var letter = myArray[x][2];

    // create array for this letter if it doesn't exist
    if ( ! object[letter] )
    {
        object[letter] = [];
    }

    object[ myArray[x][2] ].push[ myArray[x] ];
}

Demo fiddle here.

Answer 2

Good ol' ES5 Array Extras are great.

var other = {};
myArray.forEach(function(n, i, ary){
    other[n[2]] = n.slice(0,2);
});

Answer 3

You shouldn't use arrays with non-integer indexes. Your other variable should be a plain object rather than an array. (It does work with arrays, but it's not the best option.)

// assume myArray is already declared and populated as per the question

var other = {},
    letter,
    i;

for (i=0; i < myArray.length; i++) {
   letter = myArray[i][2];
   // if other doesn't already have a property for the current letter
   // create it and assign it to a new empty array
   if (!(letter in other))
      other[letter] = [];

   other[letter].push(myArray[i]);
}

Given an item in myArray [1,"The Melting Pot","A"], your example doesn't make it clear whether you want to store that whole thing in other or just the string field in the second array position - your example output only has strings but they don't match your strings in myArray. My code originally stored just the string part by saying other[letter].push(myArray[i][1]);, but some anonymous person has edited my post to change it to other[letter].push(myArray[i]); which stores all of [1,"The Melting Pot","A"]. Up to you to figure out what you want to do there, I've given you the basic code you need.

Answer 4

This code will work for your example.

var other = Object.create(null),  // you can safely use in opeator.
    letter,
    item,
    max,
    i;

for (i = 0, max = myArray.length; i < max; i += 1) {
   item = myArray[i];
   letter = myArray[2];

   // If the letter does not exist in the other dict,
   // create its items list
   other[letter] = other[letter] || [];
   other.push(item);
}

Answer 5

Try -

var myArray = new Array();
myArray[0] = [1,"The Melting Pot,A,3,Sake House","B"];
myArray[1] = [5,"Mama's MexicanKitchen","C"];
myArray[2] = [6,"Wingdome","D"];
myArray[3] = [7,"Piroshky Piroshky","D"];
myArray[4] = [4,"Crab Pot","F"];
myArray[5] = [2,"Ipanema Grill","G"];
myArray[6] = [0,"Pan Africa Market","Z"];

var map = new Object();
for(i =0 ; i < myArray.length; i++){
    var key = myArray[i][2];
    if(!map[key]){
       var array = new Array();        
        map[key] = array;
    }
    map[key].push(myArray[i]);

}

Answer 6

Try groupBy function offered by http://underscorejs.org/#groupBy

_.groupBy([1.3, 2.1, 2.4], function(num){ return Math.floor(num); });

Result => {1: [1.3], 2: [2.1, 2.4]}