array problem in javascript : Logic

Question

i have this two arrays of equal length.

array1 = {a,a,a,b,b,b,b,b,c,c,c,c,c,c,d,d,d,d,e,e}
array2 = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}

i want to create an object.

var obj = {
'a': {1,2,3} 
'b': {4,5,6,7,8}
'c': {9,10,11,12,13,14}
 .....
 .....
}

can somebody help me with the logic.

In the example you are creating two objects for array1, array2, typo? — supakeen
– supakeen, Commented Jul 21, 2011 at 6:53
Please consider this as a symbolic array. They are arrays i am sorry for syntax. — sushil bharwani
– sushil bharwani, Commented Jul 21, 2011 at 6:56
Have you tried to write an algorithm for this? What went wrong? — Aleksi Yrttiaho
– Aleksi Yrttiaho, Commented Jul 21, 2011 at 6:57

supakeen · Accepted Answer · 2011-07-21 08:54:16Z

4

Assuming you mean arrays in your example (names + question title):

var combined = {};

for(var i = 0; i < array1.length; i++) {
    var key = array1[i];

    if(!(key in combined)) {
        combined[key] = [];
    }

    combined[key].push(array2[i]);
}

edited Jul 21, 2011 at 8:54

answered Jul 21, 2011 at 6:58

supakeen

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4 Comments

Paul Phillips Over a year ago

Maybe I am missing something, but you define the function "include" and do not call it anywhere.

Tomalak Over a year ago

It's nice that the include function comes from somewhere, but what do you use it for in this example? ;)

QuentinUK Over a year ago

Instead of "!key in combined" use !(key in combined)

supakeen Over a year ago

Oh deary me, I was thinking I needed it and then didn't actually need it. Let's just keep it at the early time in my timezone. I'll edit it out.

Tomalak · Accepted Answer · 2011-07-21 07:02:15Z

3

var group = {};

if (array1.length == array2.length) {
  for (var i=0, j=array1.length; i<j; i++) {
    if ( !(array1[i] in group) ) group[array1[i]] = [];
    group[array1[i]].push(array2[i]);
  }
}

answered Jul 21, 2011 at 7:02

Tomalak

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Comments

Justin Poliey · Accepted Answer · 2011-07-21 07:04:07Z

2

var array1 = ['a','a','a','b','b','b','b','b','c','c','c','c','c','c','d','d','d','d','e','e'];
var array2 = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20];
var obj = {};
for (var i = 0, key; (key = array1[i]); i++) {
  if (!obj[key]) {
    obj[key] = [];
  }
  obj[key].push(array2[i]);
}

Should do the trick.

answered Jul 21, 2011 at 7:04

Justin Poliey

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4 Comments

Tomalak Over a year ago

+1 Nice variation on the for loop, but (!obj[key]) is too sloppy.

Tomalak Over a year ago

Thinking about it, (key = array1[i]) also is too sloppy. What if array1[i] is something that would be a valid object key value but evaluates to false, like the empty string?

Justin Poliey Over a year ago

This wasn't written for the general case with arbitrary keys and values. I see where you're calling slopiness and it's true, but only for the general case. If he's never going to use a falsy value, why worry about them?

Tomalak Over a year ago

Well, do you know it? All the OP said was "two arrays of equal length" and a generic sample. After all, if you wrapped that up in a function, it would have a bug.

Suman Basyal · Accepted Answer · 2016-07-21 15:59:03Z

var array1 = ['a','a','a','b','b','b','b','b','c','c','c','c','c','c','d','d','d','d','e','e']
var array2 = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]
var travesedArrayIndex=[];
var Component=function(name, count)
{
  this.name=name;
  this.number=count;
}
var componentList=[];
for(i=0;i<array1.length; i++)
{
  var count=0;
  var valueToBeComparedWith=array1[i];
  for(j=0;j<array1.length;j++)
  {
    if(!AlreadyTraversed(j))
    {
      if(valueToBeComparedWith==array1[j])
      {
        count=count+1;
        travesedArrayIndex.push(j);
      }
    }
  }
  if(count>0)
  {
    var comp= new Component(valueToBeComparedWith,count);
    componentList.push(comp);  
  }
}

 var FinalComponent=function()
 {
  this.item;
  this.numberArray=[];
 }
 var FinalComponentList=[];
 var j=0;
 for(i=0;i<componentList.length;i++)
 {
    var finalComp= new FinalComponent();
    finalComp.item=componentList[i].name;
    var sizeOfArray=componentList[i].number + j;
    for(j;j<sizeOfArray;j++)
    {
      finalComp.numberArray.push(array2[j]);
    }
    FinalComponentList.push(finalComp);
 }
console.log(FinalComponentList);
function AlreadyTraversed(index)
{
  for(k=0;k<travesedArrayIndex.length;k++)
  {
    if(index==travesedArrayIndex[k])
    {
      return true;
    }
  }
  return false;
}

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