ActionScript - Comparing & Removing Duplicates of Complex Arrays? - actionscript-3

Question

when comparing simple arrays, i use something like the following function to concatenate and remove duplicates:
//Merge
public function merge(a1:Array, a2:Array):Array
{
var result:Array = a1.concat(a2);
var dictionary:Dictionary = new Dictionary();
for each (var item:Object in result)
dictionary[item] = true;
result = new Array();
for (var key:Object in dictionary)
result.push(key);
dictionary = null;
return result;
}
however, this approach doesn't work on complex arrays.
is there a well known algorithm, or is it even possible to write a function of recursion that can compare a Vector.<Object> with another? one that will always work even if the some objects being compared have additional key/value pairs?
[EDIT]
to be more clear, using a dictionary to determine if items in a simple array only works on primitive data types (int, number, string, etc.) or object references, so the above example works if it's passed 2 arrays resembling something like this:
var arr1:Array = new Array(1, 2, 3, 4, 5);
var arr2:Array = new Array(8, 7, 6, 5, 4);
resulting in a merged array with the following values:
1, 2, 3, 8, 7, 6, 5, 4
in contrast, i'm asking if it's possible to pass a function 2 complex arrays or Vector.<Object> all containing unique objects that may have identical key/value pairs and remove redundencies in the resulting Vector.<Object>. for example:
var vec1:Vector.<Object> = new Vector.<Object>();
vec1.push({city:"Montreal", country:"Canada"});
vec1.push({city:"Halifax", country:"Canada"});
var vec2:Vector.<Object> = new Vector.<Object>();
vec2.push({city:"Halifax", country:"Canada"});
vec2.push({city:"Toronto", country:"Canada"});
merging the above 2 vector objects would result in the following vector by determining and removing objects with identical key/value pairs:
{city:"Montreal", country:"Canada"}
{city:"Halifax", country:"Canada"}
{city:"Toronto", country:"Canada"}
i'm searching for an algorithm which could handle the removal of these similar objects without having to know about their specific key/value names or how many key/value pairs there are within the object.

Answer 1

Sure you can, you can build a similar example with any type of Vector:
public function mergeObjectVectors(v1:Vector.<Object>,
v2:Vector.<Object>):Vector.<Object>
{
var dictionary:Dictionary = new Dictionary();
var concat:Vector.<Object> = v1.concat(v2);
var result:Vector.<Object> = new Vector.<Object>();
for each(var i:Object in concat)
{
if (!dictionary[i])
{
dictionary[i] = true;
result.push(i);
}
}
return result;
}
However if you plan on accepting vectors of any type, it's different:
public function testing():void
{
var v1:Vector.<Object> = new Vector.<Object>();
v1.push({name:"Object 1"});
v1.push({name:"Object 2"});
// Vector w duplicates
var v2:Vector.<Object> = new Vector.<Object>();
var o:Object = {name:"Object 3"};
v2.push(o);
v2.push(o);
v2.push(o);
var resultVector:Vector.<Object> = mergeAnything(v1, v2, Class(Vector.<Object>));
var resultArray:Array = mergeAnything(v1, v2, Array);
var resultObject:Object = mergeAnything(v1, v2, Object);
}
public function mergeAnything(o1:Object, o2:Object, resultClass:Class):*
{
var dictionary:Dictionary = new Dictionary();
var result:Object = new resultClass();
var i:int;
for each(var o:Object in o1)
{
if (!dictionary[o])
{
dictionary[o] = true;
result[i++] = o;
}
}
for each(o in o2)
{
if (!dictionary[o])
{
dictionary[o] = true;
result[i++] = o;
}
}
return result;
}
The first example will be more resource-efficient.
EDIT:
This should do it, try it with your example:
public function mergeObjectVectors(v1:Vector.<Object>, v2:Vector.<Object>):Vector.<Object>
{
var concat:Vector.<Object> = v1.concat(v2);
var result:Vector.<Object> = new Vector.<Object>();
var n:int = concat.length;
loop:for (var i:int = 0; i < n; i++)
{
var objectToAdd:Object = concat[i];
var m:int = result.length;
for (var j:int = 0; j < m; j++)
{
var addedObject:Object = result[j];
if (this.areObjectsIdentical(objectToAdd, addedObject))
{
continue loop;
}
}
result.push(objectToAdd);
}
return result;
}
private function areObjectsIdentical(o1:Object, o2:Object):Boolean
{
var numComparisons:int = 0;
for (var s:String in o1)
{
numComparisons++;
if (o1[s] != o2[s])
{
return false;
}
}
for (s in o2)
{
numComparisons--;
}
return !numComparisons;
}