Say I have an array of [34, 35, 45, 48, 49] and another array of [48, 55]. How can I get a resulting array of [34, 35, 45, 48, 49, 55]?
Say I have an array of [34, 35, 45, 48, 49] and another array of [48, 55]. How can I get a resulting array of [34, 35, 45, 48, 49, 55]?
I'm probably wasting time on a dead thread here. I just had to implement this and went looking to see if I was wasting my time.
I really like KennyTM's answer. That's just how I would attack the problem. Merge the keys into a hash to naturally eliminate duplicates and then extract the keys. If you actually have jQuery you can leverage its goodies to make this a 2 line problem and then roll it into an extension. The each() in jQuery will take care of not iterating over items where hasOwnProperty() is false.
jQuery.fn.extend({
union: function(array1, array2) {
var hash = {}, union = [];
$.each($.merge($.merge([], array1), array2), function (index, value) { hash[value] = value; });
$.each(hash, function (key, value) { union.push(key); } );
return union;
}
});
Note that both of the original arrays are left intact. Then you call it like this:
var union = $.union(array1, array2);
If you want a custom equals function to match your elements, you can use this function in ES2015:
function unionEquals(left, right, equals){
return left.concat(right).reduce( (acc,element) => {
return acc.some(elt => equals(elt, element))? acc : acc.concat(element)
}, []);
}
It traverses the left+right array. Then for each element, will fill the accumulator if it does not find that element in the accumulator. At the end, there are no duplicate as specified by the equals function.
Pretty, but probably not very efficient with thousands of objects.