Reputation:
Let's say we have the following:
node[1].name = "apple";
node[1].color = "red";
node[2].name = "cherry";
node[2].color = "red";
node[3].name = "apple";
node[3].color = "green";
node[4].name = "orange";
node[4].color = "orange;
if I use jQuery.unique(node) I will get all the original nodes because they all have a different name OR color. What I want to do is only get the nodes with a unique name, which should return
node[1] (apple)
node[2] (cherry)
node[4] (orange)
It should not return 3 because it is the same fruit, even though we have green and red apples.
Upvotes: 4
Views: 4543
Reputation: 30330
This approach (forked from @David's) should have better performance for large inputs (because object[]
is O(1)
).
function filter(arr, attribute) {
var out = [],
seen = {}
return arr.filter(function(n) {
return (seen[n[attribute]] == undefined)
&& (seen[n[attribute]] = 1);
})
}
console.log(filter(node, 'name'));
Upvotes: 2
Reputation: 108490
Use Array.filter
and a temporary array to store the dups:
function filterByName(arr) {
var f = []
return arr.filter(function(n) {
return f.indexOf(n.name) == -1 && f.push(n.name)
})
}
Demo: http://jsfiddle.net/mbest/D6aLV/6/
Upvotes: 7
Reputation:
What about doing it like this?
var fruitNames = [];
$.each($.unique(fruits), function(i, fruit) {
if (fruitNames.indexOf(fruit.name) == -1) {
fruitNames.push(fruit.name);
$('#output').append('<div>' + fruit.name + '</div>');
}
});
Obviously, instead of output.append I could just add the current fruit to uniqueFruit[] or something.
Upvotes: 0