I'd like to filter this data for unique sub-arrays with the second property, so
var arr = [
["foo", "one"],
["bar", "one"],
["baz", "two"],
["qux", "two"]
]
should turn into
var arr = [
["foo", "one"],
["baz", "two"]
]
I could try a manual approach with two nested forEach, but it seems a inefficient. Array.prototype.filter seems to be better suited, but I fear I'm lacking experience to use it properly.
Another version (assuming your definition of arr exists):
arr = arr.filter(function(innerArr){
return !this[innerArr[1]] && (this[innerArr[1]]=true)
},{})
this[...]=true expression exists in neither the original nor the resulting array, but rather in the global object. So it appears that you could substitute this with either window or self and have the same result of polluting or overwriting global variables (at least in the environment I tested it on).thisArg and using that as my storage to track what I've seen. Therefore this in this case is actually a temp object used in the filter function only, not global or windowIf a thisArg parameter is provided to filter, it will be passed to callback when invoked, for use as its this value. Otherwise, the value undefined will be passed for use as its this value. The this value ultimately observable by callback is determined according to the usual rules for determining the this seen by a function.Store the array which you want to keep against the second element of the array, in an Object. This is to avoid duplicates.
var obj = {};
arr.forEach(function(currentArray) {
if (obj.hasOwnProperty(currentArray[1]) === false) {
obj[currentArray[1]] = currentArray;
}
});
And then collect the values alone, like this
var result = [];
for (var key in obj) {
result.push(obj[key]);
}
console.log(result);
# [ [ 'foo', 'one' ], [ 'baz', 'two' ] ]
The run time of this solution is in O(N).
Note: This solution may not gurantee the order of the items in the result.
The following does what you want I think. http://jsfiddle.net/pitaj/ktj06Ljt/
var seenit = []; // array to store properties already found
var newarr = arr.filter(function(it){
var result = seenit.indexOf(it[1]) === -1; // check if already found
seenit.push(it[1]); // add to found array
return result; // return if it was found, as per filter
});
This way will keep them in the same order as the original.
EDIT: alternative, you can do it without the external array (as Cheery pointed out):
var newarr = arr.filter(function(it){
var result = this.indexOf(it[1]) === -1;
this.push(it[1]);
return result;
}, []);
Maybe something like this:
var arr = [
["foo", "one"],
["bar", "one"],
["baz", "two"],
["qux", "two"]
];
var uniques=[
Object.create(null),
Object.create(null)
];
var result=arr.filter(function(item_set){
return item_set.every(function(item,idx){
var u=!uniques[idx][item];
uniques[idx][item]=true;
return u;
});
});
/*
result=[["foo", "one"], ["baz", "two"]];
*/
* correction: changed uniques[item] to uniques[idx][item] so each item of the pair accesses its own lookup table.
With underscore.js, you can just type code below:
_.union([1, 2, 3], [101, 2, 1, 10], [2, 1]);
=> [1, 2, 3, 101, 10]
I'm so sorry about that i misread your question, The answer above is wrong.
I write a new demo for your question: http://jsfiddle.net/abruzzi/0an5yr9c/2/
and here are codes below:
var arr = [
["foo", "one"],
["bar", "one"],
["baz", "two"],
["qux", "two"]
];
var uniq = [];
var result = arr.filter(function(sub_arr){
return sub_arr.every(function(sub_item){
if(uniq.indexOf(sub_item) < 0) {
uniq.push(sub_item);
return true;
}
return false;
});
})
result => [
["foo", "one"],
["baz", "two"]
]
mapfunction. developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…)