[1,2,3].CONTAINS([1,2]) ==> true
[1,2,3].CONTAINS([1,2,3,4]) ==> false
or
{a:1,b:2,c:3}.HASKEYS([a,b]) ==> true
{a:1,b:2,c:3}.HASKEYS([a,b,c,d]) ==> false
Is there a single function to check an array contains another array?
-
1+1 as there's nothing wrong with the question. I don't believe there is a ready one, you'll be better off writing your own.ericosg– ericosg2013-04-24 05:25:10 +00:00Commented Apr 24, 2013 at 5:25
Add a comment
|
1 Answer
No, but you can make one:
Array.prototype.contains = function(other) {
for (var i = 0; i < other.length; i++) {
if (this.indexOf(other[i]) === -1) return false;
}
return true;
}
And if order matters:
Array.prototype.contains = function(other) {
var broken;
if (!other.length) return true;
for (var i = 0; i < this.length - other.length + 1; i++) {
broken = false;
for (var j = 0; j < other.length; j++) {
if (this[i + j] !== other[j]) {
broken = true;
break;
}
}
if (!broken) return true;
}
return false;
}
The other function is similar, so I'll leave it to you to finish:
Object.prototype.has_keys = function(keys) {
...
}
3 Comments
Timothée Groleau
The question is likely not detailed enough, but from it, I was expecting "[1,2,3].contains([2,1])" should return false. Your implementation makes it return true.
Timothée Groleau
nicely done @Blender, I think another loop is needed in there to detect all occurrences of other[0]. Otherwise the following returns false: "[1,2,3,2,1].contains([2,1])"
Blender
@TimothéeGroleau: Yep, thanks. I have a feeling that there may be a nicer way of writing it.