Is there an easy way of knowing if and array contains all elements of another array?
Example:
var myarray = [1,2,3];
var searchinarray = [1,2,3,4,5];
searchinarray contains all elements of myarray, so this should return true.
Regards
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3 Answers
Here is an implementation of that function:
function contains(a, s) {
for(var i = 0, l = s.length; i < l; i++) {
if(!~a.indexOf(s[i])) {
return false;
}
}
return true;
}
This shows that it does what you describe:
> var myarray = [1,2,3];
> var searchinarray = [1,2,3,4,5];
> contains(myarray, searchinarray)
false
> contains(myarray, [1,2])
true
> contains(myarray, [2,3])
true
1 Comment
RobG
This should test that
s[i] actually exists, or use a method like forEach that only visits members of s that exist. e.g. should contains(s, [,,1]) return true or false?A contains function should only visit members of the array to be contained that exist, e.g.
var myArray = [1,,2];
var searchInArray = [1,2,3];
contains(myArray, searchInArray);
should return true, however a simple looping solution will return false as myArray[1] doesn't exist so will return undefined, which is not present in the searchInArray. To easily avoid that and not use a hasOwnProperty test, you can use the Array every method:
function contains(searchIn, array) {
return array.every(function(v){return this.indexOf(v) != -1}, searchIn);
}
so that:
var a = [1,,3];
var s = [1,2,3,4,5];
console.log(contains(s, a)); // true
You can add a contains method to all instances of Array if you wish:
if (typeof Array.prototype.contains == 'undefined') {
Array.prototype.contains = function(a) {
return a.every(function(v, i) {
return this.indexOf(v) != -1;
}, this);
}
}
console.log(s.contains(a)); // true
console.log(s.contains([1,9])); // false
You may need a polyfill for browsers that don't have .every (IE 8?).
Comments
function compare(array, contains_array) {
for(var i = 0; i < array.length; i++) {
if(contains_array.indexOf(array[i]) == -1) return false;
}
return true;
}
compare(myarray, searchinarray);
