Two faceted question:
var array = [1,2,3,4,5]
contains(array, 0) // false
var array2: NSArray = [1,2,3,4,5]
array2.containsObject(4) // true
Is there any way to search an Array for more than 1 value? ie. Can I write below to search the array for multiple values and return true if any of the values are found? Second part to the question is how can I do that for an NSArray as well?
var array = [1,2,3,4,5]
contains(array, (0,2,3)) // this doesn't work of course but you get the point
You can chain contains together with a second array:
// Swift 1.x
contains(array) { contains([0, 2, 3], $0) }
// Swift 2 (as method)
array.contains{ [0, 2, 3].contains($0) }
// and since Xcode 7 beta 2 you can pass the contains function which is associated to the array ([0, 2, 3])
array.contains([0, 2, 3].contains)
// Xcode 12
array.contains(where: [0, 2, 3].contains)
One option would be to use a Set
for the search terms:
var array = [1,2,3,4,5]
let searchTerms: Set = [0,2,3]
!searchTerms.isDisjointWith(array)
(You have to negate the value of isDisjointWith
, as it returns false
when at least one of the terms is found.)
Note that you could also extend Array
to add a shorthand for this:
extension Array where Element: Hashable {
func containsAny(searchTerms: Set<Element>) -> Bool {
return !searchTerms.isDisjointWith(self)
}
}
array.containsAny([0,2,3])
As for the NSArray
, you can use the version of contains
which takes a block to determine the match:
var array2: NSArray = [1,2,3,4,5]
array2.contains { searchTerms.contains(($0 as! NSNumber).integerValue) }
Explanation of closure syntax (as requested in comments): you can put the closure outside the ()
of method call if it's the last parameter, and if it's the only parameter you can omit the ()
altogether. $0
is the default name of the first argument to the closure ($1
would be the second, etc). And return
may be omitted if the closure is only one expression. The long equivalent:
array2.contains({ (num) in
return searchTerms.contains((num as! NSNumber).integerValue)
})
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