Removing objects from an array based on another array

The easiest way is by using the new Set container (added in Swift 1.2 / Xcode 6.3):

var setA = Set(arrayA)
var setB = Set(arrayB)

// Return a set with all values contained in both A and B
let intersection = setA.intersect(setB) 

// Return a set with all values in A which are not contained in B
let diff = setA.subtract(setB)

If you want to reassign the resulting set to arrayA, simply create a new instance using the copy constructor and assign it to arrayA:

arrayA = Array(intersection)

The downside is that you have to create 2 new data sets. Note that intersect doesn't mutate the instance it is invoked in, it just returns a new set.

There are similar methods to add, subtract, etc., you can take a look at them

I agree with Antonio's answer, however for small array subtractions you can also use a filter closure like this:

let res = arrayA.filter { !contains(arrayB, $0) }

Like this:

var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
for word in arrayB {
    if let ix = find(arrayA, word) {
        arrayA.removeAtIndex(ix)
    }
}
// now arrayA is ["Mike", "Stacey"]

@francesco-vadicamo's answer in Swift 2/3/4+

 arrayA = arrayA.filter { !arrayB.contains($0) }