Removing from array during enumeration in Swift?

Solution 1:

In Swift 2 this is quite easy using enumerate and reverse.

var a = [1,2,3,4,5,6]
for (i,num) in a.enumerate().reverse() {
    a.removeAtIndex(i)
}
print(a)

Solution 2:

You might consider filter way:

var theStrings = ["foo", "bar", "zxy"]

// Filter only strings that begins with "b"
theStrings = theStrings.filter { $0.hasPrefix("b") }

The parameter of filter is just a closure that takes an array type instance (in this case String) and returns a Bool. When the result is true it keeps the element, otherwise the element is filtered out.

Solution 3:

In Swift 3 and 4, this would be:

With numbers, according to Johnston's answer:

var a = [1,2,3,4,5,6]
for (i,num) in a.enumerated().reversed() {
   a.remove(at: i)
}
print(a)

With strings as the OP's question:

var b = ["a", "b", "c", "d", "e", "f"]

for (i,str) in b.enumerated().reversed()
{
    if str == "c"
    {
        b.remove(at: i)
    }
}
print(b)

However, now in Swift 4.2 or later, there is even a better, faster way that was recommended by Apple in WWDC2018:

var c = ["a", "b", "c", "d", "e", "f"]
c.removeAll(where: {$0 == "c"})
print(c)

This new way has several advantages:

1. It is faster than implementations with filter.
2. It does away with the need of reversing arrays.
3. It removes items in-place, and thus it updates the original array instead of allocating and returning a new array.

Solution 4:

When an element at a certain index is removed from an array, all subsequent elements will have their position (and index) changed, because they shift back by one position.

So the best way is to navigate the array in reverse order - and in this case I suggest using a traditional for loop:

for var index = array.count - 1; index >= 0; --index {
    if condition {
        array.removeAtIndex(index)
    }
}

However in my opinion the best approach is by using the filter method, as described by @perlfly in his answer.