Number of occurrences of substring in string in Swift

I'd recommend an extension to string in Swift 3 such as:

extension String {
    func countInstances(of stringToFind: String) -> Int {
        var stringToSearch = self
        var count = 0
        while let foundRange = stringToSearch.range(of: stringToFind, options: .diacriticInsensitive) {
            stringToSearch = stringToSearch.replacingCharacters(in: foundRange, with: "")
            count += 1
        }
        return count
    }
}

It's a loop that finds and removes each instance of the stringToFind, incrementing the count on each go-round. Once the searchString no longer contains any stringToFind, the loop breaks and the count returns.

Note that I'm using .diacriticInsensitive so it ignore accents (for example résume and resume would both be found). You might want to add or change the options depending on the types of strings you want to find.

A simple approach would be to split on "Swift", and subtract 1 from the number of parts:

let s = "hello Swift Swift and Swift"
let tok =  s.components(separatedBy:"Swift")
print(tok.count-1)

This code prints 3.

Edit: Before Swift 3 syntax the code looked like this:

let tok =  s.componentsSeparatedByString("Swift")

Should you want to count characters rather than substrings:

extension String {
    func count(of needle: Character) -> Int {
        return reduce(0) {
            $1 == needle ? $0 + 1 : $0
        }
    }
}

Optimising dwsolbergs solution to count faster. Also faster than componentsSeparatedByString.

extension String {
    /// stringToFind must be at least 1 character.
    func countInstances(of stringToFind: String) -> Int {
        assert(!stringToFind.isEmpty)
        var count = 0
        var searchRange: Range<String.Index>?
        while let foundRange = range(of: stringToFind, options: [], range: searchRange) {
            count += 1
            searchRange = Range(uncheckedBounds: (lower: foundRange.upperBound, upper: endIndex))
        }
        return count
    }
}

Usage:

// return 2
"aaaa".countInstances(of: "aa")

• If you want to ignore accents, you may replace options: [] with options: .diacriticInsensitive like dwsolbergs did.
• If you want to ignore case, you may replace options: [] with options: .caseInsensitive like ConfusionTowers suggested.
• If you want to ignore both accents and case, you may replace options: [] with options: [.caseInsensitive, .diacriticInsensitive] like ConfusionTowers suggested.
• If, on the other hand, you want the fastest comparison possible and you can guarantee some canonical form for composed character sequences, then you may consider option .literal and it will only perform exact matchs.