Number of words in a Swift String for word count calculation

Question

I want to make a procedure to find out how many words are there in a string, separated by space, or comma, or some other character. And then add up the total later.

I'm making an average calculator, so I want the total count of data and then add up all the words.

Answer 1

update: Xcode 10.2.x ? Swift 5 or later

Using Foundation method enumerateSubstrings(in: Range)and setting .byWords as options:

let sentence = "I want to an algorithm that could help find out how many words are there in a string separated by space or comma or some character. And then append each word separated by a character to an array which could be added up later I'm making an average calculator so I want the total count of data and then add up all the words. By words I mean the numbers separated by a character, preferably space Thanks in advance"

var words: [Substring] = []
sentence.enumerateSubstrings(in: sentence.startIndex..., options: .byWords) { _, range, _, _ in
    words.append(sentence[range])
}
print(words) // "["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I\'m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]
"
print(words.count)  // 79

Or using native Swift 5 new Character property isLetter and the split method:

let words =  sentence.split { !$0.isLetter }

print(words) // "["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I", "m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]
"

print(words.count)  // 80

---

Extending StringProtocol to support Substrings as well:

extension StringProtocol {
    var words: [SubSequence] { 
        return split { !$0.isLetter } 
    }
    var byWords: [SubSequence] {
        var byWords: [SubSequence] = []
        enumerateSubstrings(in: startIndex..., options: .byWords) { _, range, _, _ in
            byWords.append(self[range])
        }
        return byWords
    }
}

---

sentence.words  // ["I", "want", "to", "an", "algorithm", "that", "could", "help", "find", "out", "how", "many", "words", "are", "there", "in", "a", "string", "separated", "by", "space", "or", "comma", "or", "some", "character", "And", "then", "append", "each", "word", "separated", "by", "a", "character", "to", "an", "array", "which", "could", "be", "added", "up", "later", "I", "m", "making", "an", "average", "calculator", "so", "I", "want", "the", "total", "count", "of", "data", "and", "then", "add", "up", "all", "the", "words", "By", "words", "I", "mean", "the", "numbers", "separated", "by", "a", "character", "preferably", "space", "Thanks", "in", "advance"]