Word Count in Swift

Question

What is a more elegant way of writing a simple word count function in Swift?

//Returns a dictionary of words and frequency they occur in the string
func wordCount(s: String) -> Dictionary<String, Int> {
    var words = s.componentsSeparatedByString(" ")
    var wordDictionary = Dictionary<String, Int>()
    for word in words {
        if wordDictionary[word] == nil {
            wordDictionary[word] = 1
        } else {
            wordDictionary.updateValue(wordDictionary[word]! + 1, forKey: word)
        }
    }
    return wordDictionary
}

wordCount("foo foo foo bar")
// Returns => ["foo": 3, "bar": 1]

I intend to leave this question as is. The Swift community wins when we can learn about language features by reading alternative implementations. 'Elegance' is taking advantage of syntactic sugar, readability, simplicity and even performance. Also, there is plenty of precedent for allowing subjective questions that evoke lots of different answers. See [here](http://stackoverflow.com/questions/14994391/how-do-i-think-in-angularjs-if-i-have-a-jquery-background) as an example. — mmla, Jun 09 '14 at 07:47

score 11 · Accepted Answer · answered Jun 09 '14 at 00:15

Your method was pretty solid, but this makes a couple improvements. I store the value count using Swifts "if let" keyword to check for an optional value. Then I can use count when updating the dictionary. I used the shorthand notation for updateValue (dict[key] = val). I also split the original string on all whitespace instead of just a single space.

func wordCount(s: String) -> Dictionary<String, Int> {
    var words = s.componentsSeparatedByCharactersInSet(NSCharacterSet.whitespaceCharacterSet())
    var wordDictionary = Dictionary<String, Int>()
    for word in words {
        if let count = wordDictionary[word] {
            wordDictionary[word] = count + 1
        } else {
            wordDictionary[word] = 1
        }
    }
    return wordDictionary
}

In Swift 5 you can get the list of words with `var words = s.components(separatedBy: NSCharacterSet.whitespaces)`. — Roy, May 26 '21 at 12:09

drewag · Answer 2 · 2014-07-07T18:51:47.313

I don't think this is more elegant because the readability is terrible and it requires and extension on Dictionary but it was really fun to write and shows you the potential power of swift:

extension Dictionary {
    func merge(with other: [KeyType:ValueType], by merge: (ValueType, ValueType) -> ValueType) -> [KeyType:ValueType] {
        var returnDict = self
        for (key, value) in other {
            var newValue = returnDict[key] ? merge(returnDict[key]!, value) : value
            returnDict.updateValue(newValue, forKey: key)
        }
        return returnDict
    }
}

func wordCount(s: String) -> [String:Int] {
    return s.componentsSeparatedByString(" ").map {
        [$0: 1]
    }.reduce([:]) {
        $0.merge(with: $1, +)
    }
}
wordCount("foo foo foo bar")

I do think that merge extension would be useful in other circumstances though

I learned like three + things about Swift in this example. Thanks! — mmla, Jun 09 '14 at 01:07

score 1 · Answer 3 · answered Jun 09 '14 at 00:18

Couldn't find any traces of a Counter type class in the Collections library. You can improve the code slightly by using optional chaining.

func wordCount(s: String) -> Dictionary<String, Int> {
    var words = s.componentsSeparatedByString(" ")
    var wordDictionary = Dictionary<String, Int>()
    for word in words {
        if wordDictionary[word]? {
            wordDictionary[word] = wordDictionary[word]! + 1
        } else {
            wordDictionary[word] = 1
        }
    }
    return wordDictionary
}

wordCount("foo foo foo bar")

score 0 · Answer 4 · answered Jun 09 '14 at 04:27

maybe not elegant but was fun to implement

let inString = "foo foo foo bar"

func wordCount(ss: String) -> Dictionary<String, Int> {
    var dict = Dictionary<String, Int>()
    var tempString = "";var s = ss + " "
    for each in s{
        if each == " "{ if dict[tempString] {
                var tempNumber = Int(dict[tempString]!)
                dict[tempString] = Int(tempNumber+1)
                tempString = ""
            } else {dict[tempString] = 1;tempString = ""}
        } else {
            tempString += each}}
    return dict
}

println(wordCount(inString))

Word Count in Swift

4 Answers4