split the string

Question

Created Sep ’20

Replies 3

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Participants 3

Hi,
Id like to split the string = "aabbccddee". .split function requires you to use some sort of separator. Any suggestion how to make it?

The main task is actually to count how many times the letters appear in a string. I believe the solution may be related to .map.

Any help appreciated.

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Answer 1

DTS Engineer OP

Apple

Sep ’20

In Swift a String is collection of Characters, so you can do things like this:

Code Block  for ch in "aabbccddee" {
    print(ch)
}

IMPORTANT In Swift a Character is not a simple ASCII character but rather an extended grapheme cluster. See the docs for more.

This means you can apply collection methods on those characters. For example, you can use filter(_:) like so:

Code Block  "aabbccddee".filter { $0 == "a" }

One common idiom is to chain these methods together, like this:

Code Block  "aabbccddee"
    .filter { $0 == "a" }
    .count

The main task is actually to count how many times the letters appear
in a string.

My weapon of choice for that is Dictionary.init(_:uniquingKeysWith:).

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Quinn “The Eskimo!” @ Developer Technical Support @ Apple
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Answer 2

Claude31 OP

Sep ’20

I once wrote an extension to do this:

Code Block extension Sequence where Element: Hashable { //extension on sequence where element conforms to hashable protocol
    
    var frequencies: [Element: Int] { //dictionary of key element and value int
        return Dictionary(self.map{ ($0, 1)}, uniquingKeysWith: { $0 + $1} )  //creates new dictionary for key-value pairs in sequence and increments the value for duplicate keys
    }
}
extension String {
    var frequencies: [Character: Int] {
        return Array(self).frequencies
    }
}
"aabbccddee".frequencies

yields (dictionary is not ordered)

["e": 2, "a": 2, "b": 2, "c": 2, "d": 2]

Or
"hello you".frequencies
["h": 1, " ": 1, "e": 1, "l": 2, "o": 2, "u": 1, "y": 1]

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Answer 3

Claude31 OP

Oct ’20

To further detail how frequencies is built, the expanded version of

Code Block  return Dictionary(self.map{ ($0, 1)}, uniquingKeysWith: { $0 + $1} )

would be

Code Block let partiel = Array("aabbccddee").map{ ($0, 1)}  // split String into an Array of pairs with the char and 1 as value
// partiel = [("a", 1), ("a", 1), ("b", 1), ("b", 1), ("c", 1), ("c", 1), ("d", 1), ("d", 1), ("e", 1), ("e", 1)]
// Then build a Dict from this Array ; as each key must be unique, we define how to combine when we find an already used key: add the nextValue to the existing one
let counted = Dictionary(partiel, uniquingKeysWith: { (prevCountForKey, nextValue) in prevCountForKey + nextValue} )    // nextValue is 1 here
// counted = "c": 2, "b": 2, "a": 2, "d": 2, "e": 2]
// Closure can then simply be written { $0 + $1}

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