Converting an array of strings to an array of concatenated strings

Question

Im looking for a general and functional solution to concatenate x elements of an array to a new array of strings. Beginning from the first element, first with second, second with third and so on ... Sorry for the explanation not being very clear, but it has been hard for me to put on name on what I'm trying to achieve. So I will try to demonstrate it with examples.

1) If I'm trying to concatenate the letters 2 by 2 I'm using this function:

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

   func concatenateStrings(array: [String]) -> [String] {
        return array.reduce([], { (result: [String], item:String) -> [String] in
            guard !result.isEmpty else {
                return [item] //first item
            }

            return result + [ array[result.count - 1] + " \(item)"]
        })
    }

which produces me this result ["a", "a b", "b c", "c d", "d e"]

2) 3 by 3

  func concatenateStrings(array: [String]) -> [String] {
        return array.reduce([], { (result: [String], item:String) -> [String] in
            guard !result.isEmpty else {
                return [item] //first item
            }
            guard result.count != 1 else {
                return result + [array[result.count - 1] + " \(item)"] // second item
            }
            let first = array[result.count - 2]
            let second = array[result.count - 1]
            return result + [ "\(first) " + second + " \(item)"]
        })
    }

which gives me ["a", "a b", "a b c", "b c d", "c d e"]

My question is how can I generalize this method, if for example I want to produce a new array of strings by concatenating 4 elements, 5 elements of array and so on...

Thank you, and if there is still something unclear I will happily try to be more thorough.

Martin R · Accepted Answer · 2017-09-26 20:30:38Z

4

A possible solution (Swift 3+4):

func concatenateStrings(array: [String], maxLength: Int) -> [String] {

    let nestedArray = array.reduce([[String]]()) { (result, item) -> [[String]] in
        let last = result.last ?? []
        return result + [Array((last + [item]).suffix(maxLength))]
    }
    return nestedArray.map { $0.joined(separator: " ") }
}

Example:

let letters = ["a","b","c","d","e"]
print(concatenateStrings(array: letters, maxLength: 3))
// ["a", "a b", "a b c", "b c d", "c d e"]

The function first creates a nested array of strings, for example

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

by appending the current item to the last element, and using suffix() to limit the length. Then the inner arrays are concatenated.

In Swift 4 you can also use reduce(into:) which creates less intermediate arrays, see SE-0171 Reduce with inout:

func concatenateStrings(array: [String], maxLength: Int) -> [String] {

    let nestedArray = array.reduce(into: [[String]]()) { (result, item) in
        let last = result.last ?? []
        result.append(Array((last + [item]).suffix(maxLength)))
    }
    return nestedArray.map { $0.joined(separator: " ") }
}

edited Sep 26, 2017 at 20:30

answered Sep 26, 2017 at 19:53

Martin R

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5 Comments

Leo Dabus Over a year ago

you could also make the maxLength optional

func concatenateStrings(array: [String], maxLength: Int? = nil) -> [String] {     let maxLength = maxLength ?? array.count

unniverzal Over a year ago

Amazing solution Martin. Just what I needed. Thank you!

Martin R Over a year ago

@LeoDabus: Sure. One could also make the separator a parameter. I'll keep it simple for demonstration purposes.

Leo Dabus Over a year ago

@MartinR btw Nice addition to the reduce method being able to mutate the result in Swift 4

Martin R Over a year ago

@LeoDabus: Yes, that is a useful addition in Swift 4.

Luca Angeletti · Accepted Answer · 2017-09-26 20:00:14Z

1

Given youth input array

let elms = ["a","b","c","d","e"]

Here's an iterative function to achieve what you need

Swift 4 version

func concatenations(elms:[String], maxLength: Int) -> [String] {

    var result:[String] = []

    var word = ""

    for elm in elms {
        if word.count >= maxLength {
            word.removeFirst()
            word.append(elm)
        } else {
            word.append(elm)
        }
        result.append(word)
    }
    return result
}

Examples

concatenations(elms: elms, maxLenght: 2)
// ["a", "ab", "bc", "cd", "de"]


concatenations(elms: elms, maxLenght: 3)
// ["a", "ab", "abc", "bcd", "cde"]


concatenations(elms: elms, maxLenght: 4)
// ["a", "ab", "abc", "abcd", "bcde"]

edited Sep 26, 2017 at 20:00

answered Sep 26, 2017 at 19:41

Luca Angeletti

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4 Comments

Leo Dabus Over a year ago

you could just use collection method dropFirst instead of word.remove(at: word.startIndex)

Luca Angeletti Over a year ago

@LeoDabus It's definitely nicer. I updated my code. Thank you for the suggestion.

Leo Dabus Over a year ago

you could also make maxLenght parameter optional Int func concatenations(elms:[String], maxLenght: Int? = nil) and inside your method add let maxLenght = maxLenght ?? elms.count. And btw OP wants the characters separated by a space

Luca Angeletti Over a year ago

@LeoDabus Actually I think maxLength is explicitly needed to address the problem the OP described.

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