String convert to Int and replace comma to Plus sign - swift

Using Swift, I'm trying to take a list of numbers input in a text view in an app and create a sum of this list by extracting each number for a grade calculator. Also the amount of values put in by the user changes each time. An example is shown below:
String of: 98,99,97,96...
Trying to get: 98+99+97+96...
Please Help!
Thanks

Use components(separatedBy:) to break up the comma-separated string.
Use trimmingCharacters(in:) to remove spaces before and after each element
Use Int() to convert each element into an integer.
Use compactMap (previously called flatMap) to remove any items that couldn't be converted to Int.
Use reduce to sum up the array of Int.
let input = " 98 ,99 , 97, 96 "
let values = input.components(separatedBy: ",").compactMap { Int($0.trimmingCharacters(in: .whitespaces)) }
let sum = values.reduce(0, +)
print(sum) // 390

For Swift 3 and Swift 4.
Simple way: Hard coded. Only useful if you know the exact amount of integers coming up, wanting to get calculated and printed/used further on.
let string98: String = "98"
let string99: String = "99"
let string100: String = "100"
let string101: String = "101"
let int98: Int = Int(string98)!
let int99: Int = Int(string99)!
let int100: Int = Int(string100)!
let int101: Int = Int(string101)!
// optional chaining (if or guard) instead of "!" recommended. therefore option b is better
let finalInt: Int = int98 + int99 + int100 + int101
print(finalInt) // prints Optional(398) (optional)
Fancy way as a function: Generic way. Here you can put as many strings in as you need in the end. You could, for example, gather all the strings first and then use the array to have them calculated.
func getCalculatedIntegerFrom(strings: [String]) -> Int {
var result = Int()
for element in strings {
guard let int = Int(element) else {
break // or return nil
// break instead of return, returns Integer of all
// the values it was able to turn into Integer
// so even if there is a String f.e. "123S", it would
// still return an Integer instead of nil
// if you want to use return, you have to set "-> Int?" as optional
}
result = result + int
}
return result
}
let arrayOfStrings = ["98", "99", "100", "101"]
let result = getCalculatedIntegerFrom(strings: arrayOfStrings)
print(result) // prints 398 (non-optional)

let myString = "556"
let myInt = Int(myString)

Related

Swift get decimal value for String characters

Is it possible to get the decimal value for String characters in Swift?
something like:
let words:String = "1 ring to rule them all"
var value:Int = 0
for i in 0..<words.count {
let char = words[words.index(words.startIndex,offsetBy:i)]
value += Int(char.decimal)
}
where the first character in "1 ring to rule them all" is 49. Possible?
you could try this:
let words = "1 ring to rule them all"
var value: Int = 0
for i in 0..<words.count {
let char = words[words.index(words.startIndex,offsetBy:i)]
if let val = char.asciiValue {
print("----> char: \(char) val: \(val)") // char and its "decimal" value
value += Int(val)
}
}
print("\n----> value: \(value) \n") // meaningless total value
ok, looks like this is the way:
let characterString = "蜇"
let scalars = characterString.unicodeScalars
let ui32:UInt32 = scalars[scalars.startIndex].value
If you want to add up the Unicode values associated with a string, it would be:
var value = 0
for character in string {
for scalar in character.unicodeScalars {
value += Int(scalar.value)
}
}
Or, alternatively:
let value = string
.flatMap { $0.unicodeScalars }
.compactMap { $0.value }
.reduce(0, +)
While the above adds the values, as requested, if you are trying to get a numeric representation for a string, you might consider using hashValue, or checksum, or CRC, or the like. Simply summing the values will not be able to detect, for example, character transpositions. It just depends upon your use-case for this numeric representation of your string.

Swift randomizer with variables

I have two String Type values as Int number.
I like to randomize between these two.
With this code:
let random = Int.random(in: myvar1...myvar2)
it does not work. How can i fix it?
I assume your variables look like this?
let myvar1: String = "1"
let myvar2: String = "10"
Those are both Strings, so myvar1...myvar2 becomes a range of String.
However, the random(in:) method takes in a range of Int, so you'll first need to convert them to Ints.
if let myvar1Int = Int(myvar1), let myvar2Int = Int(myvar2) {
let random = Int.random(in: myvar1Int...myvar2Int)
print(random) /// Result: 6
}

What is happening during the String.init

I found this example function that takes a integer and then sorts the number from largest to smallest. Example input: 12345 output: 54321. The solution works but I am trying to understand the String.init part. Could anyone explain what is going on?
func descendingOrder(of number: Int) -> Int {
let s = String(number).sorted().reversed()
return Int(s.compactMap(String.init).joined())!
}
s is of type ReversedCollection<[String.Element]> so it makes the conversion from String.Element to String this is the meaning of String.init inside loop of compactMap
When you see DataType.init you need to know that the item being looped is a paramter for the init of that type so loop occurs over the characters of the String s to make a new [String] as a result
Developer complicates the line , it could be simplified to this for netter understanding
func descendingOrder(of number: Int) -> Int {
let s = String(number).sorted().reversed() // sort and reverse the string descendinng
let arr = s.compactMap(String.init) // convert characters to array of strings
let joined = arr.joined() // join them in 1 string
let res = Int(joined)! // convert string to int
return res
}
OR as in comments
func descendingOrder(of number: Int) -> Int {
let s = String(number).sorted().reversed() // sort and reverse the string descendinng
let arr = s.map{ String($0) } // convert characters to array of strings
let joined = arr.joined() // join them in 1 string
let res = Int(joined)! // convert string to int
return res
}

Swift String to Int Always Returns nil

I am trying to convert a String to an Int. Seems simple enough, but for somer reason it is always returning nil.
I'm just writing a simple extension to convert dollars to cents:
func dollarsToCents() -> Int {
var temp = self;
temp = temp.replacingOccurrences(of: "$", with: "")
temp = temp.replacingOccurrences(of: ",", with: "")
if let number = Int(temp) {
return number*100
}
return 0
}
I have temp set to "$250.89". number is always nil. No matter how I approach converting temp to an Int it is always nil. Anyone know what I'm doing wrong?
Problem is, that string "250.89" (after removing currency symbol) can't be converted to Int because 250.89 isn't integer. So fix your code by converting it to Double
func dollarsToCents() -> Int {
var temp = self
temp.removeAll { "$,".contains($0) }
//temp = temp.replacingOccurrences(of: "[$,]", with: "", options: .regularExpression)
return Int(((Double(temp) ?? 0) * 100).rounded())
}
or if your "number" always have two decimal places
func dollarsToCents() -> Int {
var temp = self
temp.removeAll { !("0"..."9" ~= $0) }
return Int(temp) ?? 0
}
But I think solution is much easier. Your goal should be saving price value as number (Double,...). Then you don't have to convert String to Double and you can just multiply your number. Then when you need to add currency symbol, just convert your value to String and add $ or use NumberFormatter
let price = 250.89
let formattedPrice = "$\(price)" // $250.89
let price = 250.89
let formatter = NumberFormatter()
formatter.numberStyle = .currency
formatter.currencyCode = "USD"
let formattedPrice = formatter.string(from: price as NSNumber)! // $250.89
Just adding this for the bizarre edge-case that I just encountered - be sure there aren't any whitespaces either trailing or leading in your string when converting to Int from String.
Eg.
for date in next_thirty_days {
let date_buffer: String = date.description[8...9]
date_list.append(Int(date_buffer)!)
}
This will fail if the value for date_buffer has a leading or trailing white-space.
Another edge-case is that there's a leading 0 in the string variable, eg. 07.
Hope it helps :)

Why does swift substring with range require a special type of Range

Consider this function to build a string of random characters:
func makeToken(length: Int) -> String {
let chars: String = "abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ"
var result: String = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let idxEnd = idx + 1
let range: Range = idx..<idxEnd
let char = chars.substring(with: range)
result += char
}
return result
}
This throws an error on the substring method:
Cannot convert value of type 'Range<Int>' to expected argument
type 'Range<String.Index>' (aka 'Range<String.CharacterView.Index>')
I'm confused why I can't simply provide a Range with 2 integers, and why it's making me go the roundabout way of making a Range<String.Index>.
So I have to change the Range creation to this very over-complicated way:
let idx = Int(arc4random_uniform(UInt32(chars.characters.count)))
let start = chars.index(chars.startIndex, offsetBy: idx)
let end = chars.index(chars.startIndex, offsetBy: idx + 1)
let range: Range = start..<end
Why isn't it good enough for Swift for me to simply create a range with 2 integers and the half-open range operator? (..<)
Quite the contrast to "swift", in javascript I can simply do chars.substr(idx, 1)
I suggest converting your String to [Character] so that you can index it easily with Int:
func makeToken(length: Int) -> String {
let chars = Array("abcdefghijklmnopqrstuvwxyz0123456789!?##$%ABCDEFGHIJKLMNOPQRSTUVWXYZ".characters)
var result = ""
for _ in 0..<length {
let idx = Int(arc4random_uniform(UInt32(chars.count)))
result += String(chars[idx])
}
return result
}
Swift takes great care to provide a fully Unicode-compliant, type-safe, String abstraction.
Indexing a given Character, in an arbitrary Unicode string, is far from a trivial task. Each Character is a sequence of one or more Unicode scalars that (when combined) produce a single human-readable character. In particular, hiding all this complexity behind a simple Int based indexing scheme might result in the wrong performance mental model for programmers.
Having said that, you can always convert your string to a Array<Character> once for easy (and fast!) indexing. For instance:
let chars: String = "abcdefghijklmnop"
var charsArray = Array(chars.characters)
...
let resultingString = String(charsArray)