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I have done this successfully in Java and C++ but can't figure it out in Swift. Here is my C++ code:
int count_vowels (string input)
{
int position = 0;
for (int i = 0; i < input.length(); i++)
{
char ch;
ch = input[i];
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u'){
position++;
}
}
return position;
}
Again I'm just trying to iterate through a string, counting the values that are true in the if statement and returning the position(how many there are).
Is there any way to convert this to swift syntax.
I figured it out:
for char in vowels.characters{
if char == ("a") || char == ("e") || char == ("i") || char == ("o") || char == ("u") {
count++
}
}
Thanks everybody for there help and posts I was pulling my hair out on this one.
Thanks again
Disclaimer: I don't know Swift but....
You should be able to use Swift's higher level features to accomplish this more succinctly. For example, the vowel characters don't change within your function so they can be represented as a Swift Set:
var vowelCount = 0
var vowels = Set(["a", "e", "i", "o", "u"])
Then test if each input character is a member of the set and increment the count if it is:
for ch in input {
if (vowels.contains(ch)) {
vowelCount += 1
}
}
I agree that you should show what it is you've tried in Swift so we can provide guidance on what may be broken. That being said, I did a quick conversion. It may not be the best, but here you go
import Cocoa
let vowels: [Character] = ["a", "e", "i", "o", "u"]
func count_vowels(input: String) -> Int {
var vowelCount: Int = 0
for ch in input.lowercaseString.characters {
if (vowels.contains(ch)) {
vowelCount++
}
}
return vowelCount
}
With this, print(count_vowels("Hello world")) prints out 3
EDIT:
Modified my answer based on the answer from Tom Hicks. It definitely gets rid of the ugly if statement and makes the code easier to read. I declare the vowels list outside of the function so that it isn't being generated every time you call the function.
You could do something like this:
extension String {
var numberOfVowels: Int {
let vowels = "aeiou"
let vowelsSet = NSCharacterSet(charactersInString: vowels)
let strippedComponents = lowercaseString.componentsSeparatedByCharactersInSet(vowelsSet)
let stripped = strippedComponents.joinWithSeparator("")
return characters.count - stripped.characters.count
}
}
"Hello".numberOfVowels
This will work ... but its not elegent.
let str = "test voWeL COUNT"
let v = ["A","a","E","e","i","I","o","O","u","U"]
var vowels = 0;
for i in str.characters {
print(i)
if (v.contains("\(i)")) {
vowels++
}
}
print("Vowels \(vowels)")
Related
I need to print just the vowels in this string. I have created a var for the vowels and did an if statement and hade no luck. I tried to pull each vowel by its placement value and couldn't find a way to make it work. I am very new to this and have spent 6 hours trying to look it up and see what works. I know I'm not understanding something or maybe even all of it. Please help!
let scc = "Spokane Community College Software Development"
scc.forEach { scc in
print(scc.lowercased() )
}
You can try
let arr = ["u","i","e","a","o"]
let str = "Spokane Community College Software Development"
str.forEach {
if arr.contains($0.lowercased()) {
print($0)
}
}
func vowel(_ input :String)
{
for i in input
{
if(i == "a" || i == "e" || i == "i" || i == "u" || i == "o")
{
print(i)
}
}
}
vowel("Spokane Community College Software Development")
var vowelCount = 0
var vowels = Set(["a", "e", "i", "o", "u"])
func count_vowels(input: String) -> Int {
for ch in input {
if (vowels.contains(ch.lowercased())) {
vowelCount += 1
}
else {
}
}
return vowelCount
}
count_vowels(input: "test12222 voWeL !+COUNT")
This question already has answers here:
Break A Number Up To An Array of Individual Digits
(6 answers)
Closed 5 years ago.
I was wondering if there was a way in Swift to split an Int up into it's individual digits without converting it to a String. For example:
let x: Int = 12345
//Some way to loop/iterate over x's digits
//Then map each digit in x to it's String value
//Return "12345"
For a bit of background, I'm attempting to create my own method of converting an Int to a String without using the String description property or using String Interpolation.
I've found various articles on this site but all the ones I've been able to find either start with a String or end up using the String description property to convert the Int to a String.
Thanks.
Just keep dividing by 10 and take the remainder:
extension Int {
func digits() -> [Int] {
var digits: [Int] = []
var num = self
repeat {
digits.append(num % 10)
num /= 10
} while num != 0
return digits.reversed()
}
}
x.digits() // [1,2,3,4,5]
Note that this will return all negative digits if the value is negative. You could add a special case if you want to handle that differently. This return [0] for 0, which is probably what you want.
And because everyone like pure functional programming, you can do it that way too:
func digits() -> [Int] {
let partials = sequence(first: self) {
let p = $0 / 10
guard p != 0 else { return nil }
return p
}
return partials.reversed().map { $0 % 10 }
}
(But I'd probably just use the loop here. I find sequence too tricky to reason about in most cases.)
A recursive way...
extension Int {
func createDigitArray() -> [Int] {
if self < 10 {
return [self]
} else {
return (self / 10).createDigitArray() + [self % 10]
}
}
}
12345.createDigitArray() //->[1, 2, 3, 4, 5]
A very easy approach would be using this function:
func getDigits(of number: Int) -> [Int] {
var digits = [Int]()
var x = number
repeat{
digits.insert(abs(x % 10), at: 0)
x/=10
} while x != 0
return digits
}
And using it like this:
getDigits(of: 97531) // [9,7,5,3,1]
getDigits(of: -97531) // [9,7,5,3,1]
As you can see, for a negative number you will receive the array of its digits, but at their absolute value (e.g.: -9 => 9 and -99982 => 99982)
Hope it helps!
I'm currently converting C++ code to Swift and I've gotten stuck on one part. The parameter passed into the function is a string and the area where I'm stuck is when attempting to set a variable based on the second to last character of a string to check for a certain character.
The error shows up on this line:
line[i-1]
I've tried casting this value to an Int but this didn't work:
Int(line[i - 1])
I've also tried to see if the string's startIndex function which takes a Int would work but it didn't:
line.startIndex[i - 1]
Here is the full function:
func scanStringForSpecificCharacters(line: String){
var maxOpen: Int = 0;
var minOpen: Int = 0;
minOpen = 0;
maxOpen = 0;
var i = 0
while i < line.characters.count {
for character in line.characters {
//var c: Character = line[i];
if character == "(" {
maxOpen += 1;
if i == 0 || line[i - 1] != ":" {
minOpen += 1;
}
}
else if character == ")"{
minOpen = max(0,minOpen-1);
if i == 0 || line[i-1] != ":"{
maxOpen -= 1;
}
if maxOpen < 0{
break;
}
}
}
if maxOpen >= 0 && minOpen == 0{
print("YES")
}else{
print("NO")
}
}
}
Strings in Swift aren't indexed collections and instead you can access one of four different views: characters, UTF8, UTF16, or unicodescalars.
This is because Swift supports unicode, where an individual characters may actually be composed of multiple unicode scalars.
Here's a post that really helped me wrap my head around this: https://oleb.net/blog/2016/08/swift-3-strings/
Anyway, to answer you question you'll need to create an index using index(after:), index(before:), or index(_, offsetBy:).
In your case you'd want to do something like this:
line.index(line.endIndex, offsetBy: -2) // second to last character
Also, you'll probably find it easier to iterate directly using a String.Index type rather than Int:
let line = "hello"
var i = line.startIndex
while i < line.endIndex {
print(line[i])
i = line.index(after: i)
}
// prints ->
// h
// e
// l
// l
// o
Working with Strings in Swift was changed several times during it's evolution and it doesn't look like C++ at all. You cannot subscript string to obtain individual characters, you should use index class for that. I recommend you read this article:
https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/StringsAndCharacters.html
As already pointed out in the other answers, the compiler error
is caused by the problem that you cannot index a Swift String with
integers.
Another problem in your code is that you have a nested loop which is
probably not intended.
Actually I would try to avoid string indexing at all and only
enumerate the characters, if possible. In your case, you can
easily keep track of the preceding character in a separate variable:
var lastChar: Character = " " // Anything except ":"
for char in line.characters {
if char == "(" {
maxOpen += 1;
if lastChar != ":" {
minOpen += 1;
}
}
// ...
lastChar = char
}
Or, since you only need to know if the preceding character is
a colon:
var lastIsColon = false
for char in string.characters {
if char == "(" {
maxOpen += 1;
if !lastIsColon {
minOpen += 1;
}
}
// ...
lastIsColon = char == ":"
}
Another possible approach is to iterate over the string and a shifted
view of the string in parallel:
for (lastChar, char) in zip([" ".characters, line.characters].joined(), line.characters) {
// ...
}
As others have already explained, trying to index into Swift strings is a pain.
As a minimal change to your code, I would recommend that you just create an array of the characters in your line up front:
let linechars = Array(line.characters)
And then anywhere you need to index into the line, use linechars:
This:
if i == 0 || line[i-1] != ":" {
becomes:
if i == 0 || linechars[i-1] != ":" {
This question already has answers here:
What is the best way to determine if a string contains a character from a set in Swift
(11 answers)
Closed 7 years ago.
I'm trying to check whether a specific string contains letters or not.
So far I've come across NSCharacterSet.letterCharacterSet() as a set of letters, but I'm having trouble checking whether a character in that set is in the given string. When I use this code, I get an error stating:
'Character' is not convertible to 'unichar'
For the following code:
for chr in input{
if letterSet.characterIsMember(chr){
return "Woah, chill out!"
}
}
You can use NSCharacterSet in the following way :
let letters = NSCharacterSet.letters
let phrase = "Test case"
let range = phrase.rangeOfCharacter(from: characterSet)
// range will be nil if no letters is found
if let test = range {
println("letters found")
}
else {
println("letters not found")
}
Or you can do this too :
func containsOnlyLetters(input: String) -> Bool {
for chr in input {
if (!(chr >= "a" && chr <= "z") && !(chr >= "A" && chr <= "Z") ) {
return false
}
}
return true
}
In Swift 2:
func containsOnlyLetters(input: String) -> Bool {
for chr in input.characters {
if (!(chr >= "a" && chr <= "z") && !(chr >= "A" && chr <= "Z") ) {
return false
}
}
return true
}
It's up to you, choose a way. I hope this help you.
You should use the Strings built in range functions with NSCharacterSet rather than roll your own solution. This will give you a lot more flexibility too (like case insensitive search if you so desire).
let str = "Hey this is a string"
let characterSet = NSCharacterSet(charactersInString: "aeiou")
if let _ = str.rangeOfCharacterFromSet(characterSet, options: .CaseInsensitiveSearch) {
println("true")
}
else {
println("false")
}
Substitute "aeiou" with whatever letters you're looking for.
A less flexible, but fun swift note all the same, is that you can use any of the functions available for Sequences. So you can do this:
contains("abc", "c")
This of course will only work for individual characters, and is not flexible and not recommended.
The trouble with .characterIsMember is that it takes a unichar (a typealias for UInt16).
If you iterate your input using the utf16 view of the string, it will work:
let set = NSCharacterSet.letterCharacterSet()
for chr in input.utf16 {
if set.characterIsMember(chr) {
println("\(chr) is a letter")
}
}
You can also skip the loop and use the contains algorithm if you only want to check for presence/non-presence:
if contains(input.utf16, { set.characterIsMember($0) }) {
println("contains letters")
}
In Swift, I need to create a simple for-condition-increment loop with all the multiples of 3 from 3-100. So far I have:
var multiplesOfThree: [String] = []
for var counter = 0; counter < 30; ++counter {
multiplesOfThree.append("0")
if counter == 3 {
multiplesOfThree.append("3")
} else if counter == 6 {
multiplesOfThree.append("6")
} else if counter == 9 {
multiplesOfThree.append("9")
}
println("Adding \(multiplesOfThree[counter]) to the Array.")
}
I would like to replace all the if and else if statements with something like:
if (index %3 == 0)
but I’m not sure what the proper syntax would be? Also, if I have a single IF statement do I need a .append line to add to the Array?
You are very much on the right track. A few notes:
Swift provides a more concise way to iterate over a fixed number of integers using the ..< operator (an open range operator).
Your if statement with the modulus operator is exactly correct
To make a string from an Int you can use \(expression) inside a string. This is called String Interpolation
Here is the working code:
var multiplesOfThree: [String] = []
for test in 0..<100 {
if (test % 3 == 0) {
multiplesOfThree.append("\(test)")
}
}
However, there is no reason to iterate over every number. You can simply continue to add 3 until you reach your max:
var multiplesOfThree: [String] = []
var multiple = 0
while multiple < 100 {
multiplesOfThree.append("\(multiple)")
multiple += 3
}
As rickster pointed out in the comments, you can also do this in a more concise way using a Strided Range with the by method:
var multiplesOfThree: [String] = []
for multiple in stride(from: 0, to: 100, by: 3) {
multiplesOfThree.append("\(multiple)")
}
Getting even more advanced, you can use the map function to do this all in one line. The map method lets you apply a transform on every element in an array:
let multiplesOfThree = Array(map(stride(from: 0, to: 100, by: 3), { "\($0)" }))
Note: To understand this final code, you will need to understand the syntax around closures well.