Replacement for c style loop with modulo - swift

Since C-style for statement is deprecated and will be removed in a future version of Swift, what is the best way to achieve the following:
for var i = startIndex; i != endIndex; i = (i + 1) % arrCount {
}
(I want to iterate the array, but start from somewhere at the middle and continue until I'll reach the index I started from)
Edit:
startIndex > endIndex

"while" should be the best:
var i = startIndex
while i != endIndex {
i = (i + 1) % arrCount
}

You can shift the original array pretty easily, see this example:
let array = [3,4,5,6,7]
let startIndex = 3
let endIndex = 2
let head = array[0..<endIndex] // [3, 4]
let tail = array[startIndex..<array.count] // [6, 7]
let shiftedArray = tail + head // [6, 7, 3, 4]
for element in shiftedArray {
// do something
}

Here's a solution using a custom SequenceType & associated GeneratorType, as suggested by Sulthan in a comment above.
On the plus side, this sequence is really rather nicely readable and flexible. For instance:
let startIndex = 3
let endIndex = 2
let array = [3, 4, 5, 6, 7]
ModuloSequence(start: startIndex, end: endIndex, divisor: array.count).map {
array[$0]
}
On the minus side, it's rather verbose to implement (my take on it is, anyway):
struct ModuloSequenceGenerator : GeneratorType {
let end:Int, divisor:Int
private var dividend:Int?
init(start:Int, end:Int, divisor:Int) {
self.dividend = start
self.end = end
self.divisor = divisor
}
mutating func next() -> Int? {
guard let c = dividend where c != end else {
return nil
}
dividend = (c + 1) % divisor
return dividend
}
}
struct ModuloSequence : SequenceType {
let start:Int, end:Int, divisor:Int
func generate() -> ModuloSequenceGenerator {
return ModuloSequenceGenerator(start: start, end: end, divisor: divisor)
}
}

Related

TwoSum Swift Solution

I just started learning coding with swift, and was trying TwoSum.
"Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1]."
I found some solutions from GitHub that I cannot understand.
code is from https://github.com/soapyigu/LeetCode-Swift/blob/master/Array/TwoSum.swift
class TwoSum {
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
var dict = [Int: Int]()
for (i, num) in nums.enumerated() {
if let lastIndex = dict[target - num] {
return [lastIndex, i]
}
dict[num] = i
}
fatalError("No valid outputs")
}
}
Could someone be so kind to explain to codes. Thanks a lot.
The dict initialised in the method stores the numbers in the input as keys, and their indices as values. The program uses this to remember which number is where. The dict can tell you things like "the number 2 is at index 0".
For each number num at index i in the input array, we subtract num from the target to find the other number that we need, in order for them to add up to target.
Now we have the other number we need, we check to see if we have seen such a number before, by searching dict. This is what the if let lastIndex = dict[target - num] part is doing. If the dict knows what index the other number is at, we return that index, and i.
If we haven't seen that number before, we record i into the dictionary under the key num, hoping that in later iterations, we can find a number that when added to num, makes 9.
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
var arr:[Int] = []
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
var toggle = false
for i in 0..<nums.count {
for j in i+1..<nums.count {
if toggle == false {
if(nums[i]+nums[j]==target){
toggle = true
arr.insert(i, at: 0)
arr.insert(j, at: 1)
break
}
}
}
}
return arr
}
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
In Sweeper's excellent answer, he explained what dict is used for: It lets you use a value from the array to find that value's index. It would be more obvious what the dictionary was used for if we called it indexes, and this code builds the same dictionary in a more explicit way:
var indexes = [Int: Int]()
for index in 0..<array.count {
let value = array[index]
indexes[value] = index
}
After that, you get a dictionary:
[2:0, 7:1, 11:2, 15:3]
You could write the function this way:
func twoSum(_ array: [Int], _ target: Int) -> [Int] {
var indexes = [Int: Int]()
for index in 0..<array.count {
let value = array[index]
indexes[value] = index
}
for index in 0..<array.count {
let value = array[index]
if let otherIndex = indexes[target - value],
index != otherIndex {
return [index, otherIndex]
}
}
fatalError("Unable to match values")
}
That is a much more long-winded (and less efficient) way of doing the same thing. It loops through the array twice instead of once, but the results should be the same.
func twoSum(array: [Int], target: Int) -> [Int] {
var dict = [Int:Int]()
for (index, number) in array.enumerated() {
let value = target - number
if let sum = dict[value] {
return [sum, index]
}
dict[number] = index
}
return [0,0]
}
/*
array=[1, 2, 3] -> target=4
enumerated() => [0,1], [1,2], [2,3]
(i, n)
v4 - 1 = 3
s[3:0]
s[3:0]
v4 - 2 = 2
s[2:0]
s[2:1]
v4 - 3 = 1
s[1:1]
s[1:2]
output [0,2]
*/
var numbers: [Int] = [1, 3, 6, 7, 7, 14, 12]
var target = 26
var result = [Int]()
for i in 0..<numbers.count {
for j in i+1..<numbers.count {
if numbers[i] + numbers[j] == target {
print(numbers[i],numbers[j])
result.append(i)
result.append(j)
}
}
}
print(Array(Set(result)))
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
var dict:[Int:Int] = [:]
for i in 0..<nums.count {
if dict[target - nums[i]] != nil {
return [dict[target - nums[i]] ?? 0, i]
} else {
dict[nums[i]] = i
}
}
return [0]
}
Here is a link to the discussion section of the TwoSum problem on Leetcode.
Lots of great Swift solutions there.
https://leetcode.com/problems/two-sum/discuss/?currentPage=1&orderBy=most_votes&query=swift.
My personal two cents -
func twoSumA(_ nums: [Int], _ target: Int) -> [Int] {
var numsHashMap: Dictionary<Int, Int> = [:]
var outputArr: [Int] = []
for index in 0..<nums.count {
let currentNum = nums[index]
if numsHashMap.keys.contains(target-currentNum) {
outputArr.append(numsHashMap[target-currentNum] ?? -1)
outputArr.append(index)
return outputArr
}
numsHashMap[currentNum] = index
}
return !outputArr.isEmpty ? outputArr : [-1, -1]
}

Split string to arrays with maximum variables in each array

I have a string of numbers (each number is separated by ,) that looks like this:
"12,3,5,75,584,364,57,88,94,4,79,333,7465,867,56,6,748,546,573,466"
I want to split the string to an array of strings, that each element is a string that has maximum 10 number in it.
For the example I've added I want to achieve something like this:
stringsArray:
Element 0: "12,3,5,75,584,364,57,88,94,4"
Element 1: "79,333,7465,867,56,6,748,546,573,466"
And so on...
I've been thinking a lot about a way to do this with Swift, but couldn't find anything...
Does anybody has an idea?
Thank you!
Step 1 - get fully separated array:
let numbers = "12,3,5".components(separatedBy: ",")
Step 2 - chunk your result to parts with ext:
extension Array {
func chunked(by chunkSize: Int) -> [[Element]] {
return stride(from: 0, to: self.count, by: chunkSize).map {
Array(self[$0..<Swift.min($0 + chunkSize, self.count)])
}
}
}
let chunkedNumbers = numbers.chunked(by: 10)
Step 3:
let stringsArray = chunkedNumbers.map { $0.joined(separator: ",") }
Result: ["12,3,5,75,584,364,57,88,94,4", "79,333,7465,867,56,6,748,546,573,466"]
Link to gist playground.
I would look at the position of 10th comma in your original string, get the prefix up to this position, remove this prefix and repeat until remaining string is empty.
This is a bit brute force, but works.
I first add extension to String for convenience.
extension String {
func startIndexesOf(_ string: String) -> [Int] {
var result: [Int] = []
var start = startIndex
while let range = range(of: string, options: .literal, range: start..<endIndex) {
result.append(range.lowerBound.encodedOffset)
start = range.upperBound
}
return result
}
subscript (r: Range<Int>) -> String {
let start = index(self.startIndex, offsetBy: r.lowerBound)
let end = self.index(self.startIndex, offsetBy: r.upperBound)
return String(self[Range(start ..< end)])
}
}
let test = "12,3,5,75,584,364,57,88,94,4,79,333,7465,867,56,6,748,546,573,466,999"
var remaining = test
var arrayOf10 : [String] = []
repeat {
let indexes = remaining.startIndexesOf(",")
if indexes.count < 10 {
arrayOf10.append(remaining) // Just add what remains
break
}
let position = indexes[9]
let endBeginning = remaining.index(test.startIndex, offsetBy: position) // Beginning of what remain to parse
let beginningSubstring = remaining[remaining.startIndex ..< endBeginning]
let beginningText = String(beginningSubstring)
arrayOf10.append(beginningText)
let startNext = remaining.index(test.startIndex, offsetBy: position+1) // What will remain to parse after taking out the beginning
let remainingSubString = remaining[startNext ..< remaining.endIndex]
remaining = String(remainingSubString)
} while remaining.count > 0
for (c, s) in arrayOf10.enumerated() { print("Element", c, ": ", s)}
This will print as desired
Element 0 : 12,3,5,75,584,364,57,88,94,4
Element 1 : 79,333,7465,867,56,6,748,546,573,466
Element 2 : 999

Swift - Convert a binary string to its ascii values

I have a string of binary values e.g. "010010000110010101111001". Is there a simple way to convert this string into its ascii representation to get (in this case) "Hey"?
Only found the other way or things for Integer:
let binary = "11001"
if let number = Int(binary, radix: 2) {
print(number) // Output: 25
}
Do someone know a good and efficient solution for this case?
A variant of #OOPer's solution would be to use a conditionally binding while loop and index(_:offsetBy:limitedBy:) in order to iterate over the 8 character substrings, taking advantage of the fact that index(_:offsetBy:limitedBy:) returns nil when you try to advance past the limit.
let binaryBits = "010010000110010101111001"
var result = ""
var index = binaryBits.startIndex
while let next = binaryBits.index(index, offsetBy: 8, limitedBy: binaryBits.endIndex) {
let asciiCode = UInt8(binaryBits[index..<next], radix: 2)!
result.append(Character(UnicodeScalar(asciiCode)))
index = next
}
print(result) // Hey
Note that we're going via Character rather than String in the intermediate step – this is simply to take advantage of the fact that Character is specially optimised for cases where the UTF-8 representation fits into 63 bytes, which is the case here. This saves heap-allocating an intermediate buffer for each character.
Purely for the fun of it, another approach could be to use sequence(state:next:) in order to create a sequence of the start and end indices of each substring, and then reduce in order to concatenate the resultant characters together into a string:
let binaryBits = "010010000110010101111001"
// returns a lazily evaluated sequence of the start and end indices for each substring
// of 8 characters.
let indices = sequence(state: binaryBits.startIndex, next: {
index -> (index: String.Index, nextIndex: String.Index)? in
let previousIndex = index
// Advance the current index – if it didn't go past the limit, then return the
// current index along with the advanced index as a new element of the sequence.
return binaryBits.characters.formIndex(&index, offsetBy: 8, limitedBy: binaryBits.endIndex) ? (previousIndex, index) : nil
})
// iterate over the indices, concatenating the resultant characters together.
let result = indices.reduce("") {
$0 + String(UnicodeScalar(UInt8(binaryBits[$1.index..<$1.nextIndex], radix: 2)!))
}
print(result) // Hey
On the face of it, this appears to be much less efficient than the first solution (due to the fact that reduce should copy the string at each iteration) – however it appears the compiler is able to perform some optimisations to make it not much slower than the first solution.
You may need to split the input binary digits into 8-bit chunks, and then convert each chunk to an ASCII character. I cannot think of a super simple way:
var binaryBits = "010010000110010101111001"
var index = binaryBits.startIndex
var result: String = ""
for _ in 0..<binaryBits.characters.count/8 {
let nextIndex = binaryBits.index(index, offsetBy: 8)
let charBits = binaryBits[index..<nextIndex]
result += String(UnicodeScalar(UInt8(charBits, radix: 2)!))
index = nextIndex
}
print(result) //->Hey
Does basically the same as OOPer's solution, but he/she was faster and has a shorter, more elegant approach :-)
func getASCIIString(from binaryString: String) -> String? {
guard binaryString.characters.count % 8 == 0 else {
return nil
}
var asciiCharacters = [String]()
var asciiString = ""
let startIndex = binaryString.startIndex
var currentLowerIndex = startIndex
while currentLowerIndex < binaryString.endIndex {
let currentUpperIndex = binaryString.index(currentLowerIndex, offsetBy: 8)
let character = binaryString.substring(with: Range(uncheckedBounds: (lower: currentLowerIndex, upper: currentUpperIndex)))
asciiCharacters.append(character)
currentLowerIndex = currentUpperIndex
}
for asciiChar in asciiCharacters {
if let number = UInt8(asciiChar, radix: 2) {
let character = String(describing: UnicodeScalar(number))
asciiString.append(character)
} else {
return nil
}
}
return asciiString
}
let binaryString = "010010000110010101111001"
if let asciiString = getASCIIString(from: binaryString) {
print(asciiString) // Hey
}
A different approach
let bytes_string: String = "010010000110010101111001"
var range_count: Int = 0
let characters_array: [String] = Array(bytes_string.characters).map({ String($0)})
var conversion: String = ""
repeat
{
let sub_range = characters_array[range_count ..< (range_count + 8)]
let sub_string: String = sub_range.reduce("") { $0 + $1 }
let character: String = String(UnicodeScalar(UInt8(sub_string, radix: 2)!))
conversion += character
range_count += 8
} while range_count < characters_array.count
print(conversion)
You can do this:
extension String {
var binaryToAscii: String {
stride(from: 0, through: count - 1, by: 8)
.map { i in map { String($0)}[i..<(i + 8)].joined() }
.map { String(UnicodeScalar(UInt8($0, radix: 2)!)) }
.joined()
}
}

Count elements of array matching condition in Swift

I'm basically looking for the swift equivalent of the follow c++ code:
std::count_if(list.begin(), list.end(), [](int a){ return a % 2 == 0; }); // counts instances of even numbers in list
My problem isn't actually searching for even numbers, of course; simply the general case of counting instances matching a criterion.
I haven't seen a builtin, but would love to hear that I simply missed it.
Like this:
let a: [Int] = ...
let count = a.filter({ $0 % 2 == 0 }).count
An alternative to Aderstedt's version
let a = [ .... ]
let count = a.reduce(0){
(count, element) in
return count + 1 - element % 2
}
My intuition says my way will be faster because it doesn't require the creation of a second array. However, you'd need to profile both methods to be sure.
Edit
Following MartinR's comment about generalisation of the function, here it is
extension SequenceType
{
func countMatchingCondition(condition: (Self.Generator.Element) -> Bool) -> Int
{
return self.reduce(0, combine: { (count, e) in count + (condition(e) ? 1 : 0) })
}
}
let a = [1, 2, 3, 3, 4, 12].countMatchingCondition { $0 % 2 == 0 }
print("\(a)") // Prints 3
Default array:
let array: [Int] = [10, 10, 2, 10, 1, 2, 3]
filter(_:) method
let countOfTen = array.filter({ $0 == 10 }).count // 3
count(where:) method
Update: This Swift 5.0 feature was withdrawn in beta testing because it was causing performance issues for the type checker.
let countOfTen = array.count(where: { $0 == 10 }) // 3
You can use Collection.lazy to have the simplicity of Aderstedt's Answer but with O(1) space.
let array = [1, 2, 3]
let count = array.lazy.filter({ $0 % 2 == 0 }).count
The most compact reduce statement that will do this is:
let a = Array(1 ... 20)
let evencount = a.reduce(0) { $0 + ($1 % 2 == 0 ? 1 : 0) }
Reduce takes two variables: starts with 0 (var $0) then for every element in Array a (var $1) if the value is divisible by 2 with no remainder then add one to your count.
This is also efficient as it does not create an additional array unlike using a.filter(){}.count .
You can also do this with reduce()
let a = Array(1 ... 20)
let evenCount = a.reduce(0) { (accumulator, value) -> Int in
guard value % 2 == 0 else { return accumulator }
return accumulator + 1
}
Almost everything you want to do with the map() and filter functions can actually be done with a reduce although it's not always the most readable.
Swift 5 or later:
public extension Sequence {
func occurrences(where predicate: (Element) throws -> Bool) rethrows -> Int {
try reduce(0) { try predicate($1) ? $0 + 1 : $0 }
}
}
public extension Sequence where Element: Equatable {
func occurrences(of element: Element) -> Int {
reduce(0) { element == $1 ? $0 + 1 : $0 }
}
}
let multiplesOf2 = [1,2,3,4,4,5,4,5].occurrences{$0.isMultiple(of: 2)} // 4
"abcdeabca".occurrences(of: "a") // 3
extension BinaryInteger {
var isOdd: Bool { !isMultiple(of: 2) }
var isEven: Bool { isMultiple(of: 2) }
}
(-4).isOdd // false
(-3).isOdd // true
(-2).isOdd // false
(-1).isOdd // true
0.isOdd // false
1.isOdd // true
2.isOdd // false
3.isOdd // true
4.isOdd // false
(-4).isEven // true
(-3).isEven // false
(-2).isEven // true
(-1).isEven // false
0.isEven // true
1.isEven // false
2.isEven // true
3.isEven // false
4.isEven // true
let odds = [1,2,3,4,4,5,5,11].occurrences(where: \.isOdd) // 5
let evens = [1,2,3,4,4,5,5,11].occurrences(where: \.isEven) // 3

How do you use String.substringWithRange? (or, how do Ranges work in Swift?)

I have not yet been able to figure out how to get a substring of a String in Swift:
var str = “Hello, playground”
func test(str: String) -> String {
return str.substringWithRange( /* What goes here? */ )
}
test (str)
I'm not able to create a Range in Swift. Autocomplete in the Playground isn’t super helpful - this is what it suggests:
return str.substringWithRange(aRange: Range<String.Index>)
I haven't found anything in the Swift Standard Reference Library that helps. Here was another wild guess:
return str.substringWithRange(Range(0, 1))
And this:
let r:Range<String.Index> = Range<String.Index>(start: 0, end: 2)
return str.substringWithRange(r)
I've seen other answers (Finding index of character in Swift String) that seem to suggest that since String is a bridge type for NSString, the "old" methods should work, but it's not clear how - e.g., this doesn't work either (doesn't appear to be valid syntax):
let x = str.substringWithRange(NSMakeRange(0, 3))
Thoughts?
You can use the substringWithRange method. It takes a start and end String.Index.
var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex, end: str.endIndex)) //"Hello, playground"
To change the start and end index, use advancedBy(n).
var str = "Hello, playground"
str.substringWithRange(Range<String.Index>(start: str.startIndex.advancedBy(2), end: str.endIndex.advancedBy(-1))) //"llo, playgroun"
You can also still use the NSString method with NSRange, but you have to make sure you are using an NSString like this:
let myNSString = str as NSString
myNSString.substringWithRange(NSRange(location: 0, length: 3))
Note: as JanX2 mentioned, this second method is not safe with unicode strings.
Swift 2
Simple
let str = "My String"
let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)]
//"Strin"
Swift 3
let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
str[startIndex...endIndex] // "Strin"
str.substring(to: startIndex) // "My "
str.substring(from: startIndex) // "String"
Swift 4
substring(to:) and substring(from:) are deprecated in Swift 4.
String(str[..<startIndex]) // "My "
String(str[startIndex...]) // "String"
String(str[startIndex...endIndex]) // "Strin"
At the time I'm writing, no extension is perfectly Swift 4.2 compatible, so here is one that covers all the needs I could think of:
extension String {
func substring(from: Int?, to: Int?) -> String {
if let start = from {
guard start < self.count else {
return ""
}
}
if let end = to {
guard end >= 0 else {
return ""
}
}
if let start = from, let end = to {
guard end - start >= 0 else {
return ""
}
}
let startIndex: String.Index
if let start = from, start >= 0 {
startIndex = self.index(self.startIndex, offsetBy: start)
} else {
startIndex = self.startIndex
}
let endIndex: String.Index
if let end = to, end >= 0, end < self.count {
endIndex = self.index(self.startIndex, offsetBy: end + 1)
} else {
endIndex = self.endIndex
}
return String(self[startIndex ..< endIndex])
}
func substring(from: Int) -> String {
return self.substring(from: from, to: nil)
}
func substring(to: Int) -> String {
return self.substring(from: nil, to: to)
}
func substring(from: Int?, length: Int) -> String {
guard length > 0 else {
return ""
}
let end: Int
if let start = from, start > 0 {
end = start + length - 1
} else {
end = length - 1
}
return self.substring(from: from, to: end)
}
func substring(length: Int, to: Int?) -> String {
guard let end = to, end > 0, length > 0 else {
return ""
}
let start: Int
if let end = to, end - length > 0 {
start = end - length + 1
} else {
start = 0
}
return self.substring(from: start, to: to)
}
}
And then, you can use:
let string = "Hello,World!"
string.substring(from: 1, to: 7)gets you: ello,Wo
string.substring(to: 7)gets you: Hello,Wo
string.substring(from: 3)gets you: lo,World!
string.substring(from: 1, length: 4)gets you: ello
string.substring(length: 4, to: 7)gets you: o,Wo
Updated substring(from: Int?, length: Int) to support starting from zero.
NOTE: #airspeedswift makes some very insightful points on the trade-offs of this approach, particularly the hidden performance impacts. Strings are not simple beasts, and getting to a particular index may take O(n) time, which means a loop that uses a subscript can be O(n^2). You have been warned.
You just need to add a new subscript function that takes a range and uses advancedBy() to walk to where you want:
import Foundation
extension String {
subscript (r: Range<Int>) -> String {
get {
let startIndex = self.startIndex.advancedBy(r.startIndex)
let endIndex = startIndex.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: startIndex, end: endIndex)]
}
}
}
var s = "Hello, playground"
println(s[0...5]) // ==> "Hello,"
println(s[0..<5]) // ==> "Hello"
(This should definitely be part of the language. Please dupe rdar://17158813)
For fun, you can also add a + operator onto the indexes:
func +<T: ForwardIndex>(var index: T, var count: Int) -> T {
for (; count > 0; --count) {
index = index.succ()
}
return index
}
s.substringWithRange(s.startIndex+2 .. s.startIndex+5)
(I don't know yet if this one should be part of the language or not.)
SWIFT 2.0
simple:
let myString = "full text container"
let substring = myString[myString.startIndex..<myString.startIndex.advancedBy(3)] // prints: ful
SWIFT 3.0
let substring = myString[myString.startIndex..<myString.index(myString.startIndex, offsetBy: 3)] // prints: ful
SWIFT 4.0
Substring operations return an instance of the Substring type, instead of String.
let substring = myString[myString.startIndex..<myString.index(myString.startIndex, offsetBy: 3)] // prints: ful
// Convert the result to a String for long-term storage.
let newString = String(substring)
It is much more simple than any of the answers here, once you find the right syntax.
I want to take away the [ and ]
let myString = "[ABCDEFGHI]"
let startIndex = advance(myString.startIndex, 1) //advance as much as you like
let endIndex = advance(myString.endIndex, -1)
let range = startIndex..<endIndex
let myNewString = myString.substringWithRange( range )
result will be "ABCDEFGHI"
the startIndex and endIndex could also be used in
let mySubString = myString.substringFromIndex(startIndex)
and so on!
PS: As indicated in the remarks, there are some syntax changes in swift 2 which comes with xcode 7 and iOS9!
Please look at this page
For example to find the first name (up to the first space) in my full name:
let name = "Joris Kluivers"
let start = name.startIndex
let end = find(name, " ")
if end {
let firstName = name[start..end!]
} else {
// no space found
}
start and end are of type String.Index here and are used to create a Range<String.Index> and used in the subscript accessor (if a space is found at all in the original string).
It's hard to create a String.Index directly from an integer position as used in the opening post. This is because in my name each character would be of equal size in bytes. But characters using special accents in other languages could have used several more bytes (depending on the encoding used). So what byte should the integer refer to?
It's possible to create a new String.Index from an existing one using the methods succ and pred which will make sure the correct number of bytes are skipped to get to the next code point in the encoding. However in this case it's easier to search for the index of the first space in the string to find the end index.
Since String is a bridge type for NSString, the "old" methods should work, but it's not clear how - e.g., this doesn't work either (doesn't appear to be valid syntax):
let x = str.substringWithRange(NSMakeRange(0, 3))
To me, that is the really interesting part of your question. String is bridged to NSString, so most NSString methods do work directly on a String. You can use them freely and without thinking. So, for example, this works just as you expect:
// delete all spaces from Swift String stateName
stateName = stateName.stringByReplacingOccurrencesOfString(" ", withString:"")
But, as so often happens, "I got my mojo workin' but it just don't work on you." You just happened to pick one of the rare cases where a parallel identically named Swift method exists, and in a case like that, the Swift method overshadows the Objective-C method. Thus, when you say str.substringWithRange, Swift thinks you mean the Swift method rather than the NSString method — and then you are hosed, because the Swift method expects a Range<String.Index>, and you don't know how to make one of those.
The easy way out is to stop Swift from overshadowing like this, by casting explicitly:
let x = (str as NSString).substringWithRange(NSMakeRange(0, 3))
Note that no significant extra work is involved here. "Cast" does not mean "convert"; the String is effectively an NSString. We are just telling Swift how to look at this variable for purposes of this one line of code.
The really weird part of this whole thing is that the Swift method, which causes all this trouble, is undocumented. I have no idea where it is defined; it is not in the NSString header and it's not in the Swift header either.
The short answer is that this is really hard in Swift right now. My hunch is that there is still a bunch of work for Apple to do on convenience methods for things like this.
String.substringWithRange() is expecting a Range<String.Index> parameter, and as far as I can tell there isn't a generator method for the String.Index type. You can get String.Index values back from aString.startIndex and aString.endIndex and call .succ() or .pred() on them, but that's madness.
How about an extension on the String class that takes good old Ints?
extension String {
subscript (r: Range<Int>) -> String {
get {
let subStart = advance(self.startIndex, r.startIndex, self.endIndex)
let subEnd = advance(subStart, r.endIndex - r.startIndex, self.endIndex)
return self.substringWithRange(Range(start: subStart, end: subEnd))
}
}
func substring(from: Int) -> String {
let end = countElements(self)
return self[from..<end]
}
func substring(from: Int, length: Int) -> String {
let end = from + length
return self[from..<end]
}
}
let mobyDick = "Call me Ishmael."
println(mobyDick[8...14]) // Ishmael
let dogString = "This 🐶's name is Patch."
println(dogString[5..<6]) // 🐶
println(dogString[5...5]) // 🐶
println(dogString.substring(5)) // 🐶's name is Patch.
println(dogString.substring(5, length: 1)) // 🐶
Update: Swift beta 4 resolves the issues below!
As it stands [in beta 3 and earlier], even Swift-native strings have some issues with handling Unicode characters. The dog icon above worked, but the following doesn't:
let harderString = "1:1️⃣"
for character in harderString {
println(character)
}
Output:
1
:
1
️
⃣
In new Xcode 7.0 use
//: Playground - noun: a place where people can play
import UIKit
var name = "How do you use String.substringWithRange?"
let range = name.startIndex.advancedBy(0)..<name.startIndex.advancedBy(10)
name.substringWithRange(range)
//OUT:
You can use this extensions to improve substringWithRange
Swift 2.3
extension String
{
func substringWithRange(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(end))
return self.substringWithRange(range)
}
func substringWithRange(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(start + location))
return self.substringWithRange(range)
}
}
Swift 3
extension String
{
func substring(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: end)
let range = startIndex..<endIndex
return self.substring(with: range)
}
func substring(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: start + location)
let range = startIndex..<endIndex
return self.substring(with: range)
}
}
Usage:
let str = "Hello, playground"
let substring1 = str.substringWithRange(0, end: 5) //Hello
let substring2 = str.substringWithRange(7, location: 10) //playground
Sample Code for how to get substring in Swift 2.0
(i) Substring from starting index
Input:-
var str = "Swift is very powerful language!"
print(str)
str = str.substringToIndex(str.startIndex.advancedBy(5))
print(str)
Output:-
Swift is very powerful language!
Swift
(ii) Substring from particular index
Input:-
var str = "Swift is very powerful language!"
print(str)
str = str.substringFromIndex(str.startIndex.advancedBy(6)).substringToIndex(str.startIndex.advancedBy(2))
print(str)
Output:-
Swift is very powerful language!
is
I hope it will help you!
Easy solution with little code.
Make an extension that includes basic subStringing that nearly all other languages have:
extension String {
func subString(start: Int, end: Int) -> String {
let startIndex = self.index(self.startIndex, offsetBy: start)
let endIndex = self.index(startIndex, offsetBy: end)
let finalString = self.substring(from: startIndex)
return finalString.substring(to: endIndex)
}
}
Simply call this with
someString.subString(start: 0, end: 6)
This works in my playground :)
String(seq: Array(str)[2...4])
Updated for Xcode 7. Adds String extension:
Use:
var chuck: String = "Hello Chuck Norris"
chuck[6...11] // => Chuck
Implementation:
extension String {
/**
Subscript to allow for quick String substrings ["Hello"][0...1] = "He"
*/
subscript (r: Range<Int>) -> String {
get {
let start = self.startIndex.advancedBy(r.startIndex)
let end = self.startIndex.advancedBy(r.endIndex - 1)
return self.substringWithRange(start..<end)
}
}
}
try this in playground
var str:String = "Hello, playground"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,8))
it will give you "ello, p"
However where this gets really interesting is that if you make the last index bigger than the string in playground it will show any strings that you defined after str :o
Range() appears to be a generic function so that it needs to know the type it is dealing with.
You also have to give it the actual string your interested in playgrounds as it seems to hold all stings in a sequence one after another with their variable name afterwards.
So
var str:String = "Hello, playground"
var str2:String = "I'm the next string"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,49))
gives "ello, playground�str���I'm the next string�str2�"
works even if str2 is defined with a let
:)
Rob Napier had already given a awesome answer using subscript. But i felt one drawback in that as there is no check for out of bound conditions. This can tend to crash. So i modified the extension and here it is
extension String {
subscript (r: Range<Int>) -> String? { //Optional String as return value
get {
let stringCount = self.characters.count as Int
//Check for out of boundary condition
if (stringCount < r.endIndex) || (stringCount < r.startIndex){
return nil
}
let startIndex = self.startIndex.advancedBy(r.startIndex)
let endIndex = self.startIndex.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: startIndex, end: endIndex)]
}
}
}
Output below
var str2 = "Hello, World"
var str3 = str2[0...5]
//Hello,
var str4 = str2[0..<5]
//Hello
var str5 = str2[0..<15]
//nil
So i suggest always to check for the if let
if let string = str[0...5]
{
//Manipulate your string safely
}
In Swift3
For ex: a variable "Duke James Thomas", we need to get "James".
let name = "Duke James Thomas"
let range: Range<String.Index> = name.range(of:"James")!
let lastrange: Range<String.Index> = img.range(of:"Thomas")!
var middlename = name[range.lowerBound..<lstrange.lowerBound]
print (middlename)
Taking a page from Rob Napier, I developed these Common String Extensions, two of which are:
subscript (r: Range<Int>) -> String
{
get {
let startIndex = advance(self.startIndex, r.startIndex)
let endIndex = advance(self.startIndex, r.endIndex - 1)
return self[Range(start: startIndex, end: endIndex)]
}
}
func subString(startIndex: Int, length: Int) -> String
{
var start = advance(self.startIndex, startIndex)
var end = advance(self.startIndex, startIndex + length)
return self.substringWithRange(Range<String.Index>(start: start, end: end))
}
Usage:
"Awesome"[3...7] //"some"
"Awesome".subString(3, length: 4) //"some"
This is how you get a range from a string:
var str = "Hello, playground"
let startIndex = advance(str.startIndex, 1)
let endIndex = advance(startIndex, 8)
let range = startIndex..<endIndex
let substr = str[range] //"ello, pl"
The key point is that you are passing a range of values of type String.Index (this is what advance returns) instead of integers.
The reason why this is necessary, is that strings in Swift don't have random access (because of variable length of Unicode characters basically). You also can't do str[1]. String.Index is designed to work with their internal structure.
You can create an extension with a subscript though, that does this for you, so you can just pass a range of integers (see e.g. Rob Napier's answer).
I tried to come up with something Pythonic.
All the subscripts here are great, but the times I really need something simple is usually when I want to count from back, e.g. string.endIndex.advancedBy(-1)
It supports nil values, for both start and end index, where nil would mean index at 0 for start, string.characters.count for end.
extension String {
var subString: (Int?) -> (Int?) -> String {
return { (start) in
{ (end) in
let startIndex = start ?? 0 < 0 ? self.endIndex.advancedBy(start!) : self.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? self.characters.count < 0 ? self.endIndex.advancedBy(end!) : self.startIndex.advancedBy(end ?? self.characters.count)
return startIndex > endIndex ? "" : self.substringWithRange(startIndex ..< endIndex)
}
}
}
}
let dog = "Dog‼🐶"
print(dog.subString(nil)(-1)) // Dog!!
EDIT
A more elegant solution:
public extension String {
struct Substring {
var start: Int?
var string: String
public subscript(end: Int?) -> String {
let startIndex = start ?? 0 < 0 ? string.endIndex.advancedBy(start!) : string.startIndex.advancedBy(start ?? 0)
let endIndex = end ?? string.characters.count < 0 ? string.endIndex.advancedBy(end!) : string.startIndex.advancedBy(end ?? string.characters.count)
return startIndex > endIndex ? "" : string.substringWithRange(startIndex ..< endIndex)
}
}
public subscript(start: Int?) -> Substring {
return Substring(start: start, string: self)
}
}
let dog = "Dog‼🐶"
print(dog[nil][-1]) // Dog!!
First create the range, then the substring. You can use fromIndex..<toIndex syntax like so:
let range = fullString.startIndex..<fullString.startIndex.advancedBy(15) // 15 first characters of the string
let substring = fullString.substringWithRange(range)
here is a example to get video-Id only .i.e (6oL687G0Iso) from the whole URL in swift
let str = "https://www.youtube.com/watch?v=6oL687G0Iso&list=PLKmzL8Ib1gsT-5LN3V2h2H14wyBZTyvVL&index=2"
var arrSaprate = str.componentsSeparatedByString("v=")
let start = arrSaprate[1]
let rangeOfID = Range(start: start.startIndex,end:start.startIndex.advancedBy(11))
let substring = start[rangeOfID]
print(substring)
let startIndex = text.startIndex
var range = startIndex.advancedBy(1) ..< text.endIndex.advancedBy(-4)
let substring = text.substringWithRange(range)
Full sample you can see here
http://www.learnswiftonline.com/reference-guides/string-reference-guide-for-swift/
shows that this works well:
var str = "abcd"
str = str.substringToIndex(1)
Well, I had the same issue and solved with the "bridgeToObjectiveC()" function:
var helloworld = "Hello World!"
var world = helloworld.bridgeToObjectiveC().substringWithRange(NSMakeRange(6,6))
println("\(world)") // should print World!
Please note that in the example, substringWithRange in conjunction with NSMakeRange take the part of the string starting at index 6 (character "W") and finishing at index 6 + 6 positions ahead (character "!")
Cheers.
You can use any of the substring methods in a Swift String extension I wrote https://bit.ly/JString.
var string = "hello"
var sub = string.substringFrom(3) // or string[3...5]
println(sub)// "lo"
If you have an NSRange, bridging to NSString works seamlessly. For example, I was doing some work with UITextFieldDelegate and I quickly wanted to compute the new string value when it asked if it should replace the range.
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
let newString = (textField.text as NSString).stringByReplacingCharactersInRange(range, withString: string)
println("Got new string: ", newString)
}
Simple extension for String:
extension String {
func substringToIndex(index: Int) -> String {
return self[startIndex...startIndex.advancedBy(min(index, characters.count - 1))]
}
}
If you don't care about performance... this is probably the most concise solution in Swift 4
extension String {
subscript(range: CountableClosedRange<Int>) -> String {
return enumerated().filter{$0.offset >= range.first! && $0.offset < range.last!}
.reduce(""){$0 + String($1.element)}
}
}
It enables you to do something like this:
let myStr = "abcd"
myStr[0..<2] // produces "ab"