extension Array where Element: Numeric {
func closest(to givenValue: Element) -> Element {
let sorted = self.sorted(by: <)
let over = sorted.first(where: { $0 >= givenValue })!
let under = sorted.last(where: { $0 <= givenValue })!
let diffOver = over - givenValue
let diffUnder = givenValue - under
return (diffOver < diffUnder) ? over : under
}
}
In line 3 of this example code, Xcode gives me the incomprehensible error message Ambiguous reference to member '<', along with this great list:
What am I supposed to do here? I just want this array to get sorted.
You have to declare Element to be Comparable:
extension Array where Element: Numeric & Comparable {
In
let sorted = self.sorted(by: <)
you're not giving a boolean function for the function to use. Maybe try replacing it with :
let sorted = self.sorted(by: { $0 < $1 })
Problem is that you have defined your Element as Numeric only where > will work with Comparable.
Do it as:
extension Array where Element: Numeric, Element: Comparable {
func closest(to givenValue: Element) -> Element {
//... your code here ...
}
}
I was just running into a similar problem stemming from the type inference when chaining multiple higher order functions (sort, map, etc), when I found this post and I couldn't help but notice the inefficiency of your function.
Here is an example of how you can implement this same function using binary search, massively reducing the time it takes to execute on large datasets:
extension Array where Element: Numeric & Comparable {
func closest2(to target: Element) -> Element {
if target <= self[0] {
return self[0]
}
if target >= self[count - 1] {
return self[count - 1]
}
var i = 0
var j = count
var mid = 0
while i < j {
mid = (i + j) / 2
if self[mid] == target {
return self[mid]
}
if target < self[mid] {
if mid > 0 && target > self[mid - 1] {
return getClosest(val1: self[mid - 1], val2: self[mid], target: target)
}
j = mid
} else {
if mid < count - 1 && target < self[mid + 1] {
return getClosest(val1: self[mid], val2: self[mid + 1], target: target)
}
i = mid + 1
}
}
return self[mid]
}
private func getClosest(val1: Element, val2: Element, target: Element) -> Element {
return target - val1 > val2 - target ? val2 : val1
}
}
Related
I am doing a simple calculator, but when performing the multiplication and division, my code doesn't make them a priority over plus and minus.
When doing -> 2 + 2 * 4, result = 16 instead of 10...
How to conform to the math logic inside my switch statement?
mutating func calculateTotal() -> Double {
var total: Double = 0
for (i, stringNumber) in stringNumbers.enumerated() {
if let number = Double(stringNumber) {
switch operators[i] {
case "+":
total += number
case "-":
total -= number
case "÷":
total /= number
case "×":
total *= number
default:
break
}
}
}
clear()
return total
}
Assuming you want a generalised and perhaps extensible algorithm for any arithmetic expression, the right way to do this is to use the Shunting Yard algorithm.
You have an input stream, which is the numbers and operators as the user typed them in and you have an output stream, which is the same numbers and operators but rearranged into reverse Polish notation. So, for example 2 + 2 * 4 would be transformed into 2 2 4 * + which is easily calculated by putting the numbers on a stack as you read them and applying the operators to the top items on the stack as you read them.
To do this the algorithm has an operator stack which can be visualised as a siding (hence "shunting yard") into which low priority operators are shunted until they are needed.
The general algorithm is
read an item from the input
if it is a number send it to the output
if the number is an operator then
while the operator on the top of the stack is of higher precedence than the operator you have pop the operator on the stack and send it to the output
push the operator you read from input onto the stack
repeat the above until the input is empty
pop all the operators on the stack into the output
So if you have 2 + 2 * 4 (NB top of the stack is on the left, bottom of the stack is on the right)
start:
input: 2 + 2 * 4
output: <empty>
stack: <empty>
step 1: send the 2 to output
input: + 2 * 4
output: 2
stack: <empty>
step 2: stack is empty so put + on the stack
input: 2 * 4
output: 2
stack: +
step 3: send the 2 to output
input: * 4
output: 2 2
stack: +
step 4: + is lower priority than * so just put * on the stack
input: 4
output: 2 2
stack: * +
step 5: Send 4 to output
input:
output: 2 2 4
stack: * +
step 6: Input is empty so pop the stack to output
input:
output: 2 2 4 * +
stack:
The Wikipedia entry I linked above has a more detailed description and an algorithm that can handle parentheses and function calls and is much more extensible.
For completeness, here is an implementation of my simplified version of the algorithm
enum Token: CustomStringConvertible
{
var description: String
{
switch self
{
case .number(let num):
return "\(num)"
case .op(let symbol):
return "\(symbol)"
}
}
case op(String)
case number(Int)
var precedence: Int
{
switch self
{
case .op(let symbol):
return Token.precedences[symbol] ?? -1
default:
return -1
}
}
var operation: (inout Stack<Int>) -> ()
{
switch self
{
case .op(let symbol):
return Token.operations[symbol]!
case .number(let value):
return { $0.push(value) }
}
}
static let precedences = [ "+" : 10, "-" : 10, "*" : 20, "/" : 20]
static let operations: [String : (inout Stack<Int>) -> ()] =
[
"+" : { $0.push($0.pop() + $0.pop()) },
"-" : { $0.push($0.pop() - $0.pop()) },
"*" : { $0.push($0.pop() * $0.pop()) },
"/" : { $0.push($0.pop() / $0.pop()) }
]
}
struct Stack<T>
{
var values: [T] = []
var isEmpty: Bool { return values.isEmpty }
mutating func push(_ n: T)
{
values.append(n)
}
mutating func pop() -> T
{
return values.removeLast()
}
func peek() -> T
{
return values.last!
}
}
func shuntingYard(input: [Token]) -> [Token]
{
var operatorStack = Stack<Token>()
var output: [Token] = []
for token in input
{
switch token
{
case .number:
output.append(token)
case .op:
while !operatorStack.isEmpty && operatorStack.peek().precedence >= token.precedence
{
output.append(operatorStack.pop())
}
operatorStack.push(token)
}
}
while !operatorStack.isEmpty
{
output.append(operatorStack.pop())
}
return output
}
let input: [Token] = [ .number(2), .op("+"), .number(2), .op("*"), .number(4)]
let output = shuntingYard(input: input)
print("\(output)")
var dataStack = Stack<Int>()
for token in output
{
token.operation(&dataStack)
}
print(dataStack.pop())
If you only have the four operations +, -, x, and ÷, you can do this by keeping track of a pendingOperand and pendingOperation whenever you encounter a + or -.
Then compute the pending operation when you encounter another + or -, or at the end of the calculation. Note that + or - computes the pending operation, but then immediately starts a new one.
I have modified your function to take the stringNumbers, operators, and initial values as input so that it could be tested independently in a Playground.
func calculateTotal(stringNumbers: [String], operators: [String], initial: Double) -> Double {
func performPendingOperation(operand: Double, operation: String, total: Double) -> Double {
switch operation {
case "+":
return operand + total
case "-":
return operand - total
default:
return total
}
}
var total = initial
var pendingOperand = 0.0
var pendingOperation = ""
for (i, stringNumber) in stringNumbers.enumerated() {
if let number = Double(stringNumber) {
switch operators[i] {
case "+":
total = performPendingOperation(operand: pendingOperand, operation: pendingOperation, total: total)
pendingOperand = total
pendingOperation = "+"
total = number
case "-":
total = performPendingOperation(operand: pendingOperand, operation: pendingOperation, total: total)
pendingOperand = total
pendingOperation = "-"
total = number
case "÷":
total /= number
case "×":
total *= number
default:
break
}
}
}
// Perform final pending operation if needed
total = performPendingOperation(operand: pendingOperand, operation: pendingOperation, total: total)
// clear()
return total
}
Tests:
// 4 + 3
calculateTotal(stringNumbers: ["3"], operators: ["+"], initial: 4)
7
// 4 × 3
calculateTotal(stringNumbers: ["3"], operators: ["×"], initial: 4)
12
// 2 + 2 × 4
calculateTotal(stringNumbers: ["2", "4"], operators: ["+", "×"], initial: 2)
10
// 2 × 2 + 4
calculateTotal(stringNumbers: ["2", "4"], operators: ["×", "+"], initial: 2)
8
// 17 - 2 × 3 + 10 + 7 ÷ 7
calculateTotal(stringNumbers: ["2", "3", "10", "7", "7"], operators: ["-", "×", "+", "+", "÷"], initial: 17)
22
First you have to search in the array to see if there is a ÷ or × sign.
Than you can just sum or subtract.
mutating func calculateTotal() -> Double {
var total: Double = 0
for (i, stringNumber) in stringNumbers.enumerated() {
if let number = Double(stringNumber) {
switch operators[i] {
case "÷":
total /= number
case "×":
total *= number
default:
break
}
//Remove the number from the array and make another for loop with the sum and subtract operations.
}
}
clear()
return total
}
This will work if you are not using complex numbers.
If you don't care speed, as it's running by a computer and you may use the machine way to handle it. Just pick one feasible calculate to do it and then repeat until every one is calculated.
Just for fun here. I use some stupid variable and function names.
func evaluate(_ values: [String]) -> String{
switch values[1] {
case "+": return String(Int(values[0])! + Int(values[2])!)
case "-": return String(Int(values[0])! - Int(values[2])!)
case "×": return String(Int(values[0])! * Int(values[2])!)
case "÷": return String(Int(values[0])! / Int(values[2])!)
default: break;
}
return "";
}
func oneTime(_ string: inout String, _ strings: [String]) throws{
if let first = try NSRegularExpression(pattern: "(\\d+)\\s*(\(strings.map{"\\\($0)"}.joined(separator: "|")))\\s*(\\d+)", options: []).firstMatch(in: string , options: [], range: NSMakeRange(0, string.count)) {
let tempResult = evaluate((1...3).map{ (string as NSString).substring(with: first.range(at: $0))})
string.replaceSubrange( Range(first.range(at: 0), in: string)! , with: tempResult)
}
}
func recursive(_ string: inout String, _ strings: [String]) throws{
var count : Int!
repeat{ count = string.count ; try oneTime(&string, strings)
} while (count != string.count)
}
func final(_ string: inout String, _ strings: [[String]]) throws -> String{
return try strings.reduce(into: string) { (result, signs) in
try recursive(&string, signs)
}}
var string = "17 - 23 + 10 + 7 ÷ 7"
try final(&string, [["×","÷"],["+","-"]])
print("result:" + string)
Using JeremyP method and the Shunting Yard algorithm was the way that worked for me, but I had some differences that had to do with the Operator Associativity(left or right priority) so I had to work with it and I developed the code, which is based on JeremyP answer but uses arrays.
First we have the array with the calculation in Strings, e.g.:
let testArray = ["10","+", "5", "*" , "4", "+" , "10", "+", "20", "/", "2"]
We use the function below to get the RPN version using the Shunting Yard algorithm.
func getRPNArray(calculationArray: [String]) -> [String]{
let c = calculationArray
var myRPNArray = [String]()
var operandArray = [String]()
for i in 0...c.count - 1 {
if c[i] != "+" && c[i] != "-" && c[i] != "*" && c[i] != "/" {
//push number
let number = c[i]
myRPNArray.append(number)
} else {
//if this is the first operand put it on the opStack
if operandArray.count == 0 {
let firstOperand = c[i]
operandArray.append(firstOperand)
} else {
if c[i] == "+" || c[i] == "-" {
operandArray.reverse()
myRPNArray.append(contentsOf: operandArray)
operandArray = []
let uniqOperand = c[i]
operandArray.append(uniqOperand)
} else if c[i] == "*" || c[i] == "/" {
let strongOperand = c[i]
//If I want my mult./div. from right(eg because of parenthesis) the line below is all I need
//--------------------------------
// operandArray.append(strongOperand)
//----------------------------------
//If I want my mult./div. from left
let lastOperand = operandArray[operandArray.count - 1]
if lastOperand == "+" || lastOperand == "-" {
operandArray.append(strongOperand)
} else {
myRPNArray.append(lastOperand)
operandArray.removeLast()
operandArray.append(strongOperand)
}
}
}
}
}
//when I have no more numbers I append the reversed operant array
operandArray.reverse()
myRPNArray.append(contentsOf: operandArray)
operandArray = []
print("RPN: \(myRPNArray)")
return myRPNArray
}
and then we enter the RPN array in the function below to calculate the result. In every loop we remove the numbers and the operand used before and we import the previous result and two "p" in the array so in the end we are left with the solution and an array of "p".
func getResultFromRPNarray(myArray: [String]) -> Double {
var a = [String]()
a = myArray
print("a: \(a)")
var result = Double()
let n = a.count
for i in 0...n - 1 {
if n < 2 {
result = Double(a[0])!
} else {
if a[i] == "p" {
//Do nothing else. Calculations are over and the result is in your hands!!!
} else {
if a[i] == "+" {
result = Double(a[i-2])! + Double(a[i-1])!
a.insert(String(result), at: i-2)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.insert("p", at: 0)
a.insert("p", at: 0)
} else if a[i] == "-" {
result = Double(a[i-2])! - Double(a[i-1])!
a.insert(String(result), at: i-2)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.insert("p", at: 0)
a.insert("p", at: 0)
} else if a[i] == "*" {
result = Double(a[i-2])! * Double(a[i-1])!
a.insert(String(result), at: i-2)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.insert("p", at: 0)
a.insert("p", at: 0)
} else if a[i] == "/" {
result = Double(a[i-2])! / Double(a[i-1])!
a.insert(String(result), at: i-2)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.remove(at: i - 1)
a.insert("p", at: 0)
a.insert("p", at: 0)
} else {
// it is a number so do nothing and go the next one
}
}//no over yet
}//n>2
}//iterating
return result
}//Func
I have code which is basically like this:
func arrayHalvesEqual(data:[UInt8]) -> Bool {
let midPoint = data.count / 2
for i in 0..<midPoint {
let b = data[i]
let b2 = data[i + midPoint]
if b != b2 {
return false
}
}
return true
}
This works fine, but sometimes I want to pass in Arrays, and other times ArraySlice. I thought I'd change it to use generics and the CollectionType protocol, which converts as follows:
func arrayHalvesEqual<ByteArray : CollectionType where ByteArray.Generator.Element == UInt8>(data:ByteArray) -> Bool {
let midPoint = data.count / 2
for i in 0..<midPoint {
let b = data[i]
let b2 = data[i + midPoint]
if b != b2 {
return false
}
}
return true
}
However, I get the following compiler error:
error: binary operator '..<' cannot be applied to operands of type 'Int' and 'ByteArray.Index.Distance'
for i in 0..<midPoint {
I can switch the for loop to for i in data.indices which makes that compile, but then I can no longer divide it by 2 to get the midPoint, as data.indices returns the abstract CollectionType.Index whereas / 2 is an Int.
Is it possible to do something like this in Swift? Can I bridge between an abstract protocol Index type and some real type I can do maths on?
P.S: I've seen and found other examples for iterating over the whole collection by using indices and enumerate, but I explicitly only want to iterate over half the collection which requires some sort of division by 2
Thanks
You can restrict the method to collections which are indexed
by Int:
func arrayHalvesEqual<ByteArray : CollectionType where ByteArray.Index == Int, ByteArray.Generator.Element == UInt8>
(data:ByteArray) -> Bool { ... }
This covers both Array and ArraySlice.
And if you use indices.startIndex instead of 0 as initial index
then it suffices to restrict the index type to IntegerType.
Also the data type UInt8 can be replaced by a generic Equatable,
and the entire method shortened to
func arrayHalvesEqual<ByteArray : CollectionType where ByteArray.Index : IntegerType, ByteArray.SubSequence.Generator.Element : Equatable>
(data:ByteArray) -> Bool {
let midPoint = (data.indices.endIndex - data.indices.startIndex)/2
let firstHalf = data[data.indices.startIndex ..< midPoint]
let secondHalf = data[midPoint ..< data.indices.endIndex]
return !zip(firstHalf, secondHalf).contains { $0 != $1 }
}
Imagine we have this code which works perfectly for n >= 0.
func fibonacci(n: Int) -> Int {
var memo = [0,1]
for var i = 2; i <= n; i++ {
memo.append(memo[i-1] + memo[i-2])
}
return memo[n]
}
If I remove the C-style for loop due to upcoming changes to Swift 3.0, I get something like this:
func fibonacci(n: Int) -> Int {
var memo = [0,1]
for i in 2...n {
memo.append(memo[i-1] + memo[i-2])
}
return memo[n]
}
While this works fine for n >= 2, it fails for the numbers 0 and 1 with this error message:
fatal error: Can't form Range with end < start
What's the most concise way to fix this code so it works properly for 0 and 1?
(Note: It's okay, and even desirable, for negative numbers to crash the app.)
Note: I realize I could add a guard statement:
guard n >= 2 else { return memo[n] }
... but I'm hoping there is a better way to fix just the faulty part of the code (2...n).
For example, if there was a concise way to create a range that returns zero elements if end < start, that would be a more ideal solution.
To do this in a way that works for n < 2, you can use the stride method.
let startIndex = 2
let endIndex = n
for i in stride(from: startIndex, through: endIndex, by: 1) {
memo.append(memo[i-1] + memo[i-2])
}
You can easily create a valid range with the max() function:
for i in 2 ..< max(2, n+1) {
memo.append(memo[i-1] + memo[i-2])
}
This evaluates to an empty range 2 ..< 2 if n < 2.
It is important to use the ..< operator which excludes the upper bound because 2 ... 1 is not a valid range.
But in this function I would simply treat the special cases first
func fibonacci(n: Int) -> Int {
// Let it crash if n < 0:
precondition(n >= 0, "n must not be negative")
// Handle n = 0, 1:
if n <= 1 {
return n
}
// Handle n >= 2:
var memo = [0,1]
for i in 2 ... n {
memo.append(memo[i-1] + memo[i-2])
}
return memo[n]
}
(Note that your memo array is set to the initial value [0, 1]
for each function call, so the values are not really "memoized".
Without memoization you don't need an array, it would suffice to keep the last two numbers to compute the next.)
As it turns out, the variable i will always be equal to the count of the memoizing array, so you can just use that as your loop condition:
func fibonacci(n: Int) -> Int {
var memo = [0,1]
while n >= memo.count {
memo.append(memo[memo.count-1] + memo[memo.count-2])
}
return memo[n]
}
Alternatively, you could express the loop as a recursive function:
func fibonacci(n: Int) -> Int {
var memo = [0,1]
func rec(i: Int) -> Int {
if i >= memo.count { memo.append(rec(i-2) + rec(i-1)) }
return memo[i]
}
return rec(n)
}
Really, though, if is the best solution here. Ranges don't allow the end to be smaller than the beginning by design. The extra line for:
func fibonacci(n: Int) -> Int {
if n < 2 { return n }
var memo = [0,1]
for i in 2...n {
memo.append(memo[i-1] + memo[i-2])
}
return memo[n]
}
Is readable and understandable. (To my eye, the code above is better than the for ;; version)
#Marc's answer is great: https://stackoverflow.com/a/34324032/1032900
But the stride syntax is too long for frequent usage, so I made it a little more pleasant for the common i++ usages...
extension Strideable {
#warn_unused_result
public func stride(to end: Self) -> StrideTo<Self> {
return stride(to: end, by: 1)
}
}
extension Strideable {
#warn_unused_result
public func stride(thru end: Self) -> StrideThrough<Self> {
return stride(through: end, by: 1)
}
}
So use like this:
for i in startPos.stride(to: endPos) {
print("pos at: \(i)")
}
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
I have a for loop that checks if a number is a factor of a number, then checks if that factor is prime, and then it adds it to an array. Depending on the original number, I will get an error saying
fatal error: Can't form range with end < start
This happens almost every time, but for some numbers it works fine. The only numbers I have found to work with it are 9, 15, and 25.
Here is the code:
let num = 16 // or any Int
var primes = [Int]()
for i in 2...(num/2) {
if ((num % i) == 0) {
var isPrimeFactor = true
for l in 2...i-1 {
if ((i%l) == 0) {
isPrimeFactor = false;
}//end if
}//end for
if (isPrimeFactor == true) {
primes.append(i)
}//end if
}//end if
}//end for
Swift 5
If you need a loop with dynamic value-range, I suggest that using stride(to:by:) instead of ..< or ...
Basically ..< or ... will be crashed if start_index > end_index.
This will be crash:
let k = 5
for i in 10...k { print("i=\(i)") }
for i in 10..<k { print("i=\(i)") }
How to fix:
let k = 5
for i in stride(from: 10, through: k, by: 1) { print("i=\(i)") }
for i in stride(from: 10, to: k, by: 1) { print("i=\(i)") }
NOTE:
The code above won't print out anything, but it won't be crash when execution.
Also, if you want to stride from a higher number to a lower number then the by parameter needs to be changed to a negative number.
Reference:
http://michael-brown.net/2016/using-stride-to-convert-c-style-for-loops-to-swift-2.2/
http://swiftdoc.org/v2.1/protocol/Strideable/
In your second loop, i will always start at 2, which means you're looping from 2...1
SWIIFT 4
The best way to go is to use stride as by this documentation page:
Generic Function stride(from:to:by:)
for i in stride(from: 10, through: 5, by: -1) { print(i) }
and stride through if you want to include the lowerBound: Generic Function stride(from:through:by:)
Both ClosedRange and CountableRange throw this error unless start <= end. One problem with using stride(from:to:by) is that it returns a Strideable and not a Range, which doesn't work with the "is in range" operator ~=. To handle these cases, I use an extension which gives me a bounds-safe range, where invalid ranges become an empty range:
extension Int {
func until(_ end: Int) -> CountableRange<Int> {
return self <= end ? self..<end : self..<self
}
func through(_ end: Int) -> CountableRange<Int> {
return self <= end ? self..<(end + 1) : self..<self
}
}
Both return CountableRange so that an invalid through range becomes an empty range. (Side note: the name until is chosen because this "safe" range behaves the same as until ranges in Kotlin/Android).
Usage:
for i in 0.until(3) { /* do something */ }
let printRange = { (range: Range<Int>) -> Void in
for i in range {
print(i, terminator: " ")
}
print()
}
printRange(0.until(3)) // prints "0 1 2"
printRange(0.until(0)) // prints ""
printRange(3.until(0)) // prints ""
printRange(0.through(3)) // prints "0 1 2 3"
printRange(0.through(0)) // prints "0"
printRange(3.through(0)) // prints ""
print(0.until(1) ~= -1) // prints "false"
print(0.until(1) ~= 0) // prints "true"
print(0.until(1) ~= 1) // prints "false"
print(0.until(0) ~= 0) // prints "false"
print(1.until(0) ~= 0) // prints "false"
print(0.through(1) ~= -1) // prints "false"
print(0.through(1) ~= 0) // prints "true"
print(0.through(1) ~= 1) // prints "true"
print(0.through(0) ~= 0) // prints "true"
print(1.through(0) ~= 0) // prints "false"
If you want to use range where upperBound < lowerBound, you can add reversed() to it.
for eg: for number in (0...highestPossibleNumber).reversed()
Using a simple solution you can make for loop reverse using Stride and -
func reverseArray(oldarray: [String]) -> [String] {
var newArray = [String]()
let len = oldarray.count
for i in stride(from: len - 1, through: 0, by: -1)
{ newArray.append(oldarray[i])
print(i)
}
return newArray
}
input : print(reverseArray(oldarray: ["World", "Hello"]))
output : ["Hello", "World"]
The easiest solution is to apply ".reversed()" to your array.
Example: array.reversed()