Related
Writing the question and answer from here, I'm curious to know if there is any simpler way to write the following:
var nums = [1,2,3]
let sum1 = nums.reduce([Int]()){
let temp = $0
temp.append($1)
return temp
}
I know I can do:
var nums = [1,2,3]
let sum1 = nums.reduce([Int]()){
return $0 + [$1]
}
But that comes off as a hack.
To explain this better, I want to get closer to the example (from docs) below, just that it should be for an array:
let numbers = [1, 2, 3, 4]
let numberSum = numbers.reduce(0, { x, y in
x + y
})
EDIT:
Since folks asked what was I trying to achieve:
I was doing the leetcode's group Anagram's challenge.
My solution was:
struct WordTraits: Equatable{
let count: Int
let charactersSet: Set<Character>
}
struct Word: Equatable{
let string: String
let wordTraits: WordTraits
}
class Solution{
func groupAnagrams(_ strs: [String]) -> [[String]]{
var words : [Word] = []
var answers : [(traits: WordTraits, words: [Word])] = []
var count = 0
strs.forEach{ str in
count += 1
let count = str.count
let string = str
let characterSet = Set(str)
let wordTraits = WordTraits(count: count, charactersSet: characterSet)
let word = Word(string: string, wordTraits: wordTraits)
words.append(word)
}
while words.count != 0{
let word = words[0]
let traits = word.wordTraits
var isWordAdded = false
for (i, answer) in answers.enumerated(){
if answer.traits == traits{
answers[i].words.append(word)
isWordAdded = true
break
}
}
if !isWordAdded{
answers.append((traits: traits, words:[word]))
}
words.removeFirst()
}
let emptyArray : [[String]] = []
let finalAnswer = answers.reduce(emptyArray, { total, answer in
let strings : [String] = answer.words.reduce([String](), {
return $0 + [$1.string]
})
return total + [strings]
})
return finalAnswer
}
}
let s = Solution()
print(s.groupAnagrams(["ate", "eta", "beta", "abet"])) // [["ate", "eta"], ["beta", "abet"]]
reduce(..) has to know which type it is working with. To infer this it can use the return type or the type of the first argument. So you can also write:
var nums = [1,2,3]
let sum1: [Int] = nums.reduce([]){
return $0 + [$1]
}
[$1] can't be replaced with $1 because +-operator between value and collection is undefined.
Nope. But you can add it:
extension Array {
func appending(_ newElement: Element) -> Array<Element> {
return self + [newElement]
}
func appending(contentsOf sequence: Sequence) -> Array<Element> {
return self + sequence
}
}
Um, how about the + operator?
let nums = [1, 3, 5]
let more = nums + [7]
Your code is trying to convert a complex structure to an array of arrays. You can use map for this.
This should work:
let finalAnswer = answers.map { answer in
answer.words.map {
$0.string
}
}
Edit:
I was able to solve it using minimal code:
class Solution {
func groupAnagrams(_ words: [String]) -> [[String]] {
let processedWords = words.map {
(key: String($0.sorted()), value: $0)
}
return Dictionary(grouping: processedWords, by: { $0.key }).map { groupedValue in
groupedValue.value.map {
$0.value
}
}
}
}
You've greatly overcomplicated your computation of "final answers". It could just be:
return answers.map { $0.words.map { $0.string } }
Array in Swift has several instance methods for excluding elements, such as dropFirst(), dropLast(), drop(while:), etc. What about drop(at:)?
Note: I'd use remove(at:), but the array I'm working with is a let constant.
Note: I'd use remove(at:), but the array I'm working with is a constant.
I wouldn't let that stop you:
extension Array {
func drop(at:Int) -> [Element] {
var temp = self
temp.remove(at: at)
return temp
}
}
let numbers = [1, 2, 3, 4, 5]
print(numbers.drop(at: 2)) // [1, 2, 4, 5]
You can extend RangeReplaceableCollection protocol instead of Array type, this way you can use it on Strings as well:
extension RangeReplaceableCollection {
func drop(at offset: Int) -> SubSequence {
let index = self.index(startIndex, offsetBy: offset, limitedBy: endIndex) ?? endIndex
let next = self.index(index, offsetBy: 1, limitedBy: endIndex) ?? endIndex
return self[..<index] + self[next...]
}
}
var str = "Hello, playground"
str.drop(at: 5) // "Hello playground"
let numbers = [1, 2, 3, 4, 5]
print(numbers.drop(at: 2)) // "[1, 2, 4, 5]\n"
If you would like to accept also String.Index in your method:
extension RangeReplaceableCollection {
func drop(at index: Index) -> SubSequence {
let index = self.index(startIndex, offsetBy: distance(from: startIndex, to: index), limitedBy: endIndex) ?? endIndex
let next = self.index(index, offsetBy: 1, limitedBy: endIndex) ?? endIndex
return self[..<index] + self[next...]
}
}
var str = "Hello, playground"
str.drop(at: 0) // "ello, playground"
str.drop(at: str.startIndex) // "ello, playground"
The only thing I would add to your implementation is a guard statement with a useful error message:
extension Array {
func drop(at index: Int) -> ArraySlice<Element> {
guard indices.contains(index) else {
fatalError("Index out of range")
}
return self[0..<index] + self[(index+1)..<endIndex]
}
func drop(range: CountableRange<Int>) -> ArraySlice<Element> {
guard (indices.lowerBound, range.upperBound) <= (range.lowerBound, indices.upperBound) else {
fatalError("Range is out of the indices bounds")
}
return self[0..<range.lowerBound] + self[range.upperBound..<endIndex]
}
}
let a = [1,2,3]
a.drop(at: 3) //Fatal error: Index out of range
let b = [0,1,2,3,4,5]
b.drop(range: 1..<5) //[0, 5]
return self[0..<index] + self[index+1..<endIndex]
Ugly. Why not use the tools you're given?
extension Array {
func drop(at:Int) -> ArraySlice<Element> {
return prefix(upTo: at) + suffix(from: at+1)
}
}
let arr = [1,2,3,4,5]
let slice = arr.drop(at:2) // [1,2,4,5]
EDIT It seems Apple would prefer you to say (using partial ranges)
return self[..<at] + self[(at+1)...]
but personally I think that's uglier, and after all the methods I suggested are not deprecated or anything.
How about using a Swift extension to add drop(at:) to the Array structure?
extension Array {
func drop(at index: Int) -> ArraySlice<Element> {
precondition(indices.contains(index), "Index out of range")
return self[..<index] + self[(index+1)...]
}
}
It returns a slice of the original array without the element at the specified index.
let numbers = [1, 2, 3, 4, 5]
print(numbers.drop(at: 2))
// Prints "[1, 2, 4, 5]"
Note: You may also want to add drop(at:) to ArraySlice.
If I have two arrays e.g
let one = [1,3,5]
let two = [2,4,6]
I would like to merge/interleave the arrays in the following pattern [one[0], two[0], one[1], two[1] etc....]
//prints [1,2,3,4,5,6]
let comibned = mergeFunction(one, two)
print(combined)
What would be a good way to implement the combining function?
func mergeFunction(one: [T], _ two: [T]) -> [T] {
var mergedArray = [T]()
//What goes here
return mergedArray
}
If both arrays have the same length then this is a possible solution:
let one = [1,3,5]
let two = [2,4,6]
let merged = zip(one, two).flatMap { [$0, $1] }
print(merged) // [1, 2, 3, 4, 5, 6]
Here zip() enumerates the arrays in parallel and returns a sequence
of pairs (2-element tuples) with one element from each array. flatMap() creates a 2-element array from each pair and concatenates the result.
If the arrays can have different length then you append the
extra elements of the longer array to the result:
func mergeFunction<T>(one: [T], _ two: [T]) -> [T] {
let commonLength = min(one.count, two.count)
return zip(one, two).flatMap { [$0, $1] }
+ one.suffixFrom(commonLength)
+ two.suffixFrom(commonLength)
}
Update for Swift 3:
func mergeFunction<T>(_ one: [T], _ two: [T]) -> [T] {
let commonLength = min(one.count, two.count)
return zip(one, two).flatMap { [$0, $1] }
+ one.suffix(from: commonLength)
+ two.suffix(from: commonLength)
}
If you're just looking to interleave two arrays, you could just do something like:
let maxIndex = max(one.count, two.count)
var mergedArray = Array<T>()
for index in 0..<maxIndex {
if index < one.count { mergedArray.append(one[index]) }
if index < two.count { mergedArray.append(two[index]) }
}
return mergedArray
With Swift 5, you can use one of the following Playground sample codes in order to solve your problem.
#1. Using zip(_:_:) function and Collection's flatMap(_:) method
let one = [1, 3, 5, 7]
let two = [2, 4, 6]
let array = zip(one, two).flatMap({ [$0, $1] })
print(array) // print: [1, 2, 3, 4, 5, 6]
Apple states:
If the two sequences passed to zip(_:_:) are different lengths, the resulting sequence is the same length as the shorter sequence.
#2. Using an object that conforms to Sequence and IteratorProtocol protocols
struct InterleavedSequence<T>: Sequence, IteratorProtocol {
private let firstArray: [T]
private let secondArray: [T]
private let thresholdIndex: Int
private var index = 0
private var toggle = false
init(firstArray: [T], secondArray: [T]) {
self.firstArray = firstArray
self.secondArray = secondArray
self.thresholdIndex = Swift.min(firstArray.endIndex, secondArray.endIndex)
}
mutating func next() -> T? {
guard index < thresholdIndex else { return nil }
defer {
if toggle {
index += 1
}
toggle.toggle()
}
return !toggle ? firstArray[index] : secondArray[index]
}
}
let one = [1, 3, 5, 7]
let two = [2, 4, 6]
let sequence = InterleavedSequence(firstArray: one, secondArray: two)
let array = Array(sequence)
print(array) // print: [1, 2, 3, 4, 5, 6]
/// Alternates between the elements of two sequences.
/// - Parameter keepSuffix:
/// When `true`, and the sequences have different lengths,
/// the suffix of `interleaved` will be the suffix of the longer sequence.
func interleaved<Sequence: Swift.Sequence>(
with sequence: Sequence,
keepingLongerSuffix keepSuffix: Bool = false
) -> AnySequence<Element>
where Sequence.Element == Element {
keepSuffix
? .init { () -> AnyIterator<Element> in
var iterators = (
AnyIterator( self.makeIterator() ),
AnyIterator( sequence.makeIterator() )
)
return .init {
defer { iterators = (iterators.1, iterators.0) }
return iterators.0.next() ?? iterators.1.next()
}
}
: .init(
zip(self, sequence).lazy.flatMap { [$0, $1] }
)
}
let oddsTo7 = stride(from: 1, to: 7, by: 2)
let evensThrough10 = stride(from: 2, through: 10, by: 2)
let oneThrough6 = Array(1...6)
XCTAssertEqual(
Array( oddsTo7.interleaved(with: evensThrough10) ),
oneThrough6
)
XCTAssertEqual(
Array(
oddsTo7.interleaved(with: evensThrough10, keepingLongerSuffix: true)
),
oneThrough6 + [8, 10]
)
Is there a cleaner way to get the last two items of an array in Swift? In general, I try to avoid this approach since it's so easy to be off-by-one with the indexes. (Using Swift 1.2 for this example.)
// Swift -- slices are kind of a hassle?
let oneArray = ["uno"]
let twoArray = ["uno", "dos"]
let threeArray = ["uno", "dos", "tres"]
func getLastTwo(array: [String]) -> [String] {
if array.count <= 1 {
return array
} else {
let slice: ArraySlice<String> = array[array.endIndex-2..<array.endIndex]
var lastTwo: Array<String> = Array(slice)
return lastTwo
}
}
getLastTwo(oneArray) // ["uno"]
getLastTwo(twoArray) // ["uno", "dos"]
getLastTwo(threeArray) // ["dos", "tres"]
I was hoping for something closer to Python's convenience.
## Python -- very convenient slices
myList = ["uno", "dos", "tres"]
print myList[-2:] # ["dos", "tres"]
With Swift 5, according to your needs, you may choose one of the following patterns in order to get a new array from the last two elements of an array.
#1. Using Array's suffix(_:)
With Swift, objects that conform to Collection protocol have a suffix(_:) method. Array's suffix(_:) has the following declaration:
func suffix(_ maxLength: Int) -> ArraySlice<Element>
Returns a subsequence, up to the given maximum length, containing the final elements of the collection.
Usage:
let array = [1, 2, 3, 4]
let arraySlice = array.suffix(2)
let newArray = Array(arraySlice)
print(newArray) // prints: [3, 4]
#2. Using Array's subscript(_:)
As an alternative to suffix(_:) method, you may use Array's subscript(_:) subscript:
let array = [1, 2, 3, 4]
let range = array.index(array.endIndex, offsetBy: -2) ..< array.endIndex
//let range = array.index(array.endIndex, offsetBy: -2)... // also works
let arraySlice = array[range]
let newArray = Array(arraySlice)
print(newArray) // prints: [3, 4]
myList[-2:]
Yes, I have an enhancement request filed asking for negative index notation, and I suggest you file one too.
However, you shouldn't make this harder on yourself than you have to. The built-in global suffix function does exactly what you're after:
let oneArray = ["uno"]
let twoArray = ["uno", "dos"]
let threeArray = ["uno", "dos", "tres"]
let arr1 = suffix(oneArray,2) // ["uno"]
let arr2 = suffix(twoArray,2) // ["uno", "dos"]
let arr3 = suffix(threeArray,2) // ["dos", "tres"]
The result is a slice, but you can coerce it to an Array if you need to.
in swift 5 you can use suffix for get objects from the last and use prefix for get objects from the first, here is an example:
let exampleArray = ["first text", "second text", "third text"]
let arr1 = exampleArray.suffix(2) // ["second text", "third text"]
let arr2 = exampleArray.prefix(2) // ["first text", "second text"]
The result is a slice, but you can coerce it to an Array if you need to.
In Swift 2, you can extend CollectionType. Here's an example (borrowing from Rob Napier's answer):
extension CollectionType {
func last(count:Int) -> [Self.Generator.Element] {
let selfCount = self.count as! Int
if selfCount <= count - 1 {
return Array(self)
} else {
return Array(self.reverse()[0...count - 1].reverse())
}
}
}
You can use it on any CollectionType. Here's Array:
let array = ["uno", "dos", "tres"]
print(array.last(2)) // [dos, tres]
Here's CharacterView:
let string = "looking"
print(string.characters.last(4)) // [k, i, n, g]
(Note that my example returns an Array in all cases, not the original collection type.)
More generic answer ...
let a1 = [1,2,3,4,5]
let a2 = ["1","2","3","4","5"]
func getLast<T>(array: [T], count: Int) -> [T] {
if count >= array.count {
return array
}
let first = array.count - count
return Array(array[first..<first+count])
}
getLast(a1, count: 2) // [4, 5]
getLast(a2, count: 3) // ["3", "4", "5"]
the last two items of an array in Swift
EDIT: first checks that myArray.count >= 2
let myArray2:Array? = myArray.count >= 2 ? [myArray[myArray.count-2], myArray[myArray.count-1]] : nil
Here it is wrapped in a function which takes the array and either returns an array containing the last two or else returns nil if the passed array does not contain at least two items.
func getLastTwo(myArray:[String]) -> [String]? {
return myArray.count >= 2 ? [myArray[myArray.count-2], myArray[myArray.count-1]] : nil
}
I doubt it's going to make you that much happier, but the math is certainly simpler:
func getLastTwo(array: [String]) -> [String] {
if array.count <= 1 {
return array
} else {
return array.reverse()[0...1].reverse()
}
}
Note that reverse() is lazy, so this isn't particularly expensive.
let items = [0, 2, 5, 3, 7, 6, 9, 10]
let count = items.count
let last2 = items[count - 2 ..< count] // [9, 10]
Swift4 solution:
let oneArray = ["uno"]
let twoArray = ["uno", "dos"]
let threeArray = ["uno", "dos", "tres"]
let arr1 = threeArray.suffix(from: threeArray.count-2) // ["dos", "tres"]
Other examples to clarify the functionality of Swift's built in function func suffix(from start: Int) -> ArraySlice<Element> are...
let arr2 = oneArray.suffix(from: 0) // ["uno"]
let arr3 = twoArray.suffix(from: 0) // ["uno", "dos"]
let arr4 = twoArray.suffix(from: 1) // ["dos"]
let arr5 = threeArray.suffix(from: 1) // ["dos", "tres"]
let arr6 = threeArray.suffix(from: 2) // ["tres"]
Is there a function in Swift that checks whether all elements of an array have the same value? In my case, it's an array of type Int. I know I can iterate over it using a simple for loop I was just wondering if there is something that is built in and quicker.
With Swift 5, you can use one of the four following ways in order to tests if all elements of an array are equal.
#1. Using Array's allSatisfy(_:) method
allSatisfy(_:) returns a Boolean value indicating whether every element of a sequence satisfies a given predicate. You can set the predicate to test if all elements of the array are equal:
let array = [1, 1, 1]
let hasAllItemsEqual = array.dropFirst().allSatisfy({ $0 == array.first })
print(hasAllItemsEqual) // prints: true
let array = [1, 1, 3]
let hasAllItemsEqual = array.dropFirst().allSatisfy({ $0 == array.first })
print(hasAllItemsEqual) // prints: false
let array = [Int]()
let hasAllItemsEqual = array.dropFirst().allSatisfy({ $0 == array.first })
print(hasAllItemsEqual) // prints: true
#2. Using Array's reduce(_:_:) method
As an alternative to allSatisfy(_:), you can use reduce(_:_:):
let array = [1, 1, 1]
let hasAllItemsEqual = array.dropFirst().reduce(true) { (partialResult, element) in
return partialResult && element == array.first
}
print(hasAllItemsEqual) // prints: true
let array = [1, 1, 3]
let hasAllItemsEqual = array.dropFirst().reduce(true) { (partialResult, element) in
return partialResult && element == array.first
}
print(hasAllItemsEqual) // prints: false
let array = [Int]()
let hasAllItemsEqual = array.dropFirst().reduce(true) { (partialResult, element) in
return partialResult && element == array.first
}
print(hasAllItemsEqual) // prints: true
#3. Using elementsEqual(_:) method
elementsEqual(_:) returns a Boolean value indicating whether two sequences contain the same elements in the same order. Therefore you can create a new collection by repeating the first element of the initial array and compare the former with the latter:
let array = [1, 1, 1]
precondition(!array.isEmpty)
let repeated = repeatElement(array[0], count: array.count)
let hasAllItemsEqual = array.elementsEqual(repeated)
print(hasAllItemsEqual) // prints: true
let array = [1, 1, 3]
precondition(!array.isEmpty)
let repeated = repeatElement(array[0], count: array.count)
let hasAllItemsEqual = array.elementsEqual(repeated)
print(hasAllItemsEqual) // prints: false
#4. Using Set's init(_:) initalizer
If all elements of an array are equal, creating a set from this array should result in the set having only one element:
let array = [1, 1, 1]
let set = Set(array)
let hasAllItemsEqual = set.count <= 1
print(hasAllItemsEqual) // prints: true
let array = [1, 1, 3]
let set = Set(array)
let hasAllItemsEqual = set.count <= 1
print(hasAllItemsEqual) // prints: false
let array = [Int]()
let set = Set(array)
let hasAllItemsEqual = set.count <= 1
print(hasAllItemsEqual) // prints: true
Any method must iterate over all elements until a different element is found:
func allEqualUsingLoop<T : Equatable>(array : [T]) -> Bool {
if let firstElem = array.first {
for elem in array {
if elem != firstElem {
return false
}
}
}
return true
}
Instead of an explicit loop you can use the contains() function:
func allEqualUsingContains<T : Equatable>(array : [T]) -> Bool {
if let firstElem = array.first {
return !contains(array, { $0 != firstElem })
}
return true
}
If the array elements are Hashable (such as Int) then you can
create a Set (available since Swift 1.2) from the array elements and check if it has exactly one element.
func allEqualUsingSet<T : Hashable>(array : [T]) -> Bool {
let uniqueElements = Set(array)
return count(uniqueElements) <= 1
}
A quick benchmarking test revealed that the "contains" method is much faster than the "set" method
for an array of 1,000,000 integers, in particular if the elements are
not all equal. This make sense because contains() returns as soon
as a non-matching element is found, whereas Set(array) always
traverses the entire array.
Also the "contains" methods is equally fast or slightly faster than an explicit loop.
Here is some simple benchmarking code. Of course the results can vary
with the array size, the number of different elements and the elements data type.
func measureExecutionTime<T>(title: String, #noescape f : (() -> T) ) -> T {
let start = NSDate()
let result = f()
let end = NSDate()
let duration = end.timeIntervalSinceDate(start)
println("\(title) \(duration)")
return result
}
var array = [Int](count: 1_000_000, repeatedValue: 1)
array[500_000] = 2
let b1 = measureExecutionTime("using loop ") {
return allEqualUsingLoop(array)
}
let b2 = measureExecutionTime("using contains") {
allEqualUsingContains(array)
}
let b3 = measureExecutionTime("using set ") {
allEqualUsingSet(array)
}
Results (on a MacBook Pro, Release configuration):
using loop 0.000651001930236816
using contains 0.000567018985748291
using set 0.0344770550727844
With array[1_000] = 2 the results are
using loop 9.00030136108398e-06
using contains 2.02655792236328e-06
using set 0.0306439995765686
Update for Swift 2/Xcode 7: Due to various changes in the Swift
syntax, the function is now written as
func allEqual<T : Equatable>(array : [T]) -> Bool {
if let firstElem = array.first {
return !array.dropFirst().contains { $0 != firstElem }
}
return true
}
But you can now also define it as an extension method for arrays:
extension Array where Element : Equatable {
func allEqual() -> Bool {
if let firstElem = first {
return !dropFirst().contains { $0 != firstElem }
}
return true
}
}
print([1, 1, 1].allEqual()) // true
print([1, 2, 1].allEqual()) // false
Soliution for Swift 4.2/Xcode 10:
let arr = [1, 1, 1, 1]
let allItemsEqual = arr.dropLast().allSatisfy { $0 == arr.last }
print(allItemsEqual)
If your current version of Xcode is prior to 10.0 you can find the function allSatisfy of ArraySlice in Xcode9to10Preparation. You can install this library with CocoaPods.
let ints: [Int] = [1, 1, 1, 1]
print(ints.max() == ints.min())
If you have float buffers or if you already have an array of floats (or you think converting to floats beforehand is convenient):
import Accelerate
// [...]
// let floats = ints.map({ Double($0) })
print(vDSP.minimum(floats) == vDSP.maximum(floats))