How to split an array in half in Swift? - arrays

How do I split a deck of cards? I have an array made and a random card dealer, but have no idea how to split the deck.
Thanks everyone for the help! I now have a working card app, did run into other problems but they were solved quickly.

You can make an extension so it can return an array of two arrays, working with Ints, Strings, etc:
extension Array {
func split() -> [[Element]] {
let ct = self.count
let half = ct / 2
let leftSplit = self[0 ..< half]
let rightSplit = self[half ..< ct]
return [Array(leftSplit), Array(rightSplit)]
}
}
let deck = ["J", "Q", "K", "A"]
let nums = [0, 1, 2, 3, 4]
deck.split() // [["J", "Q"], ["K", "A"]]
nums.split() // [[0, 1], [2, 3, 4]]
But returning a named tuple is even better, because it enforces the fact that you expect exactly two arrays as a result:
extension Array {
func split() -> (left: [Element], right: [Element]) {
let ct = self.count
let half = ct / 2
let leftSplit = self[0 ..< half]
let rightSplit = self[half ..< ct]
return (left: Array(leftSplit), right: Array(rightSplit))
}
}
let deck = ["J", "Q", "K", "A"]
let splitDeck = deck.split()
print(splitDeck.left) // ["J", "Q"]
print(splitDeck.right) // ["K", "A"]
Note: credits goes to Andrei and Qbyte for giving the first correct answer, I'm just adding info.

You can use subscript range
let deck: [String] = ["J", "Q", "K", "A"]
// use ArraySlice only for transient computation
let leftSplit: ArraySlice<String> = deck[0 ..< deck.count / 2] // "J", "Q"
let rightSplit: ArraySlice<String> = deck[deck.count / 2 ..< deck.count] // "K", "A"
// make arrays from ArraySlice
let leftDeck: [String] = Array(leftSplit) // "J", "Q"
let rightDeck: [String] = Array(rightSplit) // "K", "A"
EDIT: above code is for Swift 2, maybe for Swift 3 is a more convenient way.

Swift
More generic solution to split the array into chunks the answer from this link
extension Array {
func chunked(into size: Int) -> [[Element]] {
return stride(from: 0, to: count, by: size).map {
Array(self[$0 ..< Swift.min($0 + size, count)])
}
}
}
let numbers = Array(1...100)
let result = numbers.chunked(into: 5)

And one more realization of previously provided ideas. Firstly, up to Swift current documentation, it is better to choose names in past simple tense for functions that produce some result and present tense for mutating ones.
As second, as for me, it is better to choose half adding count % 2 to give more uniformed result.
Here is it:
extension Array {
func devided() -> ([Element], [Element]) {
let half = count / 2 + count % 2
let head = self[0..<half]
let tail = self[half..<count]
return (Array(head), Array(tail))
}
}
And results:
let set1 = [1, 2, 3, 4, 5, 6, 7,8]
let set2 = [1, 2, 3, 4, 5]
let set3 = [1]
let set4 = [Int]()
print(set1.devided())
print(set2.devided())
print(set3.devided())
print(set4.devided())
([1, 2, 3, 4], [5, 6, 7, 8])
([1, 2, 3], [4, 5])
([1], [])
([], [])

You can create an extension on SequenceType, and create a function named divide.
This function would iterate through the elements of the sequence while placing those that match the predicate into one array (slice) and those that do not match into another array (remainder).
The function returns a tuple containing the slice and the remainder.
extension SequenceType {
/**
Returns a tuple with 2 arrays.
The first array (the slice) contains the elements of self that match the predicate.
The second array (the remainder) contains the elements of self that do not match the predicate.
*/
func divide(#noescape predicate: (Self.Generator.Element) -> Bool) -> (slice: [Self.Generator.Element], remainder: [Self.Generator.Element]) {
var slice: [Self.Generator.Element] = []
var remainder: [Self.Generator.Element] = []
forEach {
switch predicate($0) {
case true : slice.append($0)
case false : remainder.append($0)
}
}
return (slice, remainder)
}
}
This is an example
let tuple = [1, 2, 3, 4, 5].divide({ $0 >= 3 })
tuple.slice // [3, 4, 5]
tuple.remainder // [1, 2]

Related

Swift Array Extension to replace value of index n by the sum of the n-previous values

I am trying to write an extension for Array Types that sums the n-previous indexes in the index n.
let myArray = [1, 2, 3, 4, 5]
let mySumArray = myArray.sumNIndex()
print(mySumArray)
// returns [1,3,6,10,15]
I have tried various approaches which all failed at some point.
For instance, the example hereafter triggers a compile error
"Cannot invoke 'reduce' with an argument list of type '(Int, _)'":
extension Array {
mutating func indexSum() {
var tempArray = [Any]()
for index in 1...self.count - 1 {
self[index] += self[.prefix(index + 2).reduce(0, +)]
}
}
}
This other attempt triggers another compile error:
"Binary operator '+=' cannot be applied to two 'Element' operands"
extension Array {
mutating func indexSum() {
var tempArray = [Any]()
for index in 1...self.count - 1 {
self[index] += self[index - 1]
}
}
}
Any idea is welcome!
Thank you very much for your help!
EDIT: Many thanks to #Martin and #Carpsen who figured it out in 2 different ways
#Martin using map method:
extension Array where Element: Numeric {
func cumulativeSum() -> [Element] {
var currentSum: Element = 0
return map {
currentSum += $0
return currentSum
}
}
}
#Carpsen using reduce method:
extension Array where Element: Numeric {
func indexSum() -> [Element] {
return self.reduce(into: [Element]()) {(acc, element) in
return acc + [(acc.last ?? 0) + element]
})
}
}
The main problem is that the addition operator + is not defined for elements
of arbitrary arrays. You need to restrict the extension method, e.g. to
arrays of Numeric elements.
Also there is no need to use Any.
Here is a possible implementation as a non-mutating method:
extension Array where Element: Numeric {
func cumulativeSum() -> [Element] {
var currentSum: Element = 0
return map {
currentSum += $0
return currentSum
}
}
}
Examples:
let intArray = [1, 2, 3, 4, 5]
print(intArray.cumulativeSum()) // [1, 3, 6, 10, 15]
let floatArray = [1.0, 2.5, 3.25]
print(floatArray.cumulativeSum()) [1.0, 3.5, 6.75]
In a similar fashion we can “cumulatively join” the elements of a
string array. enumerated() is now used to provide the current element
index together with the element, and that is used to decide whether to
insert the separator or not:
extension Array where Element == String {
func cumulativeJoin(separator: String) -> [Element] {
var currentJoin = ""
return enumerated().map { (offset, elem) in
if offset > 0 { currentJoin.append(separator) }
currentJoin.append(elem)
return currentJoin
}
}
}
Examples:
let stringArray = ["a", "b", "c"]
print(stringArray.cumulativeJoin()) // ["a", "ab", "abc"]
print(stringArray.cumulativeJoin(separator: ":")) // ["a", "a:b", "a:b:c"]
Try this:
let myArray = [1, 2, 3, 4, 5]
myArray.reduce([Int](), {accumulator, element in
return accumulator + [(accumulator.last ?? 0) + element]
})
//[1, 3, 6, 10, 15]
What this reduce does is:
Start with an empty array
With each element from myArray it calculates its sum with the last element in the accumulator
Return the previous array plus the last sum
Here is a simpler, but longer version:
let myArray = [1, 2, 3, 4, 5]
let newArray = myArray.reduce([Int](), {accumulator, element in
var tempo = accumulator
let lastElementFromTheAccumulator = accumulator.last ?? 0
let currentSum = element + lastElementFromTheAccumulator
tempo.append(currentSum)
return tempo
})
print(newArray) //[1, 3, 6, 10, 15]
A more efficient solution, as suggested by Martin R in the comments, uses reduce(into:):
myArray.reduce(into: [Int]()) { (accumulator, element) in
accumulator += [(accumulator.last ?? 0) + element]
}
//[1, 3, 6, 10, 15]
And you could have it as an extension:
extension Array where Element: Numeric {
func indexSum() -> [Element] {
return self.reduce([Element](), {acc, element in
return acc + [(acc.last ?? 0) + element]
})
}
}
myArray.indexSum() //[1, 3, 6, 10, 15]
Here a solution that will work with strings too:
extension Array where Element == String {
func indexSum() -> [String] {
return self.reduce(into: [String]()) {(accumulator, element) in
accumulator += [(accumulator.last ?? "") + element]
}
}
}
["a", "b", "c", "d"].indexSum() //["a", "ab", "abc", "abcd"]
If you'd like to have a separator between the elements of the initial array elements, you could use this extension:
extension Array where Element == String {
func indexSum(withSparator: String) -> [String] {
return self.reduce(into: [String]()) {(accumulator, element) in
var previousString = ""
if let last = accumulator.last {
previousString = last + " "
}
accumulator += [previousString + element]
}
}
}
["a", "b", "c", "d"].indexSum(withSparator: " ") //["a", "a b", "a b c", "a b c d"]

Divide array into little arrays with the same elements

let's say I have an array this way : [1,4,7,4,2,2,4,7,1,2].
I need a function that divides this array into arrays with the same elements so it shows a result as this in swift :
result = [[1,1],[4,4,4],[7,7],[2,2,2]] .
How to do that in swift ? Thanks in advance
You can use a helper dictionary to categorize the values of your array into the appropriate bins. E.g.:
let arr = [1, 4, 7, 4, 2, 2, 4, 7, 1, 2]
var dict: [Int: [Int]] = [:]
arr.forEach { dict[$0] = (dict[$0] ?? []) + [$0] }
let inBins = dict.map{ $1 }.sorted{ $0.first ?? 0 < $1.first ?? 0 }
print(inBins) // [[1, 1], [2, 2, 2], [4, 4, 4], [7, 7]]
Or, make use of the general Sequence extension for the categorising part, as described in the accepted answer in thread linked to by #Hamish:
How to group by the elements of an array in Swift
E.g.:
/* from https://stackoverflow.com/a/39388832/4573247:
#mientus's Swift 3 translation of #oisdk's accepted answer */
public extension Sequence {
func categorise<U : Hashable>(_ key: (Iterator.Element) -> U) -> [U:[Iterator.Element]] {
var dict: [U:[Iterator.Element]] = [:]
for el in self {
let key = key(el)
if case nil = dict[key]?.append(el) { dict[key] = [el] }
}
return dict
}
}
let arr = [1, 4, 7 ,4, 2, 2, 4, 7, 1, 2]
let inBins = arr.categorise{ $0 }.map{ $1 }.sorted{ $0.first ?? 0 < $1.first ?? 0 }
print(inBins) // [[1, 1], [2, 2, 2], [4, 4, 4], [7, 7]]
No need for the bins to be sorted (as above)? The two options above are then reduced to (simply dropping the last sortin part):
// ... first alternative above
let inBins = dict.map{ $1 }
// ... 2nd alternative above
let inBins = arr.categorise{ $0 }.map{ $1 }
Another option could be to create an NSCountedSet:
let array = [1,4,7,4,2,2,4,7,1,2]
let countedSet = NSCountedSet(array: array)
You could then easily get the count of each unique element:
let countForOne = countedSet.count(for: 1)
As far as I know there is no native Swift equivalent of NSCountedSet yet.

Nested arrays using Any/AnyObject

According to Apple's official book "The Swift Programming Language (Swift 2.1)":
AnyObject can represent an instance of any class type.
Any can represent an instance of any type at all, including function types.
Knowing that, I wanted to emulate Python-like nested list using an Any array.
let array = [1, 2, [3, 4], [5, 6, [7, 8]], 1]
Since Int and Array are value types, I supposed that array would be typed as [Any]
But this wasn't the case here :
func flatten(list:[Any]) -> [Any] {
var new = [Any]()
for element in list {
if let array = element as? [Any] {
// this code never runs
let flattened = flatten(array)
for x in flattened {
new.append(y)
}
}else{
new.append(element)
}
}
return new
}
Note: calling this function gave me EXC_BAD_INSTRUCTION error at first until I did this tweak :
let array : [Any] = [1, 2, [3, 4], [5, 6, [7, 8]], 1]
Output : [1, 2, [3, 4], [5, 6, [7, 8]], 1]
Expected : [1, 2, 3, 4, 5, 6, 7, 8, 1]
Unexplained Solution:
I replaced everywhere in this example Any by AnyObject
Numbers now are of type NSNumber
func flatten(list:[AnyObject]) -> [AnyObject] {
var new = [AnyObject]()
for element in list {
if let array = element as? [AnyObject] {
let flattened = flatten(array)
for x in flattened {
new.append(y)
}
}else{
new.append(element)
}
}
return new
}
Output : [1, 2, 3, 4, 5, 6, 7, 8, 1]
Expected : [1, 2, 3, 4, 5, 6, 7, 8, 1]
Question :
Why is it working with [AnyObject] / NSArray and not with [Any], despite the fact that integer literals are mainly of type Int not NSNumber and array literals are of type Array and not NSArray? Is something wrong with my example?
When you use Any, Swift will wrap your nested array elements into NSArray and not into Swift arrays (which can't hold all Objective-C types). So your test if let array = element as? [Any] { will not be true if you use Any because the resulting array isn't of type [Any] but NSArray.
So if you want a function that manages Any you can change your test for:
if let nestedArray = element as? NSArray {
You then have to define a new flattenArray() function with an NSArray prototype:
func flattenArray(array: NSArray) -> [Any] {
var flattenedArray: [Any] = []
for element in array {
// We have an array
if let nestedArray = element as? NSArray {
flattenedArray = flattenedArray + flattenArray(nestedArray)
} else {
// We have a single element
flattenedArray.append(element)
}
}
return flattenedArray
}
func flattenArray(array: [Any]) -> [Any] {
var flattenedArray: [Any] = []
for element in array {
// We have an array
if let nestedArray = element as? NSArray {
flattenedArray = flattenedArray + flattenArray(nestedArray)
} else {
// We have a single element
flattenedArray.append(element)
}
}
return flattenedArray
}
and it will do the trick.
We could probably imagine a more elegant solution but with this example you get the idea behind this issue.

In Swift, what's the cleanest way to get the last two items in an Array?

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"]

Removing objects from an array based on another array

I have two arrays like this:
var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
As you can see, James and Steve match and I want to be able to remove them from arrayA. How would I write this?
#francesco-vadicamo's answer in Swift 2/3/4+
arrayA = arrayA.filter { !arrayB.contains($0) }
The easiest way is by using the new Set container (added in Swift 1.2 / Xcode 6.3):
var setA = Set(arrayA)
var setB = Set(arrayB)
// Return a set with all values contained in both A and B
let intersection = setA.intersect(setB)
// Return a set with all values in A which are not contained in B
let diff = setA.subtract(setB)
If you want to reassign the resulting set to arrayA, simply create a new instance using the copy constructor and assign it to arrayA:
arrayA = Array(intersection)
The downside is that you have to create 2 new data sets.
Note that intersect doesn't mutate the instance it is invoked in, it just returns a new set.
There are similar methods to add, subtract, etc., you can take a look at them
Like this:
var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
for word in arrayB {
if let ix = find(arrayA, word) {
arrayA.removeAtIndex(ix)
}
}
// now arrayA is ["Mike", "Stacey"]
I agree with Antonio's answer, however for small array subtractions you can also use a filter closure like this:
let res = arrayA.filter { !contains(arrayB, $0) }
matt and freytag's solutions are the ONLY ones that account for duplicates and should be receiving more +1s than the other answers.
Here is an updated version of matt's answer for Swift 3.0:
var arrayA = ["Mike", "James", "Stacey", "Steve"]
var arrayB = ["Steve", "Gemma", "James", "Lucy"]
for word in arrayB {
if let ix = arrayA.index(of: word) {
arrayA.remove(at: ix)
}
}
Original answer
This can also be implemented as a minus func:
func -<T:RangeReplaceableCollectionType where T.Generator.Element:Equatable>( lhs:T, rhs:T ) -> T {
var lhs = lhs
for element in rhs {
if let index = lhs.indexOf(element) { lhs.removeAtIndex(index) }
}
return lhs
}
Now you can use
arrayA - arrayB
Updated implementation for Swift 5
func -<T: RangeReplaceableCollection>(lhs: T, rhs: T) -> T where T.Iterator.Element: Equatable {
var lhs = lhs
for element in rhs {
if let index = lhs.firstIndex(of: element) { lhs.remove(at: index) }
}
return lhs
}
Using the Array → Set → Array method mentioned by Antonio, and with the convenience of an operator, as freytag pointed out, I've been very satisfied using this:
// Swift 3.x/4.x
func - <Element: Hashable>(lhs: [Element], rhs: [Element]) -> [Element]
{
return Array(Set<Element>(lhs).subtracting(Set<Element>(rhs)))
}
For smaller arrays I use:
/* poormans sub for Arrays */
extension Array where Element: Equatable {
static func -=(lhs: inout Array, rhs: Array) {
rhs.forEach {
if let indexOfhit = lhs.firstIndex(of: $0) {
lhs.remove(at: indexOfhit)
}
}
}
static func -(lhs: Array, rhs: Array) -> Array {
return lhs.filter { return !rhs.contains($0) }
}
}
Remove elements using indexes array:
Array of Strings and indexes
let animals = ["cats", "dogs", "chimps", "moose", "squarrel", "cow"]
let indexAnimals = [0, 3, 4]
let arrayRemainingAnimals = animals
.enumerated()
.filter { !indexAnimals.contains($0.offset) }
.map { $0.element }
print(arrayRemainingAnimals)
//result - ["dogs", "chimps", "cow"]
Array of Integers and indexes
var numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
let indexesToRemove = [3, 5, 8, 12]
numbers = numbers
.enumerated()
.filter { !indexesToRemove.contains($0.offset) }
.map { $0.element }
print(numbers)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
Remove elements using element value of another array
Arrays of integers
let arrayResult = numbers.filter { element in
return !indexesToRemove.contains(element)
}
print(arrayResult)
//result - [0, 1, 2, 4, 6, 7, 9, 10, 11]
Arrays of strings
let arrayLetters = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
let arrayRemoveLetters = ["a", "e", "g", "h"]
let arrayRemainingLetters = arrayLetters.filter {
!arrayRemoveLetters.contains($0)
}
print(arrayRemainingLetters)
//result - ["b", "c", "d", "f", "i"]

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