Swift: Set<Object> fundamental operations result in copies or pointers? - swift

I'm currently looking for some reference, outside apple's swift programming reference for the memory space of Set types and resulting fundamental operations (union, intersection, exclusion, subtraction etc)
Given the below pseudo code:
var entities = Set<GKEntity>()
var subSetA = Set<GKEntity>()
var subSetB = Set<GKEntity>()
Each subset will have a subclass of GKEntity which will be called on some routines I will use elswhere in my application.
When I use the union of these subsets, IE: I will have set as the superset of all subsets, or the union of all subsets.
Does this mean that the superset will be a copy of the elements in the subset or will they be pointers only?
I ask this for memory space usage as if the operation requires copying or allocating new memory adresses, I will need to use a different strategy of storing my elements.
From a purely structural standpoint, I assume these will be "shallow" copies meaning they will be pointers to memory adresses, but once I have created the superset as the union of all subsets, I want any removal or addition to be reflected on the superset, or down on the particular subset if the operation is made on the superset.
Hope this question is valid

Note, I'm assuming you mean this GKEntity.
Since GKEntity is a class, the sets you created will store references (pointers) to those actual GKEntity objects. So any changes to the objects in a subset will be reflected in the superset.
Here is a short piece of code that demonstrates this:
class A: IntegerLiteralConvertible, Hashable, CustomStringConvertible {
var x: Int
required init(integerLiteral value: Int) {
self.x = value
}
var hashValue: Int {
return x
}
var description: String {
return String(x)
}
}
func ==(lhs: A, rhs: A) -> Bool {
return lhs.hashValue == rhs.hashValue
}
let setA: Set<A> = [1, 2, 3, 4]
let setB: Set<A> = [5, 6, 7, 8]
print(setA) // [2, 3, 1, 4]
print(setB) // [5, 6, 7, 8]
let union = setA.union(setB)
print(union) // [2, 4, 5, 6, 7, 3, 1, 8]
setA.first!.x = 30
print(union) // [30, 4, 5, 6, 7, 3, 1, 8]
As you can see, I made a change (changed x from 2 to 30) to the first item in setA. Then I printed out union, which did contain a 30 in it.

Related

Swift - Declaring a Set using named variable

I am trying to understand Sets in Swift and how to declare them correctly but I have found the following a little confusing.
I understand that this works:
let elements = [ 1, 2, 3, 4, 5, 1, 2, 6, 7]
let setFromElements = Set(elements)
But I don't understand why the following doesn't:
let setFromElements : Set = elements
Or even:
let setFromElements : Set<Int> = elements
When the following is valid:
let setFromArray : Set = [ 1, 2, 4, 5]
Can someone please help me understand why this is the case?
let setFromArray: Set = [ 1, 2, 4, 5] works because Set conforms to ExpressibleByArrayLiteral and hence has an initializer that takes an ArrayLiteral. See Set.init(arrayLiteral:). This conformance gives syntactic sugar for not having to explicitly call the init.
On the other hand, once you save the array literal into a variable using let elements = [ 1, 2, 3, 4, 5, 1, 2, 6, 7], elements becomes an Array, not an ArrayLiteral and hence another initializer of Set has to be called that takes an Array. This init does not provide syntactic sugar like ExpressibleByArrayLiteral does, so you explicitly have to call the init by doing Set(array).
Set has an initializer that takes an array, and that makes a set, by taking the unique items in the array. But a set is not an array, two different types, so you can't just use = to assign one to the other.

Array with multiple values per index?

I'm learning swift, and I do the sololearn course to get some knowledge, but I bumped into something that I don't understand.
It is about modifying an array's values. The questionable part states the following:
In the following example, the elements with index 1, 2, 3 are replaced with two new values.
shoppingList[1...3] = [“Bananas”, “Oranges”]
How can an one dimensional array take more than one value per index? And how do I access them? Am I misunderstanding something?
What this code does is replacing the element of shoppingList in the 1...3 range using Array.subscript(_:)
That means considering this array:
var shoppingList = ["Apples", "Strawberries", "Pears", "Pineaples"]
that with:
shoppingList[1...3] = ["Bananas", "Oranges"]
Strawberries, Pears and Pineaples will be replaced by Bananas and Oranges.
so the resulting array will be: Apples, Bananas, Oranges
When you assign to a range of indices in an array (array[1...3]), those elements are removed from the array and the new elements are 'slotted in' in their place. This can result in the array growing or shrinking.
var array = Array(0...5)
// [0, 1, 2, 3, 4, 5]
array[1...3] = [-1, -2]
// [0, -1, -2, 3, 4]
Notice how our array's length is now one element shorter.
You could use a tuple (Value, Value), or create a struct to handle your values there, in fact if you plan to reuse this pair or value, a struct is the way to go.
By the way, there's no need to add [1..3], just put the values inside the brackets.
struct Value {
var name: String
var lastName: String
}
let values = [Value(name: "Mary", lastName: "Queen"), Value(name: "John", lastName: "Black")]
// Access to properties
let lastName = values[1].lastName
// OR
let tuples = [("Mary", "Queen"), ("John", "Black")]
let lastNameTuple = tuples[1].1
Hope you're enjoying Swift!

How to create a new array of even numbers from an existing array?

I am trying to create a function that takes an array of Int, and returns a new array of all of the even numbers in the original array.
I have been fumbling around with this code (I am a very new beginner)
let numberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var newArray: [Int] = []
func newInt(newEven: Int, newArray: [Int]) -> Int {
for newEven in numberArray{
var index = 0
index += 1
if newEven % 2 == 0 {
newArray.insert(newEven, at:[index])
}
return newEven
}
}
print(newArray)
This is a good start! Here are some pointers:
1. Formatting
The formatting needs some work. Generally, every new scope ({ ... }) should introduce a new layer of indentation, like so:
let numberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var newArray: [Int] = []
func newInt(newEven: Int, newArray: [Int]) -> Int {
for newEven in numberArray{
var index = 0
index += 1
if newEven % 2 == 0 {
newArray.insert(newEven, at:[index])
}
return newEven
}
}
print(newArray)
Now we can make some observations:
1. index is scoped to the for loop body, and will always have the same value of 0, and then 1 after the next line.
2. The return statement is within the for loop body, and unconditional. This function will always return the value of the first element of numberArray
3. The return type of this function is Int. But in your question, you state that you want this to return an array of all of the even numbers. So your return type will have to be Array<Int> (a.k.a. [Int]), not just Int.
2. Compilation issues
This function has several errors that will prevent compilation:
The return statement is within a loop body. If numberArray is empty, and the for loop body is never entered, then you don't hit the return statement. Once control reaches the end of the function, what value should be returned? It's not defined, so that's an error. In Swift, all code paths through a function must return a value. (with the exception of Void functions, which implicitly return nothing at the end)
You're trying to call Array.insert(_:at:) with a second argument of [index], which is an array literal of type Array<Int>. It should just be index.
3. Logical issues
Your function introduces a parameter called newArray, which shadows the global variable newArray on the line before it. The global variable isn't necessary, and you should delete it.
Your function operates over numberArray, but doesn't explicitly take it as input via a parameter. Rather than hardcoding a reference to a global variable like numberArray, you should use a parameter.
The parameter newEven is unused, and is shadowed by the local variable of the for loop
Your function name newInt(newEven:newArray:) doesn't describe what the function does. Consider a function signature like func allEvens(in input: [Int]) -> [Int]
You never actually call this function. You declare it, but never told the program to run it.
You don't need to use Array.insert(_:at:). You can simply use Array.append, which will automatically append elements to the end of the array.
4. Recommendations
Fix the method signature. You want the function to take some numbers, and output only the even numbers. Model that in code: func allEvens(in input: [Int]) -> [Int]
Create a new empty array locally (within the function), into which the even numbers will be stored. As you loop over the input array, check every number if it's even, and if so, append it to the evens array.
Return the evens array.
let numberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var newArray = numberArray.filter({$0 % 2 == 0})
This should return a new array with even numbers.
As LeoDabus mentioned in his comment, the functionality you're seeking is already contained within Swift's Standard Library, so it's not really necessary to write a dedicated function to accomplish that task. Here's how you would do it:
let numberArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
let newArray = numberArray.filter { $0.isMultiple(of: 2) }
print(newArray) // [2, 4, 6, 8, 10]
In this, you're using the filter whether $0 (where $0 is an element in your array) is a multiple of the number you specified in the parameter, in this case, 2.
As you see in the documentation, isMultiple(of:) returns a Bool (true or false). This is the signature:
func isMultiple(of other: Self) -> Bool
I would recommend having a peek at this post covering the topics of map, filter, and reduce. These are useful things to know when starting out in Swift.
Additionally, I've found Dash to be extremely helpful in navigating documetation for Swift.
Update
I should have read your question more thoroughly, as I missed the part where you've gotta do it with a loop. Swift has a cool method called forEach, which I'm a huge fan of. Using that methodology, it would look something like this:
func filter(array: [Int], forMultiplesOf multiple: Int) -> [Int] {
// Create a landing spot for whatever is a multiple...it starts as empty
var newArray: [Int] = []
// This is not the most compact way, but it satisfies the loop constraint
array.forEach { number in
if number % multiple == 0 {
newArray.append(number)
}
}
// Once you're done with the loop, then return the new array you declared at the beginning
return newArray
}
And you'd call it like so:
let newArrayUsingFunction = filter(array: numberArray, forMultiplesOf: 2)
What you're doing here is passing in 2 parameters to the function (array & multiple) to return an array of Ints. See comments in code for what's going on

How to transpose an array more Swiftly?

I want to transpose an array of arrays like this:
Original array:
[
[1,2,3],
[4,5,6],
[7,8,9]
]
Result:
[
1,4,7,
2,5,8,
3,6,9
]
[1,4,7,2,5,8,3,6,9]
Assume all the subarrays have the same length.
If you haven't noticed already, the first three items in the result is the first item of the three subarrays. The fourth, fifth and sixth items in the result is the second item of each subarray.
If you still don't understand, maybe this will help:
At the moment, I have this:
func flatten(array: [[Int]]) -> [Int] {
var flat = [Int]()
for i in 0..<array[0].count {
for subarray in array {
flat.append(subarray[i])
}
}
return flat
}
I don't think that is very swfity. How can I do this in a swifty way?
Just to avoid being an XY problem, here's why I want to do this.
I am developing a board game. I am using HLGridNode (It's basically a bunch of squares in a grid-like layout) from HLSpriteKit as the board game's board. To edit the contents of the grid node, I need to pass in an 1D array of sprite nodes, not a 2D array.
To make my life easier, I stored the model objects in a 2D array. This way, I can refer to the sqaure 5 squares from the left and 2 squares from the top just by doing:
modelObjects[5][2]
If I flatten the 2D array using .flatMap { $0 } and pass the result to the grid node, modelObjects[5][2] would appear to be 2 squares from the left and 5 squares from the top.
This is not a duplicate of this because that question seems to have definite number of arrays to work with. Although I can put my 2D array into a loop, and do those enumerate().map {...} stuff, it seems like a really long-winded approach. I think there must be a simpler to do this with 2D arrays.
Here's an improvement on Shadow Of's answer:
extension Collection where Self.Iterator.Element: RandomAccessCollection {
// PRECONDITION: `self` must be rectangular, i.e. every row has equal size.
func transposed() -> [[Self.Iterator.Element.Iterator.Element]] {
guard let firstRow = self.first else { return [] }
return firstRow.indices.map { index in
self.map{ $0[index] }
}
}
}
let matrix = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
]
matrix.transposed().forEach{ print($0) }
You can receive result you wanted by transpose your 2d matrix, using, for example, this function:
func matrixTranspose<T>(_ matrix: [[T]]) -> [[T]] {
if matrix.isEmpty {return matrix}
var result = [[T]]()
for index in 0..<matrix.first!.count {
result.append(matrix.map{$0[index]})
}
return result
}
and applying flatten (joined in swift 3) then.
let arr = [[1,2,3],[4,5,6],[7,8,9]]
print(matrixTranspose(arr))
// [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
print(matrixTranspose(arr).flatMap{$0})
// [1, 4, 7, 2, 5, 8, 3, 6, 9]
Extension version:
extension Collection where Self.Iterator.Element: Collection {
var transpose: Array<Array<Self.Iterator.Element.Iterator.Element>> {
var result = Array<Array<Self.Iterator.Element.Iterator.Element>>()
if self.isEmpty {return result}
var index = self.first!.startIndex
while index != self.first!.endIndex {
var subresult = Array<Self.Iterator.Element.Iterator.Element>()
for subarray in self {
subresult.append(subarray[index])
}
result.append(subresult)
index = self.first!.index(after: index)
}
return result
}
}
with usage
let arr = [[1,2,3],[4,5,6],[7,8,9]]
print(arr.transpose)
// [[1, 4, 7], [2, 5, 8], [3, 6, 9]]

How to add integer to array (with explicite int index) in swift?

I read swift handbook and was trying to do some exercises. But I run into a problem and I do not know if I do something wrong or if xCode 6 beta is just buggy.
// Playground - noun: a place where people can play
import Cocoa
let interestingNumbers = [
"Prime": [2, 3, 5, 7, 11, 13],
"Fibonacci": [1, 1, 2, 3, 5, 8],
"Square": [1, 4, 9, 16, 25],
]
var largest = 0
var lastLargest = Integer[]()
var index = 0
for (kind, numbers) in interestingNumbers {
for number in numbers {
if number > largest {
//lastLargest[index] = number
index++
largest = number
}
}
}
index
lastLargest
largest
As soon as I uncomment lastLargest[index] = number I do not get any results on right side in playground. Nor I get any infos about index, lastLargest or largest.
Following example does not work either:
var index2 = 0
var lastLargest2 = Integer[]()
lastLargest2[index2] = 1
index2++
lastLargest2[index2] = 2
You are appending using an out of bound array-index. Don't do that. Instead, use append:
lastLargest.append(number)
From Apple's documentation:
You can’t use subscript syntax to append a new item to the end of an array. If you try to use subscript syntax to retrieve or set a value for an index that is outside of an array’s existing bounds, you will trigger a runtime error.
When you're using explicit indexes (subscript notation) to set values in a mutable array, some value must already exist in that array at that index. When you use subscript notation, you're essentially using a 'set', rather than a 'set and add if necessary'.
As a result, you should be using:
lastLargest.insert(number, atIndex: index)
If you want to insert a new item. This will let you insert an item at the specified index, assuming your collection's size is already greater than or equal to the index you're trying to replace.