Using swift built in partition to manage elements in an array - swift

iOS 14, Swift 5.x
I watched this excellent WWDC from 2018
https://developer.apple.com/videos/play/wwdc2018/223/
And I wrote a shapes editor... and have been trying to use partition as Dave in the video says you should. I got the first three to work, but the last one I had to use a loop- cannot for the life of me figure out how to get it to work with partition.
Can someone see how I might do this?
The first method moves the selected object to the end of the list, works perfectly.
func bringToFrontEA() {
let subset = objects.partition(by: { $0.selected })
let selected = objects[subset...]
let unselected = objects[..<subset]
let reordered = unselected + selected
objects = Array(reordered)
}
The second method moves the selected object to the front of the list. Works prefectly.
func sendToBackEA() {
let subset = objects.partition(by: { !$0.selected })
let selected = objects[subset...]
let unselected = objects[..<subset]
let reordered = unselected + selected
objects = Array(reordered)
}
The third method moves the element just one element back in the list. Works perfectly.
func sendBackEA() {
if let i = objects.firstIndex(where: { $0.selected }) {
if i == 0 { return }
let predecessor = i - 1
let shapes = objects[predecessor...].partition(by: { !$0.selected })
let slice = objects[predecessor...]
let row = objects[..<predecessor]
let selected = Array(slice[..<shapes])
let unselected = Array(slice[shapes...])
objects = row + selected + unselected
}
}
The last method moves the element forward in the list, works perfectly... but unlike the other methods it will not scale as described in the WWDC video.
func bringForwardEA() {
let indexes = objects.enumerated().filter { $0.element.selected == true }.map{$0.offset}
for i in indexes {
if objects[i+1].unused {
return
}
objects.swapAt(i+1, i)
}
}
Objects is an array of shapes with a property indicating if it is selected or not. I want to exchange the loop in the last method by using a partition as I did in the first three. It needs to work for one or more selected shapes.

Looking at the WWDC video, it appears that what you are calling sendBackEA is what WWDC calls bringForward, and what you are calling bringForwardEA is what WWDC calls sendBack.
Just like how you move the first selected element forward one index (index decreases) in sendBackEA, then move all the other selected elements to immediately after that first selected element. bringForwardEA should do the reverse: move the last selected element backward one index (index increases), then move all the other selected elements to immediately before the last selected element. (See circa 19:10 in the video)
You seem to have confused yourself by trying to increase the indices of all the selected index by 1. This obviously cannot be done with a partition in general.
Also note that partition(by:) already modifies the collection, you don't need to get each partition, then recombine.
Your 4 methods can be written like this:
func bringToFrontEA() {
objects.partition(by: { $0.selected })
}
func sendToBackEA() {
objects.partition(by: { !$0.selected })
}
func sendBackEA() {
if let i = objects.indices.first(where: { objects[$0].selected }) {
if i == 0 { return }
let predecessor = i - 1
objects[predecessor...].partition(by: { !$0.selected })
}
}
func bringForwardEA() {
if let i = objects.indices.last(where: { objects[$0].selected }) {
if i == objects.indices.last { return }
let successor = i + 1
objects[...successor].partition(by: { !$0.selected })
}
}
Notice the symmetry between sendBackEA and bringForwardEA.

Related

Moving Table rows in SwiftUI

Is it possible to support moving rows in a SwiftUI Table view?
I know there's List which can optionally support selection and drag-and-drop to move single or multiple rows. Since it seems similar, I would like to do this with a Table too, but I wasn't able to find any way to do this. Is this possible in SwiftUI? And if it is, what's the best way to do it?
Where I started to figure this out was the WWDC 2021 session "SwiftUI on the Mac: Finishing Touches". I highly recommend this video, as well as the preceding one "SwiftUI on the Mac: Build the Fundamentals". The code for both sessions is available.
Since you didn't include your code to show what you want to do, I have to use my code. I have a table based on an array of an Identifiable struct called Channel. Among a number of fields which are irrelevant to this problem, there is a field "id" of type UUID.
Following the model of the WWDC video, I made an extension to Channel:
import UniformTypeIdentifiers
extension Channel {
static var draggableType = UTType(exportedAs: "com.yourCompany.yourApp.channel")
// define your own type here. don't forget to include it in your info.plist as an exported type
static func fromItemProviders(_ itemProviders: [NSItemProvider], completion: #escaping ([Channel]) -> Void) {
let typeIdentifier = Self.draggableType.identifier
let filteredProviders = itemProviders.filter {
$0.hasItemConformingToTypeIdentifier(typeIdentifier)
}
let group = DispatchGroup()
var result = [Int: Channel]()
for (index, provider) in filteredProviders.enumerated() {
group.enter()
provider.loadDataRepresentation(forTypeIdentifier: typeIdentifier) { (data, error) in
defer { group.leave() }
guard let data = data else { return }
let decoder = JSONDecoder()
guard let channel = try? decoder.decode(Channel.self, from: data)
else { return }
result[index] = channel
}
}
group.notify(queue: .global(qos: .userInitiated)) {
let channels = result.keys.sorted().compactMap { result[$0] }
DispatchQueue.main.async {
completion(channels)
}
}
}
var itemProvider: NSItemProvider {
let provider = NSItemProvider()
provider.registerDataRepresentation(forTypeIdentifier: Self.draggableType.identifier, visibility: .all) {
let encoder = JSONEncoder()
do {
let data = try encoder.encode(self)
$0(data, nil)
} catch {
$0(nil, error)
}
return nil
}
return provider
}
}
This makes an item in the table draggable. Of course, that does no good if there's nothing that will accept the drag. So, you have to make a change to your Table.
Table(selection: $selection, sortOrder: $sortOrder) {
// for clarity, I've removed the table columns
} rows: {
ForEach(document.channels) { channel in
TableRow(channel)
.itemProvider { channel.itemProvider }
}
.onInsert(of: [Channel.draggableType]) { index, providers in
Channel.fromItemProviders(providers) { channels in
document.channels.insert(contentsOf: channels, at: newIndex)
}
}
}
}
Now that will enable you to drag item or items from one window to another. You can, of course, drag within a table now, too. Unfortunately, you will end up making a copy in the new place. Not what you want to do in most cases. How to fix this? Delete the original copy! Of course, you can also run into the problem of indexing in the right place, and if you drag more than one item (from a discontinuous selection, even worse!), the results become, shall we say, undefined.
I still wanted to be able to drag multiple items from another table, so the final onInsert becomes a little more complex (Which I'm sure could be cleaned up a bot further):
Channel.fromItemProviders(providers) { channels in
var newIndex = index
let intraTableDrag = document.channels.contains(where: {$0.id == channels[0].id})
if intraTableDrag && channels.count == 1 {
let oldIndex = document.channels.firstIndex(where: {$0.id == channels[0].id})
if newIndex > oldIndex! {
newIndex -= 1
}
for channel in channels {
let channelID = channel.id
removeChannel(withID: channelID)
}
let maxIndex = document.channels.count
if index > maxIndex {
newIndex = maxIndex
}
}
if (intraTableDrag && channels.count == 1) || !intraTableDrag {
document.channels.insert(contentsOf: channels, at: newIndex)
document.setChannelLocationToArrayOrder()
}
}
}
I hope this is enough to get you started. Good luck!

"Attemped to add a SKNode which already has a parent:" in Repeat Loop. Any simple work around?

I am pretty Newbie to programming. And I am trying to pile up the random blocks dynamically till it hits the upper frame. But it seems that Swift doesn't let me to do so. Did I miss anything please? Any input are appreciated.
let blocks =[block1,block2,block3,block4,block5,block6,block7,block8,block9,block10,block11,block12]
var block:SKSpriteNode!
let blockX:Double = 0.0
var blockY:Double = -(self.size.height/2)
repeat{
block = blocks.randomBlock()
block.zPosition = 2
block.position = CGPoint(x:blockX, y:blockY)
block.size.height = 50
block.size.width = 50
self.addChild(block)
blockY += 50
} while( block.position.y < self.size.height)
extension Array {
func randomBlock()-> Element {
let randint = Int(arc4random_uniform(UInt32(self.count)))
return self[randint]
}
}
you need to have someway of tracking which blocks have been selected and ensure that they don't get selected again. The method below uses an array to store the indexes of selected blocks and then uses recursion to find a cycle through until an unused match is found.
private var usedBlocks = [Int]()
func randomBlock() -> Int {
guard usedBlocks.count != blocks.count else { return -1 }
let random = Int(arc4random_uniform(UInt32(blocks.count)))
if usedBlocks.contains(random) {
return randomBlock()
}
usedBlocks.append(random)
return random
}
in your loop change your initializer to
let index = randomBlock()
if index > -1 {
block = blocks[index]
block.zPosition = 2
block.position = CGPoint(x:blockX, y:blockY)
}
remember that if you restart the game or start a new level, etc. you must clear all of the objects from usedBlocks
usedBlocks.removeAll()

In XCUITests, how to wait for existence of either of two ui elements

Looking at XCTWaiter().wait(...) I believe we can wait for multiple expectations to become true using this code
let notHittablePredicate = NSPredicate(format: "hittable == false")
let myExpectation = XCTNSPredicateExpectation(predicate: notHittablePredicate, object: element)
let result = XCTWaiter().wait(for: [myExpectation], timeout: timeout)
//for takes array of expectations
But this uses like AND among the supplied expectations. Is there a way to do OR among the supplied expectations.
Like i have a use case at login that after tapping submit, i want to wait for one of two elements. First element is "You are already logged in on another device. If you continue any unsaved data on your other device will be lost?". And second element is the main screen after login. So any one can appear. Currently I'm first waiting for first element until timeout occurs and then for the second element. But I want to optimize time here and move on as soon as any of two elements exist==true. Then i'll check if element1 exists then tap YES and then wait for main screen otherwise just assert existence of element2.
Please comment if something isn't clear in the question. Thanks
Inspired by http://masilotti.com/ui-testing-tdd/, you don't have to rely on XCTWaiter. You can simply run a loop and test whether one of them exists.
/// Waits for either of the two elements to exist (i.e. for scenarios where you might have
/// conditional UI logic and aren't sure which will show)
///
/// - Parameters:
/// - elementA: The first element to check for
/// - elementB: The second, or fallback, element to check for
/// - Returns: the element that existed
#discardableResult
func waitForEitherElementToExist(_ elementA: XCUIElement, _ elementB: XCUIElement) -> XCUIElement? {
let startTime = NSDate.timeIntervalSinceReferenceDate
while (!elementA.exists && !elementB.exists) { // while neither element exists
if (NSDate.timeIntervalSinceReferenceDate - startTime > 5.0) {
XCTFail("Timed out waiting for either element to exist.")
break
}
sleep(1)
}
if elementA.exists { return elementA }
if elementB.exists { return elementB }
return nil
}
then you could just do:
let foundElement = waitForEitherElementToExist(elementA, elementB)
if foundElement == elementA {
// e.g. if it's a button, tap it
} else {
// element B was found
}
lagoman's answer is absolutely correct and great. I needed wait on more than 2 possible elements though, so I tweaked his code to support an Array of XCUIElement instead of just two.
#discardableResult
func waitForAnyElement(_ elements: [XCUIElement], timeout: TimeInterval) -> XCUIElement? {
var returnValue: XCUIElement?
let startTime = Date()
while Date().timeIntervalSince(startTime) < timeout {
if let elementFound = elements.first(where: { $0.exists }) {
returnValue = elementFound
break
}
sleep(1)
}
return returnValue
}
which can be used like
let element1 = app.tabBars.buttons["Home"]
let element2 = app.buttons["Submit"]
let element3 = app.staticTexts["Greetings"]
foundElement = waitForAnyElement([element1, element2, element3], timeout: 5)
// do whatever checks you may want
if foundElement == element1 {
// code
}
NSPredicate supports OR predicates too.
For example I wrote something like this to ensure my application is fully finished launching before I start trying to interact with it in UI tests. This is checking for the existence of various landmarks in the app that I know are uniquely present on each of the possible starting states after launch.
extension XCTestCase {
func waitForLaunchToFinish(app: XCUIApplication) {
let loginScreenPredicate = NSPredicate { _, _ in
app.logInButton.exists
}
let tabBarPredicate = NSPredicate { _, _ in
app.tabBar.exists
}
let helpButtonPredicate = NSPredicate { _, _ in
app.helpButton.exists
}
let predicate = NSCompoundPredicate(
orPredicateWithSubpredicates: [
loginScreenPredicate,
tabBarPredicate,
helpButtonPredicate,
]
)
let finishedLaunchingExpectation = expectation(for: predicate, evaluatedWith: nil, handler: nil)
wait(for: [finishedLaunchingExpectation], timeout: 30)
}
}
In the console while the test is running there's a series of repeated checks for the existence of the various buttons I want to check for, with a variable amount of time between each check.
t = 13.76s Wait for com.myapp.name to idle
t = 18.15s Checking existence of "My Tab Bar" Button
t = 18.88s Checking existence of "Help" Button
t = 20.98s Checking existence of "Log In" Button
t = 22.99s Checking existence of "My Tab Bar" Button
t = 23.39s Checking existence of "Help" Button
t = 26.05s Checking existence of "Log In" Button
t = 32.51s Checking existence of "My Tab Bar" Button
t = 16.49s Checking existence of "Log In" Button
And voila, now instead of waiting for each element individually I can do it concurrently.
This is very flexible of course, since you can add as many elements as you want, with whatever conditions you want. And if you want a combination of OR and AND predicates you can do that too with NSCompoundPredicate. This can easily be adapted into a more generic function that accepts an array of elements like so:
func wait(for elements: XCUIElement...) { … }
Could even pass a parameter that controls whether it uses OR or AND.
Hey other alternative that works for us. I hope help others too.
XCTAssert(
app.staticTexts["Hello Stack"]
.waitForExistence(timeout: 10) || app.staticTexts["Hi Stack"]
.waitForExistence(timeout: 10)
)

Very slow minesweeper recursive algorithm in Swift

I'm working with Swift 3 and Xcode.
I'm creating an iOS game that is basically a Minesweeper, but there are no squares but hexagons, so each hexagon can have up to 6 mines in their surrounding.
I created a recursive algorithm, so that when the player touches an hexagon, if it's not a bomb, then it call a recursive function called "reveal" which :
- if one ore more mine in the surrounding and the touched hexagon is still hidden (by hidden I mean we don't know if it's a mine or not), reveal the hexagon & set the number of surrounding mine's label, and stop the function
- if no mine in the surrounding, for each nearby hexagon that is hidden, call the reveal function.
So here's what my code looks like :
class Hexagon: SKShapeNode
{
var mine: Bool
var hide: Bool
var proximityMines: Int
init(mine: Bool = false, proximityMines: Int = 0, hide: Bool = true)
{
self.mine = mine // if it's a mine
self.proximityMines = proximityMines // number of surrounding mines (that I calculated using a function after I generated the level)
self.hide = hide // if the hexagon is still hidden
super.init()
}
required init?(coder aDecoder: NSCoder) {
fatalError("init(coder:) has not been implemented")
}
}
func reveal(hexagon: Hexagon)
{
if hexagon.proximityMines == 0 && hexagon.hide == true // if there are no mines in the surrounding
{
hexagon.hide = false // we update the value of this hexagon
setNumberLabel(hexagon: hexagon) // we set the .proximityMines number as a label (here 0)
for proxyHexagon in proximityHexagons(hexagon: hexagon) // for each surrounding hexagon ...
{
if proxyHexagon.hide == true // ... that is still hidden
{
reveal(hexagon: proxyHexagon) // we call this function again
}
}
}
else if hexagon.proximityMines != 0 && hexagon.hide == true // else if there are mines in the surrounding
{
hexagon.hide = false // update
setNumberLabel(hexagon: hexagon) // set label
}
}
the proximityHexagons(hexagon: Hexagon) function returns an array containing all surrounding hexagons of a given hexagon.
So I really checked my algorithm again and again, and I really think it's the good one.
But the fact is that when I create a level with 0 or a really low amount of mine, and I click on an hexagon, it takes something like 2 seconds for the recursive function to update all the empty hexagons.
My map contains more or less 260 hexagons, and I debugged the number of calls of reveal() and it's about the same amount.
So why is it taking so much time ? I don't think the iPhone 6 can't handle this amount of operations ! I tried it on my iPhone, not an emulator.
Do you have any idea ?
Ok I've been thinking about this because it sounds like a fun problem. I didn't look up any minesweeper solvers, so I might be way out in left field, but here is how I would approach your problem.
First you have to give every mine an index, and you need to know the pattern of that index such that you can do a little math to get the surrounding indices of every mine. If the rows have identical numbers, and the numbering is sequential across rows, then the surrounding indices are:
[index - 1, index + 1,
index - rowCount, index - rowCount - 1,
index + rowCount, index + rowCount + 1]
Then I would make a class that holds a set of all the safe spots on the map that you had when you built the puzzle. I'll call it SafetyManager.
class SafetyManager {
var safeSpots: Set<Int> = all your safe spots
func indices(surrounding index: Int) -> Set<Int> {
return [index - 1, index + 1,
index - rowCount, index - rowCount - 1,
index + rowCount, index + rowCount + 1]
}
func safePlaces(around hexagon: Int) -> Set<Int> {
let allIndices = indices(surrounding: hexagon)
let safe = allIndices.intersection(safeSpots)
safeSpots.subtract(safe)
return safe
}
}
It's got two important functions, one calculates the surrounding indices, the second filters the safe spots. I'm using sets so we can quickly determine the intersection between the safe spots and the surrounding spots.
Next we need a class that would be instantiated when a move is made so we can do the recursion. Lets call it CheckManager.
class CheckManager {
var checked : [Int]
var unchecked : Set<Int>
init(firstHex: Hexagon, surroundingSafeSpots: Set<Int>) {
checked = [firstHex.index]
unchecked = surroundingSafeSpots
}
func nextUnchecked() -> Int? {
guard !unchecked.isEmpty else { return nil }
let next = unchecked.removeFirst()
checked += [next]
return next
}
func pleaseTake(these indices: Set<Int>) {
unchecked.formUnion(indices)
}
}
You initialize it with your first hexagon, or hex index, and the surrounding safespots that the safety manager would give you, if you get no safe spots from the SafetyManager, no need to instantiate.
It keeps a set of checked spots and unchecked spots. Two important functions, the second you use to give it newly acquired safe spots from the safety manager to be added to the unchecked list. The other returns an optional Int? of the next safe spot to check the surroundings of.
Then to do the recursion, something like this..
func check(spot: Hexagon) {
let safe = safetyMan.safePlaces(around: spot.index)
guard safe.count > 0 else { .. }
let checkMan = CheckManager(firstHex: spot, surroundingSafeSpots: safe)
while let i = checkMan.nextUnchecked() {
let safeSpots = safetyMan.safePlaces(around: i)
checkMan.pleaseTake(these: safeSpots)
} // goes until unchecked is empty
for spot in checkMan.checked {
// get the hex and reveal
}
}
You could keep a dictionary of [Int: Hexagon] to quickly grab the hex for a given index. I haven't tested this so I'm not sure if it works well, or at all or has some improper syntax. It would also probably be a lot faster to use multithreading. Fun problem. Good luck.
Okay, I managed to solve my problem.
The problem was the proximityHexagons function that was taking a lot of time. In fact, each time I called this function, he made 6 complex calculations and added the surrounding hexagons in an array, so it was taking a lot of time.
Here's what it looked like :
func proximityHexagons(hexagon: Hexagon) -> Array<Hexagon>
{
var array = [Hexagon]()
var nodeArray = [[Hexagon]]()
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x, y: hexagon.position.y + hexagon.height)).filter({$0 is Hexagon}) as! [Hexagon])
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x + hexagon.width * 3/4, y: hexagon.position.y + hexagon.height / 2)).filter({$0 is Hexagon}) as! [Hexagon])
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x + hexagon.width * 3/4, y: hexagon.position.y - hexagon.height / 2)).filter({$0 is Hexagon}) as! [Hexagon])
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x, y: hexagon.position.y - hexagon.height)).filter({$0 is Hexagon}) as! [Hexagon])
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x - hexagon.width * 3/4, y: hexagon.position.y - hexagon.height / 2)).filter({$0 is Hexagon}) as! [Hexagon])
nodeArray.append(nodes(at: CGPoint(x: hexagon.position.x - hexagon.width * 3/4, y: hexagon.position.y + hexagon.height / 2)).filter({$0 is Hexagon}) as! [Hexagon])
// first, for each 6 directions, I'm adding in an array every nodes that are Hexagon, and then adding all of theses arrays in another bigger one
for node in nodeArray // for each hexagon array in the big array
{
if node.count != 0 // if there is an hexagon
{
array.append(node.first!) // we set the hexagon in the final array
}
}
return array // we return the array containing all surrounding hexagons
}
I prefer checking the surrounding hexagons with the nodes(at: Point) function because my levels aren't always regular maps, they can have a weird positioning and twiz_'s func indices(surrounding index: Int) function could not work.
So I kept my function, but I call it once at the beginning of the level and store in a new variable in my hexagon class all the surrounding hexagons of each hexagon:
class Hexagon: SKShapeNode
{
var mine: Bool
var hide: Bool
var proximityMines: Int
var proxyHexagons: [Hexagon] // here
init(mine: Bool = false, proximityMines: Int = 0, hide: Bool = true, proxyHexagons: [Hexagon] =
[Hexagon]())
{
self.mine = mine
self.proximityMines = proximityMines
self.hide = hide
self.proxyHexagons = proxyHexagons
super.init()
}
required init?(coder aDecoder: NSCoder) {
fatalError("init(coder:) has not been implemented")
}
}
And then, in the reveal function, instead of calling the proximityHexagons function, I use the .proxyHexagons array of the hexagon, like this :
func reveal(hexagon: Hexagon)
{
if hexagon.proximityMines == 0 && hexagon.hide == true
{
hexagon.hide = false
setNumberLabel(hexagon: hexagon)
for proxyHexagon in hexagon.proxyHexagons // here
{
if proxyHexagon.hide == true
{
reveal(hexagon: proxyHexagon)
}
}
}
else if hexagon.proximityMines != 0 && hexagon.hide == true
{
hexagon.hide = false
setNumberLabel(hexagon: hexagon)
}
}
And now my function is way faster, I manage to reveal all 260 hexagons in 0.001 secs instead of the old 2.81 secs.

Swift Enumerate out of Range

I'm working on making a tile-based adventure game and I'm trying to generate tile maps. I put together some code that removes inland tiles from the list of possible new tile locations. I have a method that detects if a tile is coastal.
for (index, tile) in landTiles.enumerate() {
let coastal = isCoastal(tile.y, x: tile.x)
if coastal {
coastTiles.append(tile)
} else {
print(landTiles.count)
print(index)
landTiles.removeAtIndex(index)
}
}
When I run this code I'm getting an error: index out of range message. I believe this is because the indexes are getting confused when I delete an item from landTiles. How could I fix this?
You are removing elements from an array while you enumerate it.
And this is an anti-pattern.
What should you do instead?
The Tile class
First of all, the Tile class should have a isCoastal computed property. Something like this
class Tile {
let x: Int
let y: Int
var isCoastal: Bool { /* your logic goes here */ }
init(x:Int, y:Int) {
self.x = x
self.y = y
}
}
Filtering
Now given an array of Tile(s)
var landTiles: [Tile] = ...
you can extract the ones having isCoastal true
let coastTiles = landTiles.filter { $0.isCoastal }
and overwrite the original array with the ones having isCoastal false.
landTiles = landTiles.filter { !$0.isCoastal }
That's it.